ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration is an important study material for the students from the exam perspective. Students can refer to the solutions designed by the subject matter experts at BYJUâ€™S having vast conceptual knowledge. The main aim of providing ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration PDF is to boost the exam preparation of students.

Chapter 16 consists of problems on finding the area of various figures like triangles, polygons, etc. The ML Aggarwal Solutions help students understand the concepts and their applications in a wide range. Each problem is solved after conducting wide research on the concepts to help students, irrespective of their intelligence quotient.

## ML Aggarwal Solutions for Class 9 Maths Chapter 16: Mensuration Download PDF

## Access ML Aggarwal Solutions for Class 9 Maths Chapter 16: Mensuration

Exercise 16.1

**1. Find the area of a triangle whose base is 6 cm and corresponding height is 4 cm.**

**Solution:**

It is given that

Base of triangle = 6 cm

Height of triangle = 4 cm

We know that

Area of triangle = Â½ Ã— base Ã— height

Substituting the values

= Â½ Ã— 6 Ã— 4

By further calculation

= 6 Ã— 2

= 12 cm^{2}

**2. Find the area of a triangle whose sides are**

**(i) 3 cm, 4 cm and 5 cm**

**(ii) 29 cm, 20 cm and 21 cm**

**(iii) 12 cm, 9.6 cm and 7.2 cm**

**Solution:**

(i) Consider a = 3 cm, b = 4 cm and c = 5 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (3 + 4 + 5)/ 2

= 12/2

= 6 cm

Here

= 6 cm^{2}

(ii) Consider a = 29 cm, b = 20 cm and c = 21 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (29 + 20 + 21)/ 2

= 70/2

= 35 cm

Here

So we get

= 7 Ã— 5 Ã— 3 Ã— 2

= 210 cm^{2}

(iii) Consider a = 12 cm, b = 9.6 cm and c = 7.2 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (12 + 9.6 + 7.2)/ 2

= 28.8/2

= 14.4 cm

Here

So we get

= 2.4 Ã— 2.4 Ã— 6

= 34.56 cm^{2}

**3. Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. hence, find the length of the altitude corresponding to the shortest side.**

**Solution:**

Consider 34 cm, 20 cm and 42 cm as the sides of triangle

a = 34 cm, b = 20 cm and c = 42 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (34 + 20 + 42)/ 2

= 96/2

= 48 cm

Here

So we get

= 14 Ã— 6 Ã— 4

= 336 cm^{2}

Here the shortest side of the triangle is 20 cm

Consider h cm as the corresponding altitude

Area of triangle = Â½ Ã— base Ã— height

Substituting the values

336 = Â½ Ã— 20 Ã— h

By further calculation

h = (336 Ã— 2)/ 20

So we get

h = 336/10

h = 33.6 cm

Hence, the required altitude of the triangle is 33.6 cm.

**4. The sides of a triangular field are 975 m, 1050 m and 1125 m. If this field is sold at the rate of Rs 1000 per hectare, find its selling price. (1 hectare = 10000 m ^{2})**

**Solution:**

It is given that

a = 975 m, b = 1050 m and c = 1125 m

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (975 + 1050 + 1125)/ 2

= 3150/2

= 1575 cm

Here

So we get

= 525 Ã— 450 Ã— 2

It is given that

1 hectare = 10000 m^{2}

= (525 Ã— 900)/ 10000

By further calculation

= (525 Ã— 9)/ 100

= 4725/100

= 47.25 hectares

We know that

Selling price of 1 hectare field = Rs 1000

Selling price of 47.25 hectare field = 1000 Ã— 47.25 = Rs 47250

**5. The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter.**

**Solution:**

It is given that

ABC is a right angled triangle

BC = 12 cm and AB = 13 cm

Using the Pythagoras theorem

AB^{2} = AC^{2} + BC^{2}

Substituting the values

13^{2} = AC^{2} + 12^{2}

By further calculation

AC^{2} = 13^{2} – 12^{2}

So we get

AC^{2} = 169 â€“ 144 = 25

AC = âˆš25 = 5 cm

We know that

Area of triangle ABC = Â½ Ã— base Ã— height

Substituting the values

= Â½ Ã— 12 Ã— 5

= 30 cm^{2}

Similarly

Perimeter of triangle ABC = AB + BC + CA

Substituting the values

= 13 + 12 + 5

= 30 cm

**6. Find the area of an equilateral triangle whose side is 8 m. Give your answer correct to two decimal places.**

**Solution:**

It is given that

Side of equilateral triangle = 8 m

We know that

Area of equilateral triangle = âˆš3/4 (side)^{2}

Substituting the values

= âˆš3/4 Ã— 8 Ã— 8

By further calculation

= âˆš3 Ã— 2 Ã— 8

= 1.73 Ã— 16

= 27.71 m^{2}

**7. If the area of an equilateral triangle is 81âˆš3 cm ^{2} find its perimeter.**

**Solution:**

We know that

Area of equilateral triangle = âˆš3/4 (side)^{2}

Substituting the values

81 âˆš3 = âˆš3/4 (side)^{2}

By further calculation

Side^{2} = (81 âˆš3 Ã— 4)/ âˆš3

So we get

(side)^{2} = 81 Ã— 4

side = âˆš(81 Ã— 4)

side = 9 Ã— 2 = 18 cm

So the perimeter of equilateral triangle = 3 Ã— side

= 3 Ã— 18

= 54 cm

**8. If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.**

**Solution:**

We know that

Perimeter of an equilateral triangle = 3 Ã— side

Substituting the values

36 = 3 Ã— side

By further calculation

side = 36/3 = 12 cm

So AB = BC = CA = 12 cm

Here

Area of equilateral triangle = âˆš3/4 (side)^{2}

Substituting the values

= âˆš3/4 (12)^{2}

By further calculation

= âˆš3/4 Ã— 12 Ã— 12

So we get

= âˆš3 Ã— 3 Ã— 12

= 1.73 Ã— 36

= 62.4 cm^{2}

In triangle ABD

Using Pythagoras Theorem

AB^{2} = AD^{2} + BD^{2}

Here BD = 12/2 = 6 cm

Substituting the values

12^{2} = AD^{2} + 6^{2}

By further calculation

144 =AD^{2} + 36

AD^{2} = 144 â€“ 36 = 108

So we get

AD = âˆš108 = 10.4

Therefore, the required height is 10.4 cm.

**9. (i) If the length of the sides of a triangle are in the ratio 3: 4: 5 and its perimeter is 48 cm, find its area.**

**(ii) The sides of a triangular plot are in the ratio 3: 5: 7 and its perimeter is 300 m. Find its area.**

**Solution:**

(i) Consider ABC as the triangle

Ratio of the sides are 3x, 4x and 5x

Take a = 3x cm, b = 4x cm and c = 5x cm

We know that

a + b + c = 48

Substituting the values

3x + 4x + 5x = 48

12x = 48

So we get

x = 48/12 = 4

Here

a = 3x = 3 Ã— 4 = 12 cm

b = 4x = 4 Ã— 4 = 16 cm

c = 5x = 5 Ã— 4 = 20 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (12 + 16 + 20)/ 2

= 48/2

= 24 cm

Here

So we get

= 12 Ã— 4 Ã— 2

= 96 cm^{2}

(ii) It is given that

Sides of a triangle are in the ratio = 3: 5: 7

Perimeter = 300 m

We know that

First side = (300 Ã— 3)/ sum of ration

Substituting the values

= (300 Ã— 3)/ (3 + 5 + 7)

By further calculation

= (300 Ã— 3)/ 15

= 60 m

Second side = (300 Ã— 5)/ 15 = 100 m

Third side = (300 Ã— 7)/ 15 = 140 m

Here

S = perimeter/2 = 300/2 = 150 m

So we get

We get

= 1500 Ã— 1.732

= 2598 m^{2}

**10. ABC is a triangle in which AB = AC = 4 cm and âˆ A = 90 ^{0}. Calculate the area of â–³ABC. Also find the length of perpendicular from A to BC.**

**Solution:**

It is given that

AB = AC = 4 cm

Using the Pythagoras theorem

BC^{2} = AB^{2} + AC^{2}

Substituting the values

BC^{2} = 4^{2} + 4^{2}

By further calculation

BC^{2} = 16 + 16 = 32

BC = âˆš32 = 4âˆš2 cm

We know that

Area of â–³ABC = Â½ Ã— BC Ã— h

Substituting the values

8 = Â½ Ã— 4âˆš2 Ã— h

By further calculation

h = (8 Ã— 2)/ 4âˆš2

We can write it as

h = (2 Ã— 2)/ âˆš2 Ã— âˆš2/âˆš2

So we get

h = 4âˆš2/2 = 2 Ã— âˆš2

h = 2 Ã— 1.41 = 2.82 cm

**11. Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.**

**Solution:**

Consider ABC as the isosceles triangles

Here AB = AC = 12cm

Perimeter = 30 cm

So BC = 30 â€“ (12 + 12) = 30 â€“ 24 = 6 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= 30/ 2

= 15 cm

Here

We can write it as

= 9 Ã— 3.873

= 34.857

= 34.86 cm^{2}

**12. Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.**

**Solution:**

It is given that

Base = 6 cm

Perimeter = 16 cm

Consider ABC as an isosceles triangle in which

AB = AC = x

So BC = 6 cm

We know that

Perimeter of â–³ABC = AB + BC + AC

Substituting the values

16 = x + 6 + x

By further calculation

16 = 2x + 6

16 â€“ 6 = 2x

10 = 2x

So we get

x = 10/2 = 5

Here AB = AC = 5 cm

BC = Â½ Ã— 6 = 3 cm

In â–³ABD

AB^{2} = AD^{2} + BD^{2}

Substituting the values

5^{2} = AD^{2} + 3^{2}

25 = AD^{2} + 9

By further calculation

AD^{2} = 25 â€“ 9 = 16

So we get

AD = 4 cm

Here

Area of â–³ABC = Â½ Ã— base Ã— height

Substituting the values

= Â½ Ã— 6 Ã— 4

= 3 Ã— 4

= 12 cm^{2}

**13. The sides of a right angled triangle containing the right angle are 5x cm and (3x â€“ 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm ^{2}.**

**Solution:**

Consider ABC as a right angled triangle

AB = 5x cm and BC = (3x â€“ 1) cm

We know that

Area of â–³ABC = Â½ Ã— AB Ã— BC

Substituting the values

60 = Â½ Ã— 5x (3x â€“ 1)

By further calculation

120 = 5x (3x â€“ 1)

120 = 15x^{2} â€“ 5x

It can be written as

15x^{2} â€“ 5x â€“ 120 = 0

Taking out the common terms

5 (3x^{2} â€“ x â€“ 24) = 0

3x^{2} â€“ x â€“ 24 = 0

3x^{2} â€“ 9x + 8x â€“ 24 = 0

Taking out the common terms

3x (x â€“ 3) + 8 (x â€“ 3) = 0

(3x + 8) (x â€“ 3) = 0

Here

3x + 8 = 0 or x â€“ 3 = 0

We can write it as

3x = -8 or x = 3

x = -8/3 or x = 3

x = -8/3 is not possible

So x = 3

AB = 5 Ã— 3 = 15 cm

BC = (3 Ã— 3 â€“ 1) = 9 â€“ 1 = 8 cm

In right angled â–³ABC

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

AC^{2} = 15^{2} + 8^{2}

By further calculation

AC^{2} = 225 + 64 = 289

AC^{2} = 17^{2}

So AC = 17 cm

Therefore, the hypotenuse of the right angled triangle is 17 cm.

**14. In â–³ABC, âˆ B = 90 ^{0}, AB = (2A + 1) cm and BC = (A + 1) cm. If the area of the â–³ABC is 60 cm^{2}, find its perimeter. **

**Solution:**

It is given that

AB = (2x + 1) cm

BC = (x + 1) cm

We know that

Area of â–³ABC = Â½ Ã— AB Ã— BC

Substituting the values

60 = Â½ Ã— (2x + 1) (x + 1)

By cross multiplication

60 Ã— 2 = (2x + 1) (x + 1)

By further calculation

120 = 2x^{2} + 3x + 1

We can write it as

0 = 2x^{2} + 3x + 1 â€“ 120

0 = 2x^{2} + 3x – 119

So we get

2x^{2} + 3x â€“ 119 = 0

2x^{2} + 17x â€“ 14x â€“ 119 = 0

Taking out the common terms

x (2x + 17) â€“ 7 (2x + 17) = 0

(x â€“ 7) (2x + 17) = 0

Here

x â€“ 7 = 0 or 2x + 17 = 0

x = 7 or 2x = – 17

x = 7 or x = -17/2

AB = (2x + 1) = 2 Ã— 7 + 1

AB = 14 + 1 = 15 cm

BC = (x + 1) = 7 + 1 = 8 cm

In right angled â–³ABC

Using Pythagoras Theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

AC^{2} = 15^{2} + 8^{2}

AC^{2} = 225 + 64

AC^{2} = 289

So we get

AC = 17 cm

So the perimeter = AB + BC + AC

Substituting the values

= 15 + 8 + 17

= 40 cm

**15. If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.**

**Solution:**

We know that

Perimeter of a right angled triangle = 60 cm

Hypotenuse = 25 cm

Here the sum of two sides = 60 â€“ 25 = 35 cm

Consider base = x cm

Altitude = (35 â€“ x) cm

Using the Pythagoras theorem

x^{2} + (35 â€“ x)^{2} = 25^{2}

By further calculation

x^{2} + 1225 + x^{2} â€“ 70x = 625

2x^{2} â€“ 70x + 1225 â€“ 625 = 0

2x^{2} â€“ 70x + 600 = 0

Dividing by 2

x^{2} â€“ 35x + 300 = 0

x^{2} â€“ 15x â€“ 20x + 300 = 0

Taking out the common terms

x (x â€“ 15) â€“ 20 (x â€“ 15) = 0

(x â€“ 15) (x â€“ 20) = 0

Here

x â€“ 15 = 0

So we get

x = 15

Similarly

x â€“ 20 = 0

So we get

x = 20 cm

So 15 cm and 20 cm are the sides of the triangle

Area = Â½ Ã— base Ã— altitude

Substituting the values

= Â½ Ã— 15 Ã— 20

= 150 cm^{2}

**16. The perimeter of an isosceles triangle is 40 cm. The base is two third of the sum of equal sides. Find the length of each side.**

**Solution:**

It is given that

Perimeter of an isosceles triangle = 40 cm

Consider x cm as each equal side

We know that

Base = 2/3 (2x) = 4/3 x

So according to the sum

2x + 4/3 x = 40

By further calculation

6x + 4x = 120

10x = 120

By division

x = 120/10 = 12

Therefore, the length of each equal side is 12 cm.

**17. If the area of an isosceles triangle is 60 cm ^{2} and the length of each of its equal sides is 13 cm, find its base.**

**Solution:**

It is given that

Area of isosceles triangle = 60 cm^{2}

Length of each equal side = 13 cm

Consider base BC = x cm

Construct AD perpendicular to BC which bisects BC at D

So BD = DC = x/2 cm

In right â–³ABD

AB^{2} = BD^{2} + AD^{2}

Substituting the values

13^{2} = (x/2)^{2} + AD^{2}

By further calculation

169 = x^{2}/4 + AD^{2}

AD^{2} = 169 â€“ x^{2}/4 â€¦â€¦ (1)

We know that

Area = 60 cm^{2}

AD = (area Ã— 2)/ base

Substituting the values

AD = (60 Ã— 2)/ x = 120/x â€¦â€¦ (2)

Using both the equations

169 â€“ x^{2} /4 = (120/x)^{2}

By further calculation

(676 â€“ x^{2})/ 4 = 14400/x^{2}

By cross multiplication

676x^{2} â€“ x^{4} = 57600

We can write it as

x^{4} â€“ 676x^{2} + 57600 = 0

x^{4} â€“ 576x^{2} â€“ 100x^{2} + 57600 = 0

Taking out the common terms

x^{2}Â (x^{2} â€“ 576) â€“ 100 (x^{2} â€“ 576) = 0

(x^{2} â€“ 576) (x^{2} â€“ 100) = 0

Here

x^{2} â€“ 576 = 0 where x^{2} = 576

So x = 24

Similarly

x^{2} â€“ 100 = 0 where x^{2} = 100

So x = 10

Hence, the base is 10 cm or 24 cm.

**18. The base of a triangular field is 3 times its height if the cost of cultivating the field at the rate of Rs 25 per 100 m ^{2} is Rs 60000; find its base and height.**

**Solution:**

It is given that

Cost of cultivating the field at the rate of Rs 25 per 100 m^{2} = Rs 60000

Here the cost of cultivating the field of Rs 25 for 100 m^{2}

So the cost of cultivating the field of Rs 1 = 100/25 m^{2}

Cost of cultivating the field of Rs 60000 = 100/25 Ã— 60000

= 4 Ã— 60000

= 240000 m^{2}

So the area of field = 240000 m^{2}

Â½ Ã— base Ã— height = 240000 â€¦.. (1)

Consider

Height of triangular field = h m^{2}

Base of triangular field = 3h m^{2}

Substituting the values in equation (1)

Â½ Ã— 3h Ã— h = 240000

By further calculation

Â½ Ã— 3h^{2} = 240000

h^{2} = (240000 Ã— 2)/ 3

h^{2} = 80000 Ã— 2

h^{2} = 160000

So we get

h = âˆš160000 = 400

Here the height of triangular field = 400 m

Base of triangular field = 3 Ã— 400 = 1200 m^{2}

**19. A triangular park ABC has sides 120 m, 80 m and 50 m (as shown in the given figure). A gardener Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3 m wide for a gate on one side.**

**Solution:**

It is given that

ABC is a triangular park with sides 120 m, 80 m and 50 m.

Here the perimeter of triangle ABC = 120 + 80 + 50 = 250 m

We know that

Portion at which a gate is build = 3m

Remaining perimeter = 250 â€“ 3 = 247 m

So the length of fence around it = 247 m

Rate of fencing = Rs 20 per m

Total cost of fencing = 20 Ã— 247 = Rs 4940

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= 250/2

= 125 cm

Here

**20. An umbrella is made by stitching 10 triangular pieces of cloth of two different colors (shown in the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each color is required for the umbrella?**

**Solution:**

It is given that

An umbrella is made by stitching 10 triangular pieces of cloth of two different colors

20 cm, 50 cm, 50 cm are the measurement of each triangle

So we get

s/2 = (20 + 50 + 50)/ 2

s/2 = 120/2 = 60

We know that

Here the area of 5 triangular piece of first color = 5 Ã— 200 âˆš6 = 1000 âˆš6 cm^{2}

Area of triangular piece of second color = 1000 âˆš6 cm^{2}

**21. (a) In the figure (1) given below, ABC is an equilateral triangle with each side of length 10 cm. In â–³BCD, âˆ D = 90 ^{0} and CD = 6 cm.**

**Find the area of the shaded region. Give your answer correct to one decimal place.**

**(b) In the figure (ii) given, ABC is an isosceles right angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.**

**Solution:**

(a) It is given that

ABC is an equilateral triangle of side = 10 cm

We know that

Area of equilateral triangle ABC = âˆš3/4 Ã— (side)^{2}

Substituting the values

= âˆš3/4 Ã— 10^{2}

= âˆš3/4 Ã— 100

So we get

= âˆš3Ã— 25

= 1.73 Ã— 25

= 43.3 cm^{2}

In right angled triangle BDC

âˆ D = 90^{0}

BC = 10 cm

CD = 6 cm

Using Pythagoras Theorem

BD^{2} + DC^{2} = BC^{2}

Substituting the values

BD^{2} + 6^{2} = 10^{2}

BD^{2} + 36 = 100

So we get

BD^{2} = 100 â€“ 36 = 64

BD = âˆš64 = 8 cm

We know that

Area of triangle BDC = Â½ Ã— base Ã— height

So we get

= Â½ Ã— BD Ã— DC

Substituting the values

= Â½ Ã— 8 Ã— 6

= 4 Ã— 6

= 24 cm^{2}

Here the area of shaded portion = Area of triangle ABC â€“ Area of triangle BDC

Substituting the values

= 43.3 â€“ 24

= 19.3 cm^{2}

(b) It is given that

AD = AE = 3 cm

DB = EC = 4 cm

By addition we get

AD + DB = AE + EC = (3 + 4) cm

AB = AC = 7 cm

âˆ A = 90^{0}

We know that

Area of right triangle ADE = Â½ Ã— AD Ã— AE

Substituting the values

= Â½ Ã— 3 Ã— 3

= 9/2 cm^{2}

So triangle BDG is an isosceles right triangle

Similarly

DG^{2} + BG^{2} = BD^{2}

DG^{2} + DG^{2} = 4^{2}

By further calculation

2DG^{2} = 16

DG^{2} = 16/2 = 8

DG = âˆš8 cm

Area of triangle BDG = Â½ Ã— BG Ã— DG

We can write it as

= Â½ Ã— DG Ã— DG

Substituting the values

= Â½ (âˆš8)^{2}

= Â½ Ã— 8

= 4 cm^{2}

Area of isosceles right triangle EFC = 4 cm^{2}

So the area of shaded portion = 9/2 + 4 + 4

Taking LCM

= (9 + 8 + 8)/ 2

= 25/2

= 12.5 cm^{2}

Exercise 16.2

**1. (i) Find the area of quadrilateral whose one diagonal is 20 cm long and the perpendiculars to this diagonal from other vertices are of length 9 cm and 15 cm.**

**(ii) Find the area of a quadrilateral whose diagonals are of length 18 cm and 12 cm and they intersect each other at right angles.**

**Solution:**

(i) Consider ABCD as a quadrilateral in which AC = 20 cm

We know that

BY = 9 cm and DY = 15 cm

Here

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

We can write it as

= Â½ Ã— base Ã— height + Â½ Ã— base Ã— height

So we get

= Â½ Ã— AC Ã— BX + Â½ Ã— AC Ã— DY

Substituting the values

= (Â½ Ã— 20 Ã— 9) + (Â½ Ã— 20 Ã— 15)

By further calculation

= (10 Ã— 9 + 10 Ã— 15)

= 90 + 150

= 240 cm^{2}

(ii) Consider ABCD as a quadrilateral in which the diagonals AC and BD intersect each other at M at right angles

AC = 18 cm and BD = 12 cm

We know that

Area of quadrilateral ABCD = Â½ Ã— diagonal AC Ã— diagonal BD

Substituting the values

= Â½ Ã— 18 Ã— 12

By further calculation

= 9 Ã— 12

= 108 cm^{2}

**2. Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and âˆ A = 90 ^{0}.**

**Solution:**

It is given that

ABCD is a quadrilateral field

AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and âˆ A = 90^{0}

In triangle BAD

âˆ A = 90^{0}

Using the Pythagoras Theorem

BD^{2} = BA^{2} + AD^{2}

Substituting the values

BD^{2} = 40^{2} + 9^{2}

By further calculation

BD^{2} = 1600 + 81 = 1681

So we get

BD = 41

We know that

Area of quadrilateral ABCD = Area of â–³BAD + Area of â–³BDC

It can be written as

= Â½ Ã— base Ã— height + Area of â–³BDC

Substituting the values

= Â½ Ã— 40 Ã— 9 + Area of â–³BDC

By further calculation

= 180 m^{2} + Area of â–³BDC

Determining the area of â–³BDC

Consider a = BD = 41 m, b = CD = 15 m, c = BC = 28 m

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (41 + 15 + 28)/ 2

= 42 cm

Here

So we get

= 2 Ã— 7 Ã— 3 Ã— 3

= 126 m^{2}

So the area of quadrilateral ABCD = 180 m^{2} + Area of â–³BDC

Substituting the values

= 180 + 126

= 306 m^{2}

**3. Find the area of the quadrilateral ABCD in which âˆ BCA = 90 ^{0}, AB = 13 cm and ACD is an equilateral triangle of side 12 cm.**

**Solution:**

It is given that

ABCD is a quadrilateral in which âˆ BCA = 90^{0} and AB = 13 cm

ABCD is an equilateral triangle in which AC = CD = AD = 12 cm

In right angled â–³ABC

Using Pythagoras theorem,

AB^{2} = AC^{2} + BC^{2}

Substituting the values

13^{2} = 12^{2} + BC^{2}

By further calculation

BC^{2} = 13^{2} – 12^{2}

BC^{2} = 169 â€“ 144 = 25

So we get

BC = âˆš25 = 5 cm

We know that

Area of quadrilateral ABCD = Area of â–³ABC + Area of â–³ACD

It can be written as

= Â½ Ã— base Ã— height + âˆš3/4 Ã— side^{2}

= Â½ Ã— AC Ã— BC + âˆš3/4 Ã— 12^{2}

Substituting the values

= Â½ Ã— 12 Ã— 5 + âˆš3/4 Ã— 12 Ã— 12

So we get

= 6 Ã— 5 + âˆš3 Ã— 3 Ã— 12

= 30 + 36âˆš3

Substituting the value of âˆš3

= 30 + 36 Ã— 1.732

= 30 + 62.28

= 92.28 cm^{2}

**4. Find the area of quadrilateral ABCD in which âˆ B = 90 ^{0}, AB = 6 cm, BC = 8 cm and CD = AD = 13 cm.**

**Solution:**

It is given that

ABCD is a quadrilateral in which âˆ B = 90^{0}, AB = 6 cm, BC = 8 cm and CD = AD = 13 cm

In â–³ABC

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

AC^{2} = 6^{2} + 8^{2}

By further calculation

AC^{2} = 36 + 64 = 100

So we get

AC^{2} = 10^{2}

AC = 10 cm

We know that

Area of quadrilateral ABCD = Area of â–³ABC + Area of â–³ACD

It can be written as

= Â½ Ã— base Ã— height + Area of â–³ACD

= Â½ Ã— AB Ã— BC + Area of â–³ACD

Substituting the values

= Â½ Ã— 6 Ã— 8 + Area of â–³ACD

By further calculation

= 24 cm^{2} + Area of â–³ACD â€¦â€¦ (1)

Finding the area of â–³ACD

Consider a = AC = 10 cm, b = CD = 13 cm, c = AD = 13 cm

We know that

S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= (10 + 13 + 13)/ 2

= (10 + 26)/2

= 36/2

= 18 cm

Here

So we get

= 3 Ã— 2 Ã— 2 Ã— 5

= 60 cm^{2}

Using equation (1)

Area of quadrilateral ABCD = 24 cm^{2} + Area of â–³ACD

Substituting the values

= 24 + 60

= 84 cm^{2}

**5. The perimeter of a rectangular cardboard is 96 cm; if its breadth is 18 cm, find the length and the area of the cardboard.**

**Solution:**

Consider ABCD as a rectangle

Take length = l cm

Breadth = 18 cm

Perimeter = 96 cm

We know that

2 Ã— (l + b) = 96 cm

Substituting the values

2 Ã— (l + 18) = 96 cm

By further calculation

(l + 18) = 96/2

l + 18 = 48

So we get

l = 48 â€“ 18 = 30 cm

Here

Area of rectangular cardboard = l Ã— b

Substituting the values

= 30 Ã— 18

= 540 cm^{2}

**6. The length of a rectangular hall is 5 cm more than its breadth, if the area of the hall is 594 m ^{2}, find its perimeter.**

**Solution:**

Consider ABCD is a rectangular hall

Take Breadth = x m

Length = (x + 5) m

We know that

Area of rectangular field = l Ã— b

Substituting the values

594 = x (x + 5)

By further calculation

594 = x^{2} + 5x

0 = x^{2} + 5x â€“ 594

x^{2} + 5x â€“ 594 = 0

It can be written as

x^{2} + 27x â€“ 22x â€“ 594 = 0

Taking out the common terms

x (x + 27) â€“ 22 (x + 27) = 0

So we get

(x â€“ 22) (x + 27) = 0

Here

x â€“ 22 = 0 or x + 27 = 0

We get

x = 22 m or x = -27 which is not possible

We know that

Breadth = 22 m

Length = (x + 5) = 22 + 5 = 27 m

Perimeter = 2 (l + b)

Substituting the values

= 2 (27 + 22)

By further calculation

= 2 Ã— 49

= 98 m

**7. (a) The diagram (i) given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. The width of each path is 5 metres. Find the area of the shaded portion.**

**(b) In the diagram (ii) given below, calculate the area of the shaded portion. All measurements are in centimetres.**

**Solution:**

(a) We know that

Area of shaded portion = Area of rectangle ABCD + Area of rectangle PQRS â€“ Area of square LMNO

Substituting the values

= 50 Ã— 5 + 5 Ã— 35 â€“ 5 Ã— 5

By further calculation

= 250 + 175 â€“ 25

So we get

= 250 + 150

= 400 m^{2}

(b) We know that

Area of shaded portion = Area of ABCD â€“ 5 Ã— Area of any small square

It can be written as

= l Ã— b â€“ 5 Ã— side Ã— side

Substituting the values

= 8 Ã— 6 â€“ 5 Ã— 2 Ã— 2

By further calculation

= 48 â€“ 20

= 28 cm^{2}

**8. A rectangular plot 20 m long and 14 m wide is to be covered with grass leaving 2 m all around. Find the area to be laid with grass.**

**Solution:**

Consider ABCD as a plot

Length of plot = 20 m

Breadth of plot = 14 m

Take PQRS as the grassy plot

Here

Length of grassy lawn = 20 â€“ 2 Ã— 2

By further calculation

= 20 â€“ 4

= 16 m

Breadth of grassy lawn = 14 â€“ 2 Ã— 2

By further calculation

= 14 â€“ 4

= 10 m

Area of grassy lawn = length Ã— breadth

Substituting the values

= 16 Ã— 10

= 160 m^{2}

**9. The shaded region of the given diagram represents the lawn in front of a house. On three sides of the lawn there are flower beds of width 2 m.**

**(i) Find the length and the breadth of the lawn.**

**(ii) Hence, or otherwise, find the area of the flower â€“ beds.**

**Solution:**

Consider BCDE as the lawn

(i) We know that

Length of lawn BCDE = BC

It can be written as

= AD â€“ AB â€“ CD

Substituting the values

= 30 â€“ 2 â€“ 2

By further calculation

= 30 â€“ 4

= 26 m

Breadth of lawn BDCE = BE

It can be written as

= AG â€“ GH

Substituting the values

= 12 â€“ 2

= 10 m

(ii) We know that

Area of flower beds = Area of rectangle ADFG â€“ Area of lawn BCDE

It can be written as

= AD Ã— AG â€“ BC Ã— BE

Substituting the values

= 30 Ã— 12 â€“ 26 Ã— 10

By further calculation

= 360 â€“ 260

= 100 m^{2}

**10. A foot path of uniform width runs all around the inside of a rectangular field 50 m long and 38 m wide. If the area of the path is 492 m ^{2}, find its width.**

**Solution:**

Consider ABCD as a rectangular field having

Length = 50 m

Breadth = 38 m

We know that

Area of rectangular field ABCD = l Ã— b

Substituting the values

= 50 Ã— 38

= 1900 m^{2}

Let x m as the width of foot path all around the inside of a rectangular field

Length of rectangular field PQRS = (50 â€“ x â€“ x) = (50 â€“ 2x) m

Breadth of rectangular field PQRS = (38 â€“ x â€“ x) = (38 â€“ 2x) m

Here

Area of foot path = Area of rectangular field ABCD â€“ Area of rectangular field PQRS

Substituting the values

492 = 1900 â€“ (50 â€“ 2x) (38 â€“ 2x)

It can be written as

492 = 1900 â€“ [50 (38 â€“ 2x) â€“ 2x (38 â€“ 2x)]

By further calculation

492 = 1900 â€“ (1900 â€“ 100x â€“ 76x + 4x^{2})

492 = 1900 â€“ 1900 + 100x + 76x â€“ 4x^{2}

On further simplification

492 = 176x â€“ 4x^{2}

Taking out 4 as common

492 = 4 (44x â€“ x^{2})

44x â€“ x^{2} = 492/4 = 123

We get

x^{2} â€“ 44x + 123 = 0

It can be written as

x^{2} â€“ 41x â€“ 3x + 123 = 0

Taking out the common terms

x (x â€“ 41) â€“ 3 (x â€“ 41) = 0

(x â€“ 3) (x â€“ 41) = 0

Here

x â€“ 3 = 0 or x â€“ 41 = 0

So x = 3 m or x = 41 m which is not possible

Therefore, width is 3 m.

**11. The cost of enclosing a rectangular garden with a fence all around at the rate of Rs 15 per metre is Rs 5400. If the length of the garden is 100 m, find the area of the garden.**

**Solution:**

Consider ABCD as a rectangular garden

Length = 100 m

Take breadth = x m

We know that

Perimeter of the garden = 2 (l + b)

Substituting the values

= 2 (100 + x)

= (200 + 2x) m

We know that

Cost of 1 m to enclosing a rectangular garden = Rs 15

So the cost of (200 + 2x) m to enclosing a rectangular garden = 15 (200 + 2x)

= 3000 + 30x

Given cost = Rs 5400

We get

3000 + 30x = 5400

It can be written as

30x = 5400 â€“ 3000

x = 2400/30 = 80 m

Breadth of garden = 80 m

So the area of rectangular field = l Ã— b

Substituting the values

= 100 Ã— 80

= 8000 m^{2}

**12. A rectangular floor which measures 15 m Ã— 8 m is to be laid with tiles measuring 50 cm Ã— 25 cm find the number of tiles required further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?**

**Solution:**

Consider ABCD as a rectangular field of measurement 15m Ã— 8m

Length = 15 m

Breadth = 8 m

Here the area = l Ã— b = 15 Ã— 8 = 120 m^{2}

Measurement of tiles = 50 cm Ã— 25 cm

Length = 50 cm = 50/100 = Â½ m

Breadth = 25 cm = 25/100 = Â¼ m

So the area of one tile = Â½ Ã— Â¼ = 1/8 m^{2}

No. of required tiles = Area of rectangular field/Area of one tile

Substituting the values

= 120/ (1/8)

By further calculation

= (120 Ã— 8)/ 1

= 960 tiles

Length of carpet = 15 â€“ 1 â€“ 1

= 15 â€“ 2

= 13 m

Breadth of carpet = 8 â€“ 1 â€“ 1

= 8 â€“ 2

= 6 m

Area of carpet = l Ã— b

= 13 Ã— 6

= 78 m^{2}

We know that

Area of floor which is uncovered by carpet = Area of floor â€“ Area of carpet

Substituting the values

= 120 â€“ 78

= 42 m^{2}

Fraction = Area of floor which is uncovered by carpet/ Area of floor

Substituting the values

= 42/120

= 7/20

**13. The width of a rectangular room is 3/5 of its length x metres. If its perimeter is y metres, write an equation connecting x and y. Find the floor area of the room if its perimeter is 32 m.**

**Solution:**

It is given that

Length of rectangular room = x m

Width of rectangular room = 3/5 of its length

= 3x/5 m

Perimeter = y m

We know that

Perimeter = 2 (l + b)

Substituting the values

y = 2 [(5x + 3x)/ 5]

By further calculation

y = 2 Ã— 8x/5

y = 16x/5

We get

5y = 16x

16x = 5y â€¦.. (1)

Equation (1) is the required relation between x and y

Given perimeter = 32 m

So y = 32 m

Now substituting the value of y in equation (1)

16x = 5 Ã— 32

By further calculation

x = (5 Ã— 32)/ 16

x = (5 Ã— 2)/ 1

x = 10 m

Breadth = 3/5 Ã— x

Substituting the value of x

= 3/5 Ã— 10

= 3 Ã— 2

= 6 m

Here the floor area of the room = l Ã— b

Substituting the values

= 10 Ã— 6

= 60 m^{2}

**14. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.**

**Solution:**

Consider ABCD as a rectangular garden

Length = 10 m

Breadth = 16 m

So the area of ABCD = l Ã— b

Substituting the values

= 10 Ã— 16

= 160 m^{2}

Consider x m as the width of the walk

Length of rectangular garden PQRS = 10 â€“ x â€“ x = (10 â€“ 2x) m

Breadth of rectangular garden PQRS = 16 â€“ x â€“ x = (16 â€“ 2x) m

**15. A rectangular room is 6 m long, 4.8 m wide and 3.5 m high. Find the inner surface of the four walls.**

**Solution:**

It is given that

Length of rectangular room = 6 m

Breadth of rectangular room = 4.8 m

Height of rectangular room = 3.5 m

Here

Inner surface area of four wall = 2 (l + b) Ã— h

Substituting the values

= 2 (6 + 4.8) Ã— 3.5

By further calculation

= 2 Ã— 10.8 Ã— 3.5

= 21.6 Ã— 3.5

= 75.6 m^{2}

**16. A rectangular plot of land measures 41 metres in length and 22.5 metres in width. A boundary wall 2 metres high is built all around the plot at a distance of 1.5 m from the plot. Find the inner surface area of the boundary wall.**

**Solution:**

It is given that

Length of rectangular plot = 41 metres

Breadth of rectangular plot = 22.5 metres

Height of boundary wall = 2 metre

Here

Boundary wall is built at a distance of 1.5 m

New length = 41 + 1.5 + 1.5

= 41 + 3

= 44 m

New breadth = 22.5 + 1.5 + 1.5

= 22.5 + 3

= 25.5 m

We know that

Inner surface area of the boundary wall = 2 (l + b) Ã— h

Substituting the values

= 2 (44 + 25.5) Ã— 2

By further calculation

= 2 Ã— 69.5 Ã— 2

= 2 Ã— 139

= 278 m^{2}

**17. (a) Find the perimeter and area of the figure**

**(i) given below in which all corners are right angled.**

**(b) Find the perimeter and area of the figure**

**(ii) given below in which all corners are right angles.**

**(c) Find the area and perimeter of the figure**

**(iii) given below in which all corners are right angled and all measurement in centimetres.**

**Solution:**

(a) It is given that

AB = 2m, BE = 4m, FE = 4m and FG = 1.5 m

So BD = 4 + 1.5 = 5.5 m

AC = BD = 5.5 m

CG = (4 + 2) = 6 m

We know that

Perimeter of figure (i) = AC + CG + GF + FE + EB + BA

Substituting the values

= 5.5 + 6 + 1.5 + 4 + 4 + 2

= 23 m

Here

Area of given figure = Area of ABEDC + Area of FEDG

It can be written as

= length Ã— breadth + length Ã— breadth

Substituting the values

= 2 Ã— 5.5 + 4 Ã— 1.5

= 11 + 6

= 17 m^{2}

(b) It is given that

AB = CD = 3m

HI = AC = 7m

JE = BE = 5m

GF = DE = 2m

DG = EF = 8m

GH = JI = 2m

We know that

CH = CD + DG + GH

Substituting the values

= 3 + 8 + 2

= 13 m

Perimeter of the given figure = AB + AC + CH + HI + IJ + JF + FE + BE

Substituting the values

= 3 + 7 + 13 + 7 + 2 + 5 + 8 + 5

= 50 m

Here

Area of given figure = Area of first figure + Area of second figure + Area of third figure

Substituting the values

= 7 Ã— 3 + 8 Ã— 2 + 7 Ã— 2

By further calculation

= 21 + 16 + 14

= 51 m^{2}

(c) It is given that

AB = 12 cm

AL = BC = 7 cm

JK = DE = 5 cm

HJ = GF = 3 cm

LK = HG = CD = 2 cm

We know that

Perimeter of given figure = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

Substituting the values

= 12 + 7 + 2 + 5 + 3 + 3 + 2 + 3 + 3 + 5 + 2 + 7

= 54 cm

Here

Area of given figure =Area of first part + Area of second part + Area of third part + Area of fourth part + Area of fifth part

Substituting the values

= 7 Ã— 2 + 2 Ã— 3 + (2 + 3) Ã— 2 + 2 Ã— 3 + 7 Ã— 2

By further calculation

= 14 + 6 + 10 + 6 + 14

= 50 cm^{2}

**18. The length and the breadth of a rectangle are 12 cm and 9 cm respectively. Find the height of a triangle whose base is 9 cm and whose area is one third that of rectangle.**

**Solution:**

It is given that

Length of a rectangle = 12 cm

Breadth of a rectangle = 9 cm

So the area = l Ã— b

Substituting the values

= 12 Ã— 9

= 108 cm^{2}

Using the condition

Area of triangle ABC = 1/3 Ã— area of rectangle

Substituting the values

= 1/3 Ã— 108

= 36 cm^{2}

Consider h cm as the height of triangle ABC

Area of triangle ABC = Â½ Ã— base Ã— height

Substituting the values

36 = Â½ Ã— 9 Ã— h

By further calculation

36 Ã— 2 = 9 Ã— h

h = (36 Ã— 2)/ 9

So we get

h = 4 Ã— 2

h = 8 cm

Therefore, height of triangle ABC is 8 cm.

**19. The area of a square plot is 484 m ^{2}. Find the length of its one side and the length of its one diagonal.**

**Solution:**

It is given that

ABCD is a square plot having area = 484 m^{2}

Sides of square are AB, BC, CD and AD

We know that

Area of square = side Ã— side

Substituting the values

484 = (side)^{2}

So we get

Side = âˆš484 = 22 m

AB = BC = 22 m

In triangle ABC

Using Pythagoras Theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

AC^{2} = 22^{2} + 22^{2}

AC^{2} = 484 + 484 = 968

By further calculation

AC = âˆš968 = âˆš (484 Ã— 2)

AC = 22 Ã— âˆš2

So we get

AC = 22 Ã— 1.414 = 31.11 m

Therefore, length of side is 22 m and length of diagonal is 31.11 m.

**20. A square has the perimeter 56 m. Find its area and the length of one diagonal correct up to two decimal places.**

**Solution:**

Consider ABCD as a square with side x m

Perimeter of square = 4 Ã— side

Substituting the values

56 = 4x

By further calculation

x = 56/4 = 14 m

In triangle ABC

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

AC^{2} = 14^{2} + 14^{2}

By further calculation

AC^{2} = 196 + 196 = 392

So we get

AC = âˆš392

AC = âˆš (196 Ã— 2)

AC = 14 âˆš2

Substituting the value of âˆš2

AC = 14 Ã— 1.414

AC = 19.80 m

Therefore, the side of square is 14 m and diagonal is 19.80 m.

**21. A wire when bent in the form of an equilateral triangle encloses an area of 36 âˆš3 cm ^{2}. Find the area enclosed by the same wire when bent to form:**

**(i) a square, and**

**(ii) a rectangle whose length is 2 cm more than its width.**

**Solution:**

It is given that

Area of equilateral triangle = 36 âˆš3 cm^{2}

Consider x cm as the side of equilateral triangle

We know that

Area = âˆš3/4 (side)^{2}

Substituting the values

36 âˆš3 = âˆš3/4 Ã— (x)^{2}

By further calculation

x^{2} = (36 âˆš3 Ã— 4)/ âˆš3

x^{2} = 36 Ã— 4

So we get

x = âˆš (36 Ã— 4)

x = 6 Ã— 2

x = 12 cm

Here

Perimeter of equilateral triangle = 3 Ã— side

= 3 Ã— 12

= 36 cm

(i) We know that

Perimeter of equilateral triangle = Perimeter of square

It can be written as

36 = 4 Ã— side

So we get

Side = 36/4 = 9 cm

Area of square = side Ã— side

= 9 Ã— 9

= 81 cm^{2}

(ii) We know that

Perimeter of triangle = Perimeter of rectangle â€¦. (1)

According to the condition of rectangle

Length is 2 cm more than its width

Width of rectangle = x cm

Length of rectangle = (x + 2) cm

Perimeter of rectangle = 2 (l + b)

Substituting the values

= 2 [(x + 2) + x]

By further calculation

= 2 (2x + 2)

= 4x + 4

Using equation (1)

4x + 4 = Perimeter of triangle

4x + 4 = 36

By further calculation

4x = 36 â€“ 4 = 32 cm

So we get

x = 32/4 = 8 cm

Here

Length of rectangle = 8 + 2 = 10 cm

Breadth of rectangle = 8 cm

Area of rectangle = length Ã— breadth

= 10 Ã— 8

= 80 cm^{2}

**22. Two adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the area of the parallelogram. Also find the distance between shorter sides.**

**Solution:**

Consider ABCD as a parallelogram

Longer side AB = 15 cm

Shorter side = 10 cm

Distance between longer side DM = 8 cm

Consider DN as the distance between the shorter side

Area of parallelogram ABCD = base Ã— height

We can write it as

= AB Ã— DM

Substituting the values

= 15 Ã— 8

= 120 cm^{2}

If base is AD

Area of parallelogram = AD Ã— DN

Substituting the values

120 = 10 Ã— DN

So we get

DN = 120/10 = 12 cm

Therefore, the area of parallelogram is 120 cm^{2} and the distance between shorter side is 12 cm.

**23. ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.**

**Solution:**

It is given that

ABCD is a parallelogram

AB = 12 cm, BC = 10 cm and AC = 16 cm

Area of triangle ABC

BC = a = 10 cm

AC = b = 16 cm

AB = c = 12 cm

We know that

s = (a + b + c)/ 2

Substituting the values

s = (10 + 16 + 12)/ 2

By further calculation

s = 38/2 = 19 cm

Here

So we get

= 3 âˆš399 cm^{2}

We know that

Area of parallelogram = 2 Ã— Area of triangle ABC

Substituting the values

= 2 Ã— 3 âˆš399

= 6 âˆš399

So we get

= 6 Ã— 19.96

= 119.8 cm^{2}

Consider DM as the distance between the shorter lines

Base = AD = BC = 10 cm

Area of parallelogram = AD Ã— DM

Substituting the values

119.8 = 10 Ã— DM

By further calculation

DM = 119.8/ 10

DM = 11.98 cm

Therefore, the distance between shorter lines is 11.98 cm.

**24. Diagonals AC and BD of a parallelogram ABCD intersect at O. Given that AB = 12 cm and perpendicular distance between AB and DC is 6 cm. Calculate the area of the triangle AOD.**

**Solution:**

It is given that

ABCD is a parallelogram

AC and BD are the diagonal which intersect at O

AB = 12 cm and DM = 6 cm

We know that

Area of parallelogram ABCD = AB Ã— DM

Substituting the values

= 12 Ã— 6

= 72 cm^{2}

Similarly

Area of triangle AOD = Â¼ Ã— Area of parallelogram

Substituting the values

= Â¼ Ã— 72

= 18 cm^{2}

**25. ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.**

**Solution:**

It is given that

ABCD is a parallelogram

AB = 10 cm, AC = 12 cm

AO = CO = 12/2 = 6 cm

BD = 16 cm

BO = OD = 16/2 = 8 cm

In triangle AOB

a = 10 cm, b = AO = 6 cm, c = BO = 8 cm

We know that

s = (a + b + c)/ 2

Substituting the values

s = (10 + 6 + 8)/ 2

By further calculation

s = 24/2 = 12 cm

So we get

= 12 Ã— 2

= 24 cm^{2}

We know that

Area of parallelogram ABCD = 4 Ã— Area of triangle AOB

Substituting the values

= 4 Ã— 24

= 96 cm^{2}

**26. The area of a parallelogram is p cm ^{2} and its height is q cm. A second parallelogram has equal area but its base is r cm more than that of the first. Obtain an expression in terms of p, q and r for the height h of the second parallelogram.**

**Solution:**

It is given that

Area of a parallelogram = p cm^{2}

Height of first parallelogram = q cm

We know that

Area of parallelogram = base Ã— height

Substituting the values

p = base Ã— q

Base = p/q

Here

Base of second parallelogram = (p/q + r)

Taking LCM

= (p + qr)/ q cm

Area of second parallelogram = Area of first parallelogram = p cm^{2}

It can be written as

Base Ã— height = p cm^{2}

Substituting the values

[(p + qr)/q] Ã— h = pSo we get

h = pq/ (p + qr) cm

Therefore, the height of second parallelogram is h = pq/ (p + qr) cm.

**27. What is the area of a rhombus whose diagonals are 12 cm and 16 cm?**

**Solution:**

It is given that

ABCD is a rhombus

BD = 12 cm and AC = 16 cm are diagonals

We know that

Area of rhombus ABCD = Â½ Ã— AC Ã— BD

Substituting the values

= Â½ Ã— 16 Ã— 12

By further calculation

= 8 Ã— 12

= 96 cm^{2}

**28. The area of a rhombus is 98 cm ^{2}. If one of its diagonal is 14 cm, what is the length of the other diagonal?**

**Solution:**

It is given that

Area of rhombus = 98 cm^{2}

One of its diagonal = 14 cm

We know that

Area of rhombus = Â½ Ã— product of diagonals

Substituting the values

98 = Â½ Ã— one diagonal Ã— other diagonal

98 = Â½ Ã— 14 Ã— other diagonal

By further calculation

Other diagonal = (98 Ã— 2)/ 14

= 7 Ã— 2

= 14 cm

Therefore, the other diagonal is 14 cm.

**29. The perimeter of a rhombus is 45 cm. If its height is 8 cm, calculate its area.**

**Solution:**

It is given that

ABCD is a rhombus

Consider x cm as each side

Perimeter = 45 cm

AB + BC + CD + AD = 45 cm

Substituting the values

x + x + x + x = 45

4x = 45

By division

x = 45/4 cm

We know that

Height = 8 cm

Area of rhombus = base Ã— height

Substituting the values

= 45/4 Ã— 8

= 45 Ã— 2

= 90 cm^{2}

**30. PQRS is a rhombus. If it is given that PQ = 3 cm and the height of the rhombus is 2.5 cm, calculate its area.**

**Solution:**

It is given that

PQRS is a rhombus

PQ = 3 cm

Height = 2.5 cm

Consider PQ as the base of rhombus PQRS.

SM = 2.5 cm is the height of rhombus

We know that

Area of rhombus PQRS = base Ã— height

Substituting the values

= 3 Ã— 2.5

= 7.5 cm^{2}

**31. If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.**

**Solution:**

Consider ABCD as a rhombus with AC and BD as two diagonals

Here

AC = 8 cm and BD = 6 cm

AO = 4 cm and BO = 3 cm

In triangle ABC

Using Pythagoras theorem

AB^{2} = AO^{2} + BO ^{2}

Substituting the values

AB^{2} = 4^{2} + 3^{2}

By further calculation

AB^{2} = 16 + 9 = 25

So we get

AB = âˆš25 = 5 cm

Side of rhombus ABCD = 5 cm

Here

Perimeter of rhombus = 4 Ã— side

Substituting the values

= 4 Ã— 5

= 20 cm

**32. If the sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate**

**(i) the length of the other diagonal, and**

**(ii) the area of the rhombus.**

**Solution:**

It is given that

ABCD is a rhombus with side AB, BC, CD and AD

AB = BC = CD = AD = 5 cm

AC = 8 cm and AO = 4 cm

In triangle AOB

Using Pythagoras theorem

AB^{2} = AO^{2} + BO^{2}

Substituting the values

5^{2} = 4^{2} + BO^{2}

By further calculation

25 = 16 + BO^{2}

BO^{2} = 25 â€“ 16 = 9

So we get

BO = âˆš9 = 3 cm

BD = 2 Ã— BO = 2 Ã— 3 = 6 cm

Length of other diagonal = 6 cm

Area of rhombus = Â½ Ã— product of diagonals

Substituting the values

= Â½ Ã— 8 Ã— 6

By further calculation

= 4 Ã— 6

= 24 cm^{2}

**33. (a) The diagram (i) given below is a trapezium. Find the length of BC and the area of the trapezium. Assume AB = 5 cm, AD = 4 cm, CD = 8 cm.**

**(b) The diagram (ii) given below is a trapezium. Find **

**(i) AB**

**(ii) area of trapezium ABCD**

**(c) The cross-section of a canal is shown in figure (iii) given below. If the canal is 8 m wide at the top and 6 m wide at the bottom and the area of the cross-section is 16.8 m ^{2}, calculate its depth.**

**Solution:**

(a) It is given that

ABCD is a trapezium

AB = 5 cm, AD = 4 cm and CD = 8 cm

Construct BN perpendicular to CD

Here

BN = 4 cm

CN = CD â€“ ND

CN = CD â€“ AO

CN = 8 â€“ 5 = 3 cm

In triangle BCN

Using Pythagoras theorem

BC^{2} = BN^{2} + CN^{2}

Substituting the values

BC^{2} = 4^{2} + 3^{2}

By further calculation

BC^{2} = 16 + 9 = 25

BC = âˆš25 = 5 cm

Length of BC = 5 cm

Area of trapezium = Â½ Ã— sum of parallel sides Ã— height

It can be written as

= Â½ Ã— (AB + CD) Ã— AD

Substituting the values

= Â½ Ã— (5 + 8) Ã— 4

By further calculation

= Â½ Ã— 13 Ã— 4

So we get

= 13 Ã— 2

= 26 cm^{2}

Area of trapezium = 26 cm^{2}

(b) From the figure (ii)

AD = 8 units

BC = 2 units

CD = 10 units

Construct CN perpendicular to AD

AN = 2 units

We know that

DN = AD â€“ DN

Substituting the values

= 8 â€“ 6

= 2 units

In triangle CDN

Using Pythagoras theorem

CD^{2} = DN^{2} + NC^{2}

Substituting the values

10^{2} = 6^{2} + NC^{2}

By further calculation

NC^{2} = 10^{2} – 6^{2}

NC^{2} = 100 â€“ 36 = 64

So we get

NC = âˆš64 = 8 units

From the figure NC = AB = 8 units

We know that

Area of trapezium = Â½ Ã— sum of parallel sides Ã— height

It can be written as

= Â½ Ã— (BC + AD) Ã— AB

Substituting the values

= Â½ Ã— (2 + 8) Ã— 8

By further calculation

= Â½ Ã— 10 Ã— 8

= 5 Ã— 8

= 40 sq. units

(c) Consider ABCD as the cross section of canal in the shape of trapezium.

AB = 6 m, DC = 8 m

Take AL as the depth of canal

So the area of cross-section = 16.8 m^{2}

It can be written as

Â½ Ã— sum of parallel sides Ã— depth = 16.8

Â½ Ã— (AB + DC) Ã— AL = 16.8

Substituting the values

Â½ Ã— (6 + 8) Ã— AL = 16.8

By further calculation

Â½ Ã— 14 Ã— AL = 16.8

AL = (16.8 Ã— 2)/ 14

So we get

AL = (16.8 Ã— 1)/7

AL = 2.4 m

**34. The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.**

**Solution:**

Consider ABCD as a trapezium in which AB || DC

Height CL = 12 cm

E and F are the mid-points of sides AD and BC

EF = 18 cm

We know that

EF = Â½ (AB + DC) = 18 cm

Here

Area of trapezium ABCD = Â½ (AB + DC) Ã— height

Substituting the values

= 18 Ã— 12

= 216 cm^{2}

**35. The area of a trapezium is 540 cm ^{2}. If the ratio of parallel sides is 7: 5 and the distance between them is 18 cm, find the length of parallel sides.**

**Solution:**

It is given that

Area of trapezium = 540 cm^{2}

Ratio of parallel sides = 7: 5

Consider 7x cm as one parallel side

Other parallel side = 5x cm

Distance between the parallel sides = height = 18 cm

We know that

Area of trapezium = Â½ Ã— sum of parallel sides Ã— height

Substituting the values

540 = Â½ Ã— (7x + 5x) Ã— 18

By further calculation

540 = Â½ Ã— 12x Ã— 18

540 = 6x Ã— 18

540 = 108x

x = 540/108 = 5

Here

First parallel side = 7x = 7 Ã— 5 = 35 cm

Second parallel side = 5x = 5 Ã— 5 = 25 cm

**36. The parallel sides of an isosceles trapezium are in the ratio 2: 3. If its height is 4 cm and area is 60 cn ^{2}, find the perimeter.**

**Solution:**

It is given that

ABCD is an isosceles trapezium

BC = AD

Height = 4 cm

Consider CD = 2x and AB = 3x

We know that

Area of trapezium = Â½ (sum of parallel sides) Ã— height

Substituting the values

60 = Â½ Ã— (2x + 3x) Ã— 4

By further calculation

60 = Â½ Ã— 5x Ã— 4

60 = 5x Ã— 2

60 = 10x

So we get

x = 60/10 = 6

Here

CD = 2x = 2 Ã— 6 = 12 cm

AB = 3x = 3 Ã— 6 = 18 cm

AN = BM

We can write it as

AN = AB â€“ BN

AN = AB – (MN + BM)

We know that

MN = CD

AN = AB â€“ (CD + BM)

Similarly BM = AN

AN = AB â€“ (CD + AN)

Substituting the values

AN = 18 â€“ (12 + AN)

AN = 18 â€“ 12 â€“ AN

AN + AN = 6

2AN = 6

By division

AN = 6/2 = 3

In triangle AND

Using Pythagoras theorem

AD^{2} = DN^{2} + AN^{2}

Here DN = 4 cm

AD^{2} = 4^{2} + 3^{2}

By further calculation

AD^{2} = 16 + 9 = 25

So we get

AD= âˆš25 = 5 cm

Here AD = BC = 5 cm

Perimeter of trapezium = AB + BC + CD + AD

Substituting the values

= 18 + 5 + 12 + 5

= 40 cm

**37. The area of a parallelogram is 98 cm ^{2}. If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.**

**Solution:**

It is given that

Area of parallelogram = 98 cm^{2}

Condition â€“ If one altitude is half the corresponding base

Take base = x cm

Corresponding altitude = x/2 cm

We know that

Area of parallelogram = base Ã— altitude

Substituting the values

98 = x Ã— x/2

98 = x^{2}/2

By cross multiplication

x^{2} = 98 Ã— 2 = 196

So we get

x = âˆš196 = 14 cm

Base = 14 cm

Here altitude = 14/2 = 7 cm

**38. The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.**

**Solution:**

The dimensions of rectangular garden are

Breadth = x m

Length = (x + 12) m

We know that

Area = l Ã— b

Substituting the values

Area = (x + 12) Ã— x

Area = (x^{2} + 12x) m^{2}

Perimeter = 2 (l + b)

Substituting the values

= 2 [(x + 12) + x]

= 2 [ x + 12 + x]

= 2 [2x + 12]

= (4x + 24) m

Based on the question

Numerical value of area = 4 Ã— numerical value of perimeter

x ^{2} + 12x = 4 Ã— (4x + 24)

By further calculation

x^{2 }+ 12x = 16x + 96

x^{2} + 12x â€“ 16x â€“ 96 = 0

x^{2} â€“ 4x â€“ 96 = 0

It can be written as

x^{2}Â â€“ 12x + 8x â€“ 96 = 0

Taking out the common terms

x (x â€“ 12) + 8 (x â€“ 12) = 0

(x + 8) (x â€“ 12) = 0

Here

x + 8 = 0 or x â€“ 12 = 0

So we get

x = -8 (not possible) or x = 12

Breadth of rectangular garden = 12 m

Length of rectangular garden = 12 + 12 = 24 m

**39. If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.**

**Solution:**

It is given that

Perimeter of a rectangular plot = 68 m

Length of its diagonal = 26 m

ABCD is a rectangular plot of length x m and breath y m

Perimeter = 2 (length + breadth)

Substituting the values

68 = 2 (x + y)

By further calculation

68/2 = x + y

34 = x + y

So we get

x = 34 â€“ y â€¦â€¦ (1)

In triangle ABC

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

26^{2} = x^{2} + y^{2}

x^{2}Â + y^{2} = 676

Now substituting the value of x in equation (1)

(34 â€“ y)^{2}Â + y^{2} = 676

1156 + y^{2} â€“ 68y + y^{2} = 676

By further calculation

2y^{2} â€“ 68y + 1156 â€“ 676 = 0

2y^{2} â€“ 68y â€“ 480 = 0

Taking 2 as common

2 (y^{2} â€“ 34y â€“ 240) = 0

y^{2} â€“ 34y â€“ 240 = 0

It can be written as

y^{2} â€“ 24y â€“ 10y â€“ 240 = 0

Taking out the common terms

y (y â€“ 24) â€“ 10 (y â€“ 24) = 0

(y â€“ 10) (y â€“ 24) = 0

Here

y â€“ 10 = 0 or y â€“ 24 = 0

y = 10 m or y = 24 m

Now substituting the value of y in equation (1)

y = 10 m, x = 34 â€“ 10 = 24 m

y = 24 m, x = 34 â€“ 24 = 10 m

Area in both cases = xy

= 24 Ã— 10 or 10 Ã— 24

= 240 m^{2}

Therefore, the area of the rectangular block is 240 m^{2}.

**40. A rectangle has twice the area of a square. The length of the rectangle is 12 cm greater and the width is 8 cm greater than 2 side of a square. Find the perimeter of the square.**

**Solution:**

Consider

Side of a square = x cm

Length of rectangle = (x + 12) cm

Breadth of rectangle = (x + 8) cm

We know that

Area of square = side Ã— side = x Ã— x = x^{2} cm^{2}

Area of rectangle = l Ã— b

Substituting the values

= (x + 12) (x + 8) cm^{2}

Based on the question

Area of rectangle = 2 Ã— area of square

Substituting the values

(x + 12) (x + 8) = 2 Ã— x^{2}

It can be written as

x (x + 8) + 12 (x + 8) = 2x^{2}

x^{2} + 8x + 12x + 96 = 2x^{2}

x^{2} â€“ 2x^{2} + 8x + 12x + 96 = 0

By further calculation

-x^{2} + 20x + 96 = 0

– (x^{2} â€“ 20x â€“ 96) = 0

x^{2} â€“ 20x â€“ 96 = 0

We can write it as

x^{2} â€“ 24x + 4x â€“ 96 = 0

x (x â€“ 24) + 4 (x â€“ 24) = 0

(x + 4) (x â€“ 24) = 0

Here

x + 4 = 0 or x â€“ 24 = 0

x = -4 or x = 24 cm

Side of square = 24 cm

Perimeter of square = 4 Ã— side

Substituting the values

= 4 Ã— 24

= 96 cm

**41. The perimeter of a square is 48 cm. The area of a rectangle is 4 cm ^{2} less than the area of the square. If the length of the rectangle is 4 cm greater than its breadth, find the perimeter of the rectangle.**

**Solution:**

It is given that

Perimeter of a square = 48 cm

Side = perimeter/4 = 48/4 = 12 cm

We know that

Area = side^{2} = 12^{2} = 144 cm^{2}

Area of rectangle = 144 â€“ 4 = 140 cm^{2}

Take breadth of rectangle = x cm

Length of rectangle = x + 4 cm

So the area = (x + 4) Ã— x cm^{2}

Substituting the values

(x + 4) x = 140

By further calculation

x^{2} + 4x â€“ 140 = 0

x^{2} + 14x â€“ 10x â€“ 140 = 0

Taking out the common terms

x (x + 14) â€“ 10 (x + 14) = 0

(x + 14) (x â€“ 10) = 0

Here

x + 14 = 0 where x = – 14

x â€“ 10 = 0 where x = 10

Breadth = 10 cm

Length = 10 + 4 = 14 cm

Perimeter = 2 (l + b)

Substituting the values

= 2 (14 + 10)

= 2 Ã— 24

= 48 cm

**42. In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangle; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.**

**Solution:**

It is given that

ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm

HAD and BFC are equilateral triangle with each side = 8 cm

AEB and DCG are right angled isosceles triangles with hypotenuse = 10 cm

Consider AE = EB = x cm

In triangle ABE

AE^{2} + EB^{2} = AB^{2}

Substituting the values

x^{2} + x^{2} = 10^{2}

2x^{2} = 100

By further calculation

x^{2} = 100/2 = 50

x = âˆš50 = âˆš (25 Ã— 2) = 5âˆš2 cm

We know that

Area of triangle AEB = Area of triangle GCD

It can be written as

= Â½ Ã— x Ã— x

= Â½ x^{2} cm^{2}

Substituting the value of x

= Â½ Ã— 50

= 25 cm^{2}

Area of triangle HAD = Area of BFC

It can be written as

= âˆš3/4 Ã— 8^{2}

= âˆš3/4 Ã— 64

= 16âˆš3 cm^{2}

Area of shaded portion = Area of rectangle ABCD + 2 area of triangle AEB + 2 area of triangle BFC

Substituting the values

= (10 Ã— 8 + 2 Ã— 25 + 2 Ã— 16âˆš3)

By further calculation

= 80 + 50 + 32âˆš3

So we get

= (130 + 32âˆš3) cm^{2}

Here

Perimeter of the figure = AE + EB + BF + FC + CD + GD + DH + HA

It can be written as

= 4AE + 4BF

Substituting the values

= 4 Ã— 5âˆš2 + 4 Ã— 8

So we get

= 20âˆš2 + 32

= (32 + 20âˆš2) cm

**43. (a) Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DEFG is an isosceles trapezium. All measurements are in centimetres.**

**(b) Find the area enclosed by the figure (ii) given below. AH measurements are in centimetres.**

**(c) In the figure (iii) given below, from a 24 cm Ã— 24 cm piece of cardboard, a block in the shape of letter M is cut off. Find the area of the cardboard left over, all measurements are in centimetres.**

**Solution:**

(a) It is given that

ABC is an equilateral triangle and DEFG is an isosceles trapezium

EF = GD = 5 cm

DE = 6 cm

GF = GB + BC + CF

Substituting the values

= 3 + 6 + 3

= 12 cm

AB = AC = BC = 6 cm

Join BD and CE

In right triangle CEF

CE^{2} = EF^{2} â€“ CF^{2}

Substituting the values

= 5^{2} – 3^{2}

= 25 â€“ 9

= 16

So we get

CE = âˆš16 = 4 cm

Area of triangle ABC = âˆš3/4 Ã— 6^{2}

By further calculation

= âˆš3/4 Ã— 36

= 9âˆš3 cm^{2}

Area of trapezium DEFG = Â½ (DE + GF) Ã— CE

Substituting the values

= Â½ Ã— (6 + 12) Ã— 4

By further calculation

= Â½ Ã— 18 Ã— 4

= 36 cm^{2}

So the area of figure = 9âˆš3 + 36

Substituting the values

= 9 Ã— 1.732 + 36

= 15.59 + 36

= 51.59 cm^{2}

(b) We know that

Length of rectangle = 2 + 2 + 2 + 2 = 8 cm

Width of rectangle = 2 cm

Area of rectangle = l Ã— b

Substituting the values

= 8 Ã— 2

= 16 cm^{2}

Here

Area of each trap = Â½ (2 + 2) Ã— (6 â€“ 2)

By further calculation

= Â½ Ã— 4 Ã— 4

= 8 cm^{2}

So the total area = area of rectangle + area of 2 trapezium

Substituting the values

= 16 + 8 + 8

= 32 cm^{2}

(c) We know that

Length of each rectangle = 24 cm

Width of each rectangle = 6 cm

Area of each rectangle = l Ã— b

Substituting the values

= 24 Ã— 6

= 144 cm^{2}

Base of each parallelogram = 8 cm

Height of each parallelogram = 6 cm

So the area of each parallelogram = 8 Ã— 6 = 48 cm^{2}

Here

Area of the M-shaped figure = 2 Ã— 144 + 2 Ã— 48

So we get

= 288 + 96

= 384 cm^{2}

Area of the square cardboard = 24 Ã— 24 = 576 cm^{2}

Area of the removing cardboard = 576 â€“ 384 = 192 cm^{2}

**44. (a) The figure (i) given below shows the cross-section of the concrete structure with the measurements as given. Calculate the area of cross-section.**

**(b) The figure (ii) given below shows a field with the measurements given in metres. Find the area of the field.**

**(c) Calculate the area of the pentagon ABCDE shown in fig (iii) below, given that AX = BX = 6 cm, EY = CY = 4 cm, DE = DC = 5 cm, DX = 9 cm and DX is perpendicular to EC and AB.**

**Solution:**

(a) From the figure (i)

AB = 1.8 m, CD = 0.6 m, DE = 1.2 m

EF = 0.3 m, AF = 2.4 m

Construct DE to meet AB in G

âˆ FEG = âˆ GAF = 90^{0}

So, AGEF is a rectangle

We know that

Area of given figure = Area of rectangle AGEF + Area of trapezium GBCD

It can be written as

= l Ã— b + Â½ (sum of parallel sides Ã— height)

= AF Ã— AG + Â½ (GB + CD) Ã— DG

Substituting the values

= 2.4 Ã— 0.3 + Â½ [(AB â€“ AG) + CD] Ã— (DE + EG)

Here AG = FE and using EG = AF

= 0.72 + Â½ [(1.8 â€“ 0.3) + 0.6] Ã— (1.2 + 2.4)

By further calculation

= 0.72 + Â½ [1.5 + 0.6] Ã— 3.6

= 0.72 + Â½ Ã— 2.1 Ã— 3.6

So we get

= 0.72 + 2.1 Ã— 1.8

= 0.72 + 3.78

= 4.5 m^{2}

(b) It is given that

ABCD is a pentagonal field

AX = 12 m, BX = 30 m, XZ = 15 m, CZ = 25 m,

DZ = 10 m, AD = 12 + 15 + 10 = 37 m, EY = 20 m

We know that

Area of pentagonal field ABCDE = Area of triangle ABX + Area of trapezium BCZX + Area of triangle CDZ + Area of triangle AED

It can be written as

= Â½ Ã— base Ã— height + Â½ (sum of parallel sides) Ã— height + Â½ Ã— base Ã— height + Â½ Ã— base Ã— height

= Â½ Ã— BX Ã— AX + Â½ (BX + CZ) Ã— XZ + Â½ Ã— CZ Ã— DZ + Â½ Ã— AD Ã— EY

Substituting the values

= Â½ Ã— 30 Ã— 12 + Â½ (30 + 25) Ã— 15 + Â½ Ã— 25 Ã— 10 + Â½ Ã— 37 Ã— 20

By further calculation

= 15 Ã— 12 + 7.5 Ã— 55 + 25 Ã— 5 + 37 Ã— 10

So we get

= 180 + 412.5 + 125 + 370

= 1087.5 m^{2}

(c) It is given that

ABCDE is a pentagon

AX = BX = 6 cm, EY = CY = 4 cm

DE = DC = 5 cm, DX = 9 cm

Construct DX perpendicular to EC and AB

In triangle DEY

Using Pythagoras Theorem

DE^{2} = DY^{2} + EY^{2}

Substituting the values

5^{2} = DY^{2} + 4^{2}

25 = DY^{2} + 16

DY^{2} = 25 â€“ 16 = 9

So we get

DY = âˆš9 = 3 cm

Here

Area of pentagonal field ABCDE = Area of triangle DEY + Area of triangle DCY + Area of trapezium EYXA + Area of trapezium CYXB

It can be written as

= Â½ Ã— base Ã— height + Â½ Ã— base Ã— height + Â½ Ã— (sum of parallel sides) Ã— height + Â½ Ã— (sum of parallel sides) Ã— height

= Â½ Ã— EY Ã— DY + Â½ Ã— CY Ã— DY + Â½ Ã— (EY + AX) Ã— XY + Â½ Ã— (CY + BX) Ã— XY

Substituting the values

= Â½ Ã— 4 Ã— 3 + Â½ Ã— 4 Ã— 3 + Â½ Ã— (4 + 6) Ã— (DX â€“ DY) + Â½ (4 + 6) Ã— (DX â€“ DY)

By further calculation

= 2 Ã— 3 + 2 Ã— 2 + Â½ Ã— 10 Ã— (9 â€“ 3) + Â½ Ã— 10 Ã— (9 â€“ 3)

So we get

= 6 + 6 + 5 Ã— 6 + 5 Ã— 6

= 6 + 6 + 30 + 30

= 72 cm^{2}

**45. If the length and the breadth of a room are increased by 1 metre the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres. Find the perimeter of the room.**

**Solution:**

Take length of room = x m

Breadth of room = y m

Here

Area of room = l Ã— b = xy m^{2}

We know that

Length is increased by 1 m then new length = (x + 1) m

Breadth is increased by 1 m then new breadth = (y + 1) m

So the new area = new length Ã— new breadth

Substituting the values

= (x + 1) (y + 1) m^{2}

Based on the question

xy = (x + 1) (y + 1) â€“ 21

By further calculation

xy = x (y + 1) + 1 (y + 1) â€“ 21

xy = xy + x + y + 1 â€“ 21

So we get

0 = x + y + 1 â€“ 21

0 = x + y â€“ 20

x + y â€“ 20 = 0

x + y = 20 â€¦.. (1)

Similarly

Length is increased by 1 metre then new length = (x + 1) metre

Breadth is decreased by 1 metre than new breadth = (y â€“ 1) metre

So the new area = new length Ã— new breadth

= (x + 1) (y â€“ 1) m^{2}

Based on the question

xy = (x + 1) (y â€“ 1) + 5

By further calculation

xy = x (y â€“ 1) + 1 (y â€“ 1) + 5

xy = xy â€“ x + y â€“ 1 + 5

0 = -x + y + 4

So we get

x â€“ y = 4 â€¦â€¦ (2)

By adding equations (1) and (2)

2x = 24

x = 24/2 = 12 m

Now substituting the value of x in equation (1)

12 + y = 20

y = 20 â€“ 12 = 8 m

Here

Length of room = 12 m

Breadth of room = 8 m

So the perimeter = 2 (l + b)

Substituting the values

= 2 (12 + 8)

= 2 Ã— 20

= 40 m

**46. A triangle and a parallelogram have the same base and same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.**

**Solution:**

It is given that

Sides of a triangle = 26 cm, 28 cm and 30 cm

We know that

s = (a + b + c)/ 2

Substituting the values

s = (26 + 28 + 30)/ 2

s = 84/2

s = 42 cm

Here

So we get

= 2 Ã— 4 Ã— 6 Ã— 7

= 336 cm^{2}

We know that

Base = 28 cm

Height = Area/base

Substituting the values

= 336/28

= 12 cm

**47. A rectangle of area 105 cm ^{2} has its length equal to x cm. Write down its breadth in terms of x. Given that its perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.**

**Solution:**

It is given that

Area of rectangle = 105 cm^{2}

Length of rectangle = x cm

We know that

Area = length Ã— breadth

Substituting the values

105 = x Ã— breadth

x = 105/ x cm

Perimeter of rectangle = 44 cm

So we get

2 (l + b) = 44

2 (x + 105/x) = 44

By further calculation

(x^{2} + 105)/ x = 22

By cross multiplication

x^{2} + 105 = 22x

x^{2} â€“ 22x + 105 = 0

We can write it as

x^{2} â€“ 15x â€“ 7x + 105 = 0

x (x â€“ 15) â€“ 7 (x â€“ 15) = 0

(x â€“ 7) (x â€“ 15) = 0

Here

x â€“ 7 = 0 or x â€“ 15 = 0

x = 7 cm or x = 15 cm

If x = 7 cm,

Breadth = 105/7 = 15 cm

If x = 15 cm,

Breadth = 105/15 = 7 cm

Therefore, the required dimensions of rectangle are 15 cm and 7 cm.

**48. The perimeter of a rectangular plot is 180 m and its area is 1800 m ^{2}. Take the length of plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the value of the length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.**

**Solution:**

It is given that

Perimeter of a rectangular plot = 180 m

Area of a rectangular plot = 1800 m^{2}

Take length of rectangle = x m

Here

Perimeter = 2 (length + breadth)

Substituting the values

180 = 2 (x + breadth)

180/2 = x + breadth

x + breadth = 90

Breadth = 90 â€“ x m

We know that

Area of rectangle = l Ã— b

Substituting the values

1800 = x Ã— (90 â€“ x)

90x â€“ x^{2} = 1800

It can be written as

– (x^{2}Â â€“ 90x) = 1800

x^{2} â€“ 90x = – 1800

By further calculation

x^{2} â€“ 90x + 1800 = 0

x^{2} â€“ 60x â€“ 30x + 1800 = 0

x (x â€“ 60) â€“ 30 (x â€“ 60) = 0

(x â€“ 30) (x â€“ 60) = 0

Here

x â€“ 30 = 0 or x â€“ 60 = 0

x = 30 or x = 60

If x = 30 m

Breadth = 90 â€“ 30 = 60 m

If x = 60 m

Breadth = 90 â€“ 60 = 30 m

Therefore, the required length of rectangle is 60 m and the breadth of rectangle is 30 m.

Exercise 16.3

**1. Find the length of the diameter of a circle whose circumference is 44 cm.**

**Solution:**

Consider radius of the circle = r cm

Circumference = 2 Ï€r

We know that

2 Ï€r = 44

So we get

(2 Ã— 22)/ 7 r = 44

By further calculation

r = (44 Ã— 7)/ (2 Ã— 22) = 7 cm

Diameter = 2r = 2 Ã— 7 = 14 cm

**2. Find the radius and area of a circle if its circumference is 18Ï€ cm.**

**Solution:**

Consider the radius of the circle = r

Circumference = 2 Ï€r

We know that

2 Ï€r = 18Ï€

So we get

2r = 18

r = 18/2 = 9 cm

Here

Area = Ï€r^{2}

Substituting the value of r

= Ï€ Ã— 9 Ã— 9

= 81Ï€ cm^{2}

**3. Find the perimeter of a semicircular plate of radius 3.85 cm.**

**Solution:**

It is given that

Radius of semicircular plate = 3.85 cm

\

We know that

Length of semicircular plat = Ï€r

Perimeter = Ï€r + 2r = r (Ï€ + 2)

Substituting the values

= 3.85 (22/7 + 2)

By further calculation

= 3.85 Ã— 36/7

= 0.55 Ã— 36

= 19.8 cm

**4. Find the radius and circumference of a circle whose area is 144 Ï€ cm ^{2}.**

**Solution:**

It is given that

Area of a circle = 144 Ï€ cm^{2}

Consider radius = r

Ï€ r^{2} = 144 Ï€

r^{2} = 144

So we get

r = âˆš144 = 12 cm

Here

Circumference = 2 Ï€r

So we get

= 2 Ã— 12 Ã— Ï€

= 24Ï€ cm

**5. A sheet is 11 cm long and 2 cm wide. Circular pieces 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.**

**Solution:**

It is given that

Length of sheet = 11 cm

Width of sheet = 2 cm

We have to cut the sheet to a square of side 0.5 cm

Here

Number of squares = 11/0.5 Ã— 2/0.5

Multiply and divide by 10

= (11 Ã— 10)/5 Ã— (2 Ã— 10)/5

By further calculation

= 22 Ã— 4

= 88

Hence, the number of discs will be equal to number of squares cut out = 88.

**6. If the area of a semicircular region is 77 cm ^{2}, find its perimeter.**

**Solution:**

It is given that

Area of a semicircular region = 77 cm^{2}

Consider r as the radius of the region

We know that

Area = Â½ Ï€r^{2}

Â½ Ï€r^{2} = 77

By further calculation

Â½ Ã— 22/7 r^{2} = 77

So we get

r^{2} = (77 Ã— 2 Ã— 7)/ 22 = 49 = 7^{2}

r = 7 cm

Here

Perimeter of the region = Ï€r + 2r

By further calculation

= 22/7 Ã— 7 + 2 Ã— 7

So we get

= 22 + 14

= 36 cm

**7. (a) In the figure (i) given below, AC and BD are two perpendicular diameters of a circle ABCD. Given that the area of shaded portion is 308 cm ^{2}, calculate**

**(i) the length of AC and**

**(ii) the circumference of the circle**

**(b) In the figure (ii) given below, AC and BD are two perpendicular diameters of a circle with centre O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take Ï€ = 3.14)**

**Solution:**

(a) It is given that

Area of shaded portion = Area of semicircle = 308 cm^{2}

Consider r as the radius of circle

Â½ Ï€r^{2} = 308

By further calculation

Â½ Ã— 22/7 r^{2} = 308

r^{2} = (308 Ã— 2 Ã— 7)/ 22

r^{2} = 196 = 14^{2}

So r = 14 cm

(i) AC = 2r = 2 Ã— 14 = 28 cm

(ii) We know that

Circumference of the circle = 2Ï€r

Substituting the values

= 28 Ã— 22/7

So we get

= 4 Ã— 22

= 88 cm

(b) We know that

Diameter of circle = 16 cm

Radius of circle = 16/2 = 8 cm

Here

Area of shaded part = 2 Ã— area of one quadrant

So we get

= Â½ Ï€r^{2}

Substituting the values

= Â½ Ã— 3.14 Ã— 8 Ã— 8

= 100.48 cm^{2}

We know that

Perimeter of shaded part = Â½ of circumference + 4r

It can be written as

= Â½ Ã— 2Ï€r + 4r

= Ï€r + 4r

Taking r as common

= r (Ï€ + 4)

Substituting the values

= 8 (3.14 + 4)

= 8 Ã— 7.14

= 57.12 cm

**8. A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/sec, calculate the number of complete revolutions the wheel makes in raising the bucket.**

**Solution:**

It is given that

Diameter of wheel = 77 cm

So radius of wheel = 77/2 cm

We know that

Circumference of wheel = 2Ï€r

Substituting the values

= 2 Ã— 22/7 Ã— 77/2

= 242 cm

**9. The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km/hr. Give your answer correct to the nearest km.**

**Solution:**

It is given that

Diameter of wheel = 84 cm

Radius of wheel = 84/2 = 42 cm

Here

Circumference of wheel = 2Ï€r

Substituting the values

= 2 Ã— 22/7 Ã— 42

= 264 cm

So the distance covered in 5 reductions = 264 Ã— 5 = 1320 cm

Time = 1 second

We know that

Speed of wheel = 1320/1 Ã— (60 Ã— 60)/ (100 Ã— 1000)

= 47.25 km/hr

= 48 km/hr

**10. The circumference of a circle is 123.2 cm. Calculate:**

**(i) the radius of the circle in cm.**

**(ii) the area of the circle in cm ^{2}, correct to the nearest cm^{2}.**

**(iii) the effect on the area of the circle if the radius is doubled.**

**Solution:**

It is given that

Circumference of a circle = 123.2 cm

Consider radius = r cm

(i) We know that

2Ï€r = 123.2

By further calculation

(2 Ã— 22)/ 7 r = 1232/10

So we get

r = (1232 Ã— 7)/ (10 Ã— 2 Ã— 22)

r = 19.6 cm

Therefore, radius of the circle is 19.6 cm

(ii) Here

Area of the circle = Ï€r^{2}

Substituting the values

= 22/7 Ã— 19.6 Ã— 19.6

= 1207.36

= 1207 cm^{2}

(iii) We know that

If radius is doubled = 19.6 Ã— 2 = 39.2 cm

So the area of circle = Ï€r^{2}

Substituting the values

= 22/7 Ã— 39.2 Ã— 39.2

= 4829.44 cm^{2}

Effect on area = 4829.44/1207 = 4 times

**11. (a) In the figure (i) given below, the area enclosed between the concentric circles is 770 cm ^{2}. Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.**

**(b) In the figure (ii) given below, the area enclosed between the circumferences of two concentric circles is 346.5 cm ^{2}. The circumference of the inner circle is 88 cm. Calculate the radius of the outer circle.**

**Solution:**

(a) It is given that

Radius of outer circle (R) = 21 cm

Consider r cm as the radius of inner circle

We know that

Area of the ring = Ï€ (R^{2} â€“ r^{2})

Substituting the values

= 22/7 (21^{2} â€“ r^{2})

= 22/7 (441 â€“ r^{2})

Area of the ring = 770 cm^{2}

So we get

22/7 (441 â€“ r^{2}) = 770

By further calculation

441 â€“ r^{2} = (770 Ã— 7)/ 22 = 245

r^{2} = 441 â€“ 245 = 196

r = âˆš196 = 14

Hence, the radius of inner circle is 14 cm.

(b) It is given that

Area of ring = 346.5 cm^{2}

Circumference of inner circle = 88 cm

We know that

Radius = (88 Ã— 7)/ (2 Ã— 22) = 14 cm

Consider R cm as the radius of outer circle

Area of ring = Ï€ (R^{2} â€“ r^{2})

Substituting the values

= 22/7 (R^{2} â€“ 14^{2})

= 22/7 (R^{2} â€“ 196) cm^{2}

Area of ring = 346.5 cm^{2}

By equating we get

22/7 (R^{2} â€“ 196) = 346.5

By further calculation

R^{2} â€“ 196 = (346.5 Ã— 7)/ 22 = 110.25

R^{2} = 110.25 + 196 = 306.25

So we get

R = âˆš306.25 = 17.5

Hence, the radius of outer circle is 17.5 cm.

**12. A road 3.5 m wide surrounds a circular plot whose circumference is 44m. Find the cost of paving the road at Rs 50 per m ^{2}.**

**Solution:**

It is given that

Circumference of circular plot = 44 m

Radius of circular plot = (44 Ã— 7)/ (22 Ã— 2) = 7 m

Width of the road = 3.5 m

So the radius of outer circle = 7 + 3.5 = 10.5 m

We know that

Area of road = Ï€ (R^{2} â€“ r^{2})

Substituting the values

= 22/7 (10.5^{2} â€“ 7^{2})

We can write it as

= 22/7 (10.5 + 7) (10 â€“ 7)

= 22/7 Ã— 17.5 Ã— 3.5

= 192.5 m^{2}

Here

Rate of paving the road = Rs 50 per m^{2}

Total cost = 192.5 Ã— 50 = Rs 9625

**13. The sum of diameters of two circles is 14 cm and the difference of their circumferences is 8 cm. Find the circumference of the two circles.**

**Solution:**

It is given that

Sum of the diameters of two circles = 14 cm

Consider R and r as the radii of two circles

2R + 2r = 14

Dividing by 2

R + r = 7 â€¦â€¦ (1)

We know that

Difference of their circumferences = 8 cm

2 Ï€R – 2 Ï€r = 8

Taking out the common terms

2 Ï€ (R â€“ r) = 8

(2 Ã— 22)/ 7 (R â€“ r) = 8

By further calculation

R â€“ r = (8 Ã— 7)/ (2 Ã— 22) = 14/11 â€¦.. (2)

By adding both the equations

2R = 7 + 14/11

By taking LCM

2R = (77 + 14)/ 11 = 91/11

By further calculation

R = 91/ (11 Ã— 2) = 91/22

From equation (1)

R + r = 7

Substituting the value of R

91/22 + r = 7

r = 7 â€“ 91/22

Taking LCM

r = (154 â€“ 91)/ 22 = 63/22

We know that

Circumference of first circle = 2 Ï€r

Substituting the values

= 2 Ã— 22/7 Ã— 91/22

= 26 cm

Circumference of second circle = 2 Ï€ R

Substituting the values

= 2 Ã— 22/7 Ã— 63/22

= 18 cm

**14. Find the circumference of the circle whose area is equal to the sum of the areas of three circles with radius 2 cm, 3 cm and 6 cm.**

**Solution:**

It is given that

Radius of first circle = 2 cm

Area of first circle = Ï€r^{2}

= Ï€ (2)^{2}

= 4 Ï€ cm^{2}

Radius of second circle = 3 cm

Area of first circle = Ï€r^{2}

= Ï€ (3)^{2}

= 9 Ï€ cm^{2}

Radius of second circle = 6 cm

Area of first circle = Ï€r^{2}

= Ï€ (6)^{2}

= 36 Ï€ cm^{2}

So the total area of the three circles = 4 Ï€ + 9 Ï€ + 36 Ï€ = 49 Ï€ cm^{2}

Area of the given circle = 49 Ï€ cm^{2}

We know that

Radius = âˆš (49 Ï€/ Ï€) = âˆš49 = 7 cm

Circumference = 2 Ï€r = 2 Ã— 22/7 Ã— 7 = 44 cm

**15. A copper wire when bent in the form of a square encloses an area of 121 cm ^{2}. If the same wire is bent into the form of a circle, find the area of the circle.**

**Solution:**

It is given that

Area of square = 121 cm^{2}

So side = âˆš121 = 11 cm

Perimeter = 4a = 4 Ã— 11 = 44 cm

Circumference of circle = 44 cm

Radius of circle = (44 Ã— 7)/ (2 Ã— 22) = 7 cm

We know that

Area of the circle = Ï€r^{2}

Substituting the values

= 22/7 (7)^{2}

So we get

= 22/7 Ã— 7Ã— 7

= 154 cm^{2}

**16. A copper wire when bent into an equilateral triangle has area 121 âˆš3 cm ^{2}. If the same wire is bent into the form of a circle, find the area enclosed by the wire.**

**Solution:**

It is given that

Area of equilateral triangle = 121 âˆš3 cm^{2}

Consider a as the side of triangle

Area = âˆš3/4 a^{2}

It can be written as

âˆš3/4 a^{2} = 121âˆš3

By further calculation

a^{2} = (121 Ã— âˆš3 Ã— 4)/ âˆš3

a^{2} = 484

So we get

a = âˆš484 = 22 cm

Here

Length of the wire = 66 cm

Radius of the circle = 66/2Ï€

By further calculation

= (66 Ã— 7)/ (2 Ã— 22)

= 21/2 cm

We know that

Area of the circle = Ï€r^{2}

By further calculation

= 22/7 Ã— (21/2)^{2}

= 22/7 Ã— 21/2 Ã— 21/2

So we get

= 693/2

= 346.5 cm^{2}

**17. (a) Find the circumference of the circle whose area is 16 times the area of the circle with diameter 7 cm.**

**(b) In the given figure, find the area of the unshaded portion within the rectangle. (Take Ï€ = 3.14)**

**Solution:**

(a) It is given that

Diameter of the circle = 7 cm

Radius of the circle = 7/2 cm

We know that

Area of the circle = Ï€r^{2}

Substituting the values

= 22/7 Ã— 7/2 Ã— 7/2

= 77/2 cm^{2}

Here

Area of bigger circle = 77/2 Ã— 16 = 616 cm^{2}

Consider r as the radius

Ï€r^{2} = 616

We can write it as

22/7 r^{2} = 616

By further calculation

r^{2} = (616 Ã— 7)/ 22

r^{2} = 196 cm^{2}

So we get

r = âˆš196 = 14 cm

Circumference = 2Ï€r

Substituting the values

= 2 Ã— 22/7 Ã— 14

= 88 cm

(b) It is given that

Radius of each circle = 3 cm

Diameter of each circle = 2 Ã— 3 = 6 cm

Here

Length of rectangle (l) = 6 + 6 + 3 = 15 cm

Breadth of rectangle (b) = 6 cm

So the area of rectangle = l Ã— b

Substituting the values

= 15 Ã— 6

= 90 cm^{2}

We know that

Area of 2 Â½ circles = 5/2 Ï€r^{2}

Substituting the values

= 5/2 Ã— 3.14 Ã— 3 Ã— 3

By further calculation

= 5 Ã— 1.57 Ã— 9

= 70.65 cm^{2}

So the area of unshaded portion = 90 â€“ 70.65

= 19.35 cm^{2}

**18. In the adjoining figure, ABCD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semi-circles are drawn with AD and BC as diameters. Find the area of the shaded region. Take Ï€ = 22/7.**

**Solution:**

It is given that

Side of square = 21 cm

So the area of square = side^{2} = 21^{2} = 441 cm^{2}

We know that

âˆ AOD + âˆ COD + âˆ AOB + âˆ BOC = 441

Substituting the values

x + x + x + x = 441

4x = 441

So we get

x = 441/4 = 110.25 cm^{2}

Based on the question

We should find the area of shaded portion in square ABCD which is âˆ AOD and âˆ BOC

âˆ AOD + âˆ BOC = 110.25 + 110.25 = 220.5 cm^{2}

Here

Area of two semicircle = Ï€r^{2}

Substituting the values

= 22/7 Ã— 10.5 Ã— 10.5

= 346.50 cm^{2}

So the area of shaded portion = 220.5 + 346.5 = 567 cm^{2}

**19. (a) In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.**

**(b) In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.**

**(c) In the figure (iii) given below, the diameter of the semicircle is equal to 14 cm. Calculate the area of the shaded region. Take Ï€ = 22/7.**

**Solution:**

(a) It is given that

ABCD is a square of each side (a) = 14 cm

APD and BPC are semi-circle with diameter = 14 cm

Radius of each semi-circle (a) = 14/2 = 7 cm

(i) We know that

Area of square = a^{2} = 14^{2} = 196 cm^{2}

Area of two semicircles = 2 Ã— Â½ Ï€r^{2}

= Ï€r^{2}

Substituting the values

= 22/7 Ã— 7 Ã— 7

= 154 cm^{2}

So the area of shaded portion = 196 â€“ 154 = 42 cm^{2}

(ii) Here

Length of arcs of two semicircles = 2Ï€r

Substituting the values

= 2 Ã— 22/7 Ã— 7

= 44 cm

So the perimeter of shaded portion = 44 + 14 + 14 = 72 cm

(b) It is given that

ABCD is a square whose each side (a) = 14 cm

4 circles are drawn which touch each other and the sides of squares

Radius of each circle (r) = 7/2 = 3.5 cm

(i) We know that

Area of square ABCD = a^{2} = 14^{2} = 196 cm^{2}

Area of 4 circles = 4 Ã— Ï€r^{2}

Substituting the values

= 4 Ã— 22/7 Ã— 7/2 Ã— 7/2

= 154 cm^{2}

So the area of shaded portion = 196 â€“ 154 = 42 cm^{2}

(ii) Here

Perimeter of 4 circles = 4 Ã— 2 Ï€r

Substituting the values

= 4 Ã— 2 Ã— 22/7 Ã— 7/2

= 88 cm

So the perimeter of shaded portion = perimeter of 4 circles + perimeter of square

Substituting the values

= 88 + 4 Ã— 14

So we get

= 88 + 56

= 144 cm

(c) We know that

Area of rectangle ACDE = ED Ã— AE

Substituting the values

= 14 Ã— 7

= 98 cm^{2}

Here

Area of semicircle DEF = Ï€r^{2}/2

Substituting the values

= (22 Ã— 7 Ã— 7)/ (7 Ã— 2)

= 77 cm^{2}

So the area of shaded region = 77 + (98 â€“ 2 Ã— Â¼ Ã— 22/7 Ã— 7 Ã— 7)

= 77 + 21

= 98 cm^{2}

**20. (a) Find the area and the perimeter of the shaded region in figure (i) given below. The dimensions are in centimetres.**

**(b) In the figure (ii) given below, area of â–³ABC = 35 cm ^{2}. Find the area of the shaded region.**

**Solution:**

(a) It is given that

There are 2 semicircles where the smaller is inside the larger

Radius of larger semicircles (R) = 14 cm

Radius of smaller circle (r) = 14/2 = 7 cm

(i) We know that

Area of shaded portion = Area of larger semicircle â€“ Area of smaller circle

= Â½ Ï€R^{2} â€“ Â½ Ï€r^{2}

We can write it as

= Â½ Ï€ (R^{2} â€“ r^{2})

Substituting the values

= Â½ Ã— 22/7 (14^{2} â€“ 7^{2})

By further calculation

= 11/7 (14 + 7) (14 â€“ 7)

So we get

= 11/7 Ã— 21 Ã— 7

= 231 cm^{2}

(ii) Here

Perimeter of shaded portion = circumference of larger semicircle + circumference of smaller semicircle + radius of larger semicircle

= Ï€R + Ï€r + R

Substituting the values

= 22/7 Ã— 14 + 22/7 Ã— 7 + 14

By further calculation

= 44 + 22 + 14

= 80 cm

(b) We know that

Area of â–³ABC formed in a semicircle = 3.5 cm

Altitude CD = 5 cm

So the base AB = (area Ã— 2)/ altitude

Substituting the values

= (35 Ã— 2)/ 5

= 14 cm

Here

Diameter of semicircle = 14 cm

Radius of semicircle (R) = 14/2 = 7 cm

So the area of semicircle = Â½ Ï€R^{2}

Substituting the values

= Â½ Ã— 22/7 Ã— 7 Ã— 7

= 77 cm^{2}

Area of shaded portion = Area of semicircle â€“ Area of triangle

= 77 â€“ 35

= 42 cm^{2}

**21. (a) In the figure (i) given below, AOBC is a quadrant of a circle of radius 10m. Calculate the area of the shaded portion. Take Ï€ = 3.14 and give your answer correct to two significant figures.**

**(b) In the figure (ii) given below, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.**

**Solution:**

(a) In the figure

Shaded portion = Quadrant – â–³AOB

Radius of the quadrant = 10 m

Here

Area of quadrant = Â¼ Ï€r^{2}

Substituting the values

= Â¼ Ã— 3.14 Ã— 10 Ã— 10

By further calculation

= (3.14 Ã— 100)/ 4

= 314/4

= 78.5 m^{2}

We know that

Area of â–³AOB = Â½ Ã— AO Ã— OB

Substituting the values

= Â½ Ã— 10 Ã— 10

= 50 m^{2}

So the area of shaded portion = 78.5 â€“ 50 = 28.5 m^{2}

(b) In the figure

Radius of quadrant = 3.5 cm

(i) We know that

Area of quadrant = Â¼ Ï€r^{2}

Substituting the values

= Â¼ Ã— 22/7 Ã— 3.5 Ã— 3.5

= 9.625 cm^{2}

(ii) Here

Area of â–³AOD = Â½ Ã— AO Ã— OD

Substituting the values

= Â½ Ã— 3.5 Ã— 2

= 3.5 cm^{2}

So the area of shaded portion = Area of quadrant â€“ Area of â–³AOD

Substituting the values

= 9.625 â€“ 3.6

= 6.125 cm^{2}

**22. A student takes a rectangular piece of paper 30 cm long and 21 cm wide. Find the area of the biggest circle that can be cut out from the paper. Also find the area of the paper left after cutting out the circle. (Take Ï€ = 22/7)**

**Solution:**

It is given that

Length of rectangle = 30 cm

Width of rectangle = 21 cm

We know that

Area of rectangle = l Ã— b

= 30 Ã— 21

= 630 cm^{2}

So the radius of the biggest circle = 21/2 cm

Here

Area of the circle = Ï€r^{2}

Substituting the values

= 22/7 Ã— 21/2 Ã— 21/2

So we get

= 693/2

= 346.5 cm^{2}

So the area of remaining part = 630 â€“ 346.5 = 283.5 cm^{2}

**23. A rectangle with one side 4 cm is inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.**

**Solution:**

Consider ABCD as a rectangle

AB = 4 cm

Diameter of circle AC = 2.5 Ã— 2 = 5 cm

Here

So we get

= 3 cm

We know that

Area of rectangle = AB Ã— BC

Substituting the values

= 4 Ã— 3

= 12 cm^{2}

**24. (a) In the figure (i) given below, calculate the area of the shaded region correct to two decimal places. (Take Ï€ = 3.142)**

**(b) In the figure (ii) given below, ABC is an isosceles right angled triangle with âˆ ABC = 90 ^{0}. A semicircle is drawn with AC as diameter. If AB = BC = 7 cm, find the area of the shaded region. Take Ï€ = 22/7.**

**Solution:**

(a) We know that

ABCD is a rectangle which is inscribed in a circle of length = 12 cm

Width = 5 cm

= 13 cm

Diameter of circle = AC = 13 cm

Radius of circle = 13/2 = 6.5 cm

Here

Area of circle = Ï€r^{2}

Substituting the values

= 3.142 Ã— (6.5)^{2}

By further calculation

= 3.142 Ã— 42.25

= 132.75 cm^{2}

Area of rectangle = l Ã— b

Substituting the values

= 12 Ã— 5

= 60 cm^{2}

So the area of the shaded portion = 132.75 â€“ 60 = 72.75 cm^{2}

(b) We know that

Area of â–³ABC = Â½ Ã— AB Ã— BC

Substituting the values

= Â½ Ã— 7 Ã— 7

= 49/2 cm^{2}

Here

AC^{2} = AB^{2} + BC^{2}

Substituting

= 49 + 49

So we get

AC = 7âˆš2

Radius of semi-circle = 7âˆš2/ 2 cm

Area of semi-circle = Ï€/2 Ã— (7âˆš2/ 2)^{2}

By further calculation

= Â½ Ã— 22/7 Ã— 98/4

= 77/2 cm^{2}

Area of the shaded region = Area of the semi-circle â€“ Area of â–³ABC

Substituting the values

= 77/2 â€“ 49/2

= 28/2

= 14 cm^{2}

**25. A circular field has perimeter 660 m. A plot in the shape of a square having its vertices on the circumference is marked in the field. Calculate the area of the square field.**

**Solution:**

It is given that

Perimeter of circular field = 660 m

Radius of the field = 660/2 Ï€

Substituting the values

= (660 Ã— 7)/ (2 Ã— 22)

= 105 m

Here

ABCD is a square which is inscribed in the circle where AC is the diagonal which is the diameter of the circular field

Consider a as the side of the square

AC = âˆš2 a

a = AC/âˆš2

Substituting the values

a = (105 Ã— 2)/ âˆš2

Multiply and divide by âˆš2

a = (105 Ã— 2 Ã— âˆš2)/ (âˆš2 Ã— âˆš2)

By further calculation

a = (105 Ã— 2 Ã— âˆš2)/ 2

a = 105 âˆš2 m

We know that

Area of the square = a^{2}

It can be written as

= (105âˆš2)^{2}

= 105âˆš2 Ã— 105âˆš2

= 22050 m^{2}

**26. In the adjoining figure, ABCD is a square. Find the ratio between **

**(i) the circumferences**

**(ii) the areas of the incircle and the circumcircle of the square.**

**Solution:**

Consider side of the square = 2a

Area = (2a)^{2} = 4a^{2}

Diagonal of AC = âˆš2AB

(i) We know that

Radius of the circumcircle = Â½ AC

It can be written as

= Â½ (âˆš2 Ã— AB)

Substituting the values

= âˆš2/2 Ã— 2a

= âˆš2a

Circumference = 2 Ï€r = 2 Ã— Ï€ Ã— âˆš2a = 2âˆš2 Ï€a

Radius of incircle = AB = Â½ Ã— 2a = a

Circumference = 2 Ï€r = 2 Ï€a

Here

Ratio between the circumference incircle and circumcircle = 2 Ï€a: 2âˆš2 Ï€a

= 1: âˆš2

(ii) We know that

Area of incircle = Ï€r^{2} = Ï€a^{2}

Area of circumcircle = Ï€R^{2}

= Ï€ (âˆš2a)^{2}

= Ï€2a^{2}

= 2 Ï€a^{2}

So the ratio = Ï€a^{2}: 2 Ï€a^{2} = 1: 2

**27. (a) The figure (i) given below shows a running track surrounding a grassed enclosure PQRSTU. The enclosure consists of a rectangle PQST with a semicircular region at each end.**

** PQ = 200 m, PT = 70 m**

**(i) Calculate the area of the grassed enclosure in m ^{2}.**

**(ii) Given that the track is of constant width 7 m, calculate the outer perimeter ABCDEF of the track.**

**(b) In the figure (ii) given below, the inside perimeter of a practice running track with semi-circular ends and straight parallel sides is 312 m. The length of the straight portion of the track is 90 m. If the track has a uniform width of 2m throughout, find its area.**

**Solution:**

(a) It is given that

Length of PQ = 200 m

Width PT = 70 m

(i) We know that

Area of rectangle PQST = l Ã— b

= 200 Ã— 70

= 14000 m^{2}

Radius of each semi-circular part on either side of rectangle = 70/2 = 35 m

Area of both semi-circular parts = 2 Ã— Â½ Ï€r^{2}

Substituting the values

= 22/7 Ã— 35 Ã— 35

= 3850 m^{2}

So the total area of grassed enclosure = 1400 + 3850 = 17850 m^{2}

(ii) We know that

Width of track around the enclosure = 7 m

Outer length = 200 m

So the width = 70 + 7 Ã— 2

= 70 + 14

= 84 m

Outer radius = 84/2 = 42 m

Here

Circumference of both semi-circular part = 2Ï€r

Substituting the values

= 2 Ã— 22/7 Ã— 42

= 264 m

Outer perimeter = 264 + 200 Ã— 2

= 264 + 400

= 664 m

(b) It is given that

Inside perimeter = 312 m

Total length of the parallel sides = 90 + 90 = 180 m

Circumference of two semi-circles = 312 â€“ 180 = 132 m

Here

Radius of each semi-circle = 132/2Ï€

So we get

= 66/3.14

= 21.02 m

Diameter of each semi-circle = 66/Ï€ Ã— 2

So we get

= 132/ Ï€

= 132/3.14

Multiply and divide by 100

= (132 Ã— 100)/ 314

= 42.04 m

Width of track = 2 m

Outer diameter = 42.04 + 4 = 46.04 m

Radius = 46.04/2 = 23.02 m

We know that

Area of two semi-circles = 2 Ã— Â½ Ã— Ï€R^{2}

= Ï€R^{2}

Substituting the values

= 3.14 Ã— (23.02)^{2}

= 3.14 Ã— 23.02 Ã— 23.02

= 1663.95 m^{2}

Area of rectangle = 90 Ã— 46.04

= 4143.6 m^{2}

Total area = 1663.95 + 4143.60 = 5807.55 m^{2}

Area of two inner circles = 2 Ã— Â½ Ï€r^{2}

Substituting the values

= 3.14 Ã— 21.02 Ã— 21.02

= 1387.38 m^{2}

Area of inner rectangle = 90 Ã— 42.04

= 3783.6 m^{2}

Total inner area = 3783.60 + 1387.38

= 5170.98 m^{2}

Here

Area of path = 5807.55 â€“ 5170.98

= 636.57 m^{2}

**28. (a) In the figure (i) given below, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.**

**(b) The quadrants shown in the figure (ii) given below are each of radius 7 cm. Calculate the area of the shaded portion.**

**Solution:**

(a) It is given that

AC = 8 cm

BC = AC â€“ AB = 8 â€“ 3 = 5 cm

We know that

Area of big circle of radius AC = Ï€R^{2}

Substituting the values

= 22/7 Ã— 8 Ã— 8

By further calculation

= 64 Ã— 22/7 cm^{2}

Area of smaller circle = Ï€r^{2}

Substituting the values

= 22/7 Ã— 5 Ã— 5

By further calculation

= (25 Ã— 22)/ 7 cm^{2}

Here

Area of shaded portion = (64 Ã— 22)/ 7 â€“ (25 Ã— 22)/ 7

Taking out the common terms

= 22/7 (64 â€“ 25)

By further calculation

= 22/7 Ã— 39

= 122.57 cm^{2}

(b) We know that

Radius of each quadrant = 7 cm

Here

Area of shaded region = Area of square â€“ 4 area of the quadrant

It can be written as

= (side)^{2} â€“ 4 Ã— Â¼ Ï€r^{2}

Substituting the values

= 14^{2} â€“ 22/7 Ã— 7 Ã— 7

By further calculation

= 196 â€“ 154

= 42 cm^{2}

**29. (a) In the figure (i) given below, two circular flower beds have been shown on the two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.**

**(b) In the figure (ii) given below, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (Ï€ = 3.14)**

**Solution:**

(a) It is given that

Side of square lawn ABCD (a) = 56 cm

Area = a^{2} = 5^{2} = 3136 cm^{2}

We know that

Length of the diagonal of the square = âˆš2 a

= âˆš2 Ã— 56 cm

Radius of each quadrant = (âˆš2 Ã— 56)/ 2

= 28 âˆš2 cm

So the area of each segment = Â¼ Ï€r^{2} â€“ area of â–³OBC

Substituting the values

= Â¼ Ã— 22/7 Ã— 28âˆš2 Ã— 28âˆš2 â€“ Â½ Ã— 28âˆš2 Ã— 28âˆš2

Taking out the common terms

= 28âˆš2 Ã— 28âˆš2 (1/4 Ã— 22/7 â€“ 1/2)

By further calculation

= 784 Ã— 2 (11/14 â€“ Â½)

So we get

= 784 Ã— 2 Ã— 4/14

= 448 cm^{2}

Here

Area of two segments = 448 Ã— 2 = 896 cm^{2}

So the total area of the lawn and beds = 3136 + 896 = 4032 cm^{2}

(b) In the figure

OPBQ is a quadrant and OABC is a square which is inscribed in a side of square = 20 cm

OB is joined

Here

OB = âˆš2 a = âˆš2 Ã—20 cm

Radius of quadrant = OB = 20âˆš2cm

We know that

Area of quadrant = Â¼ Ï€r^{2}

Substituting the values

= Â¼ Ã— 3.14 Ã— (20âˆš2)^{2}

By further calculation

= Â¼ Ã— 3.14 Ã— 800

So we get

= 314 Ã— 2

= 628 cm^{2}

Area of square = a^{2} = 20^{2} = 400 cm^{2}

So the area of shaded portion = 628 â€“ 400 = 228 cm^{2}

**30. (a) In the figure (i) given below, ABCD is a rectangle, AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area if the shaded portion.**

**(b) In the figure (ii) given below, O is the centre of a circle with AC = 24 cm, AB = 7 cm and âˆ BOD = 90 ^{0}. Find the area of the shaded region. (Use Ï€ = 3.14)**

**Solution:**

(a) It is given that

ABCD is a rectangle

Three semicircles are drawn with AB = 14 cm and BC = 7 cm

We know that

Area of rectangle ABCD = l Ã— b

Substituting the values

= 14 Ã— 7

= 98 cm^{2}

Radius of each outer semicircles = 7/2 cm

So the area = 2 Ã— Â½ Ï€r^{2}

Substituting the values

= 22/7 Ã— 7/2 Ã— 7/2

= 77/2

= 38.5 cm^{2}

Here

Area of semicircle drawn on CD as diameter = Â½ Ï€R^{2}

Substituting the values

= Â½ Ã— 22/7 Ã— 7^{2}

By further calculation

= 11/7 Ã— 7 Ã— 7

= 77 cm^{2}

So the area of shaded region = 98 + 38.5 â€“ 77

= 59.5 cm^{2}

(b) In the figure

AC = 24 cm

AB = 7 cm

âˆ BOD = 90^{0}

In â–³ABC

Using Pythagoras theorem

BC^{2} = AC^{2} + AB^{2}

Substituting the values

BC^{2} = 24^{2} + 7^{2}

By further calculation

BC = âˆš (576 + 49) = âˆš625 = 25 cm

So the radius of circle = 25/2 cm

We know that

Area of â–³ABC = Â½ Ã— AB Ã— AC

Substituting the values

= Â½ Ã— 7 Ã— 24

= 84 cm^{2}

Area of quadrant COD = Â¼ Ï€r^{2}

Substituting the values

= Â¼ Ã— 3.14 Ã— 25/2 Ã— 25/2

By further calculation

= 1962.5/16

= 122.66 cm^{2}

Area of circle = Ï€r^{2}

Substituting the values

= 3.14 Ã— 25/2 Ã— 25/2

By further calculation

= 1962.5/4

= 490.63 cm^{2}

Area of shaded portion = Area of circle â€“ (Area of â–³ABC + Area of quadrilateral COD)

Substituting the values

= 490.63 â€“ (84 + 122.66)

By further calculation

= 490.63 â€“ 206.66

= 283.97 cm^{2}

**31. (a) In the figure given below ABCD is a square of side 14 cm. A, B, C and D are centres of the equal circle which touch externally in pairs. Find the area of the shaded region.**

**(b) In the figure (ii) given below, the boundary of the shaded region in the given diagram consists of three semicircular arcs, the smaller being equal. If the diameter of the larger one is 10 cm, calculate.**

**(i) the length of the boundary.**

**(ii) the area of the shaded region. (Take Ï€ to be 3.14)**

**Solution:**

(a) It is given that

Side of square ABCD = 14 cm

Radius of each circle drawn from A, B, C and D and touching externally in pairs = 14/2 = 7 cm

We know that

Area of square = a^{2}

Substituting the values

= 14 Ã— 14

= 196 cm^{2}

Area of 4 sectors of 90^{0} each = 4 Ã— Ï€ Ã— r^{2}

Substituting the values

= 4 Ã— 22/7 Ã— 7 Ã— 7 Ã— Â¼

= 154 cm

Area of each sector of 270^{0} angle = 3/4 Ï€r^{2}

Substituting the values

= Â¾ Ã— 22/7 Ã— 7 Ã— 7

= 231/2

= 115.5 cm^{2}

Here

Area of 4 sectors = 115.5 Ã— 4 = 462 cm^{2}

So the area of shaded portion = area of square + area of 4 bigger sector â€“ area of 4 smaller sector

Substituting the values

= 196 + 462 â€“ 154

= 658 â€“ 154

= 504 cm^{2}

(b) We know that

Radius of big semi-circle = 10/2 = 5 cm

Radius of each smaller circle = 5/2 cm

(i) Length of boundary = circumference of bigger semi-circle + 2 circumference of smaller semi-circles

It can be written as

= Ï€R + Ï€r + Ï€r

= 3.14 (R + 2r)

Substituting the values

= 3.14 (5 + 2 Ã— 5/2)

By further calculation

= 3.14 Ã— 10

= 31.4 cm

(ii) Here

Area of shaded region = area of bigger semi-circle + area of one smaller semi-circle â€“ area of other smaller semi-circle

We know that

Area of bigger semi-circle = Â½ Ï€R^{2}

Substituting the values

= 3.14/2 Ã— 5 Ã— 5

= 1.57 Ã— 25

= 39.25 cm

**32. (a) In the figure (i) given below, the points A, B and C are centres of arcs of circles of radii 5 cm, 3 cm and 2 cm respectively. Find the perimeter and the area of the shaded region. (Take Ï€ = 3.14)**

**(b) In the figure (ii) given below, ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm, and at the centre a circle of diameter 2 cm are drawn. Find the perimeter and the area of the shaded region. Take Ï€ = 3.14**

**Solution:**

(a) It is given that

Radius of bigger circle = 5 cm

Radius of small circle (r_{1}) = 3 cm

Radius of smaller circle (r_{2}) = 2 cm

(i) We know that

Perimeter of shaded region = circumference of bigger semi-circle + circumference of small semi-circle + circumference of smaller semi-circle

It can be written as

= Ï€R + Ï€ r_{1} + Ï€ r_{2}

= Ï€ (R + r_{1} + r_{2})

Substituting the values

= Ï€ (5 + 3 + 2)

= 3.14 Ã— 10

= 31.4 cm^{2}

(ii) We know that

Area of shaded region = area of bigger semi-circle + area of smaller semi-circle â€“ area of small semicircle

It can be written as

= Â½ Ï€R^{2} + Â½ Ï€r_{2}^{2} â€“ Â½ Ï€r_{1}^{2}

= Â½ Ï€ (R^{2} + r_{2}^{2} â€“ r_{1}^{2})

Substituting the values

= Â½ Ï€ (5 ^{2} + 2^{2}Â â€“ 3^{2})

By further calculation

= Â½ Ï€ (25 + 4 â€“ 9)

= Â½ Ï€ Ã— 20

So we get

= 10 Ã— 3.14

= 31.4 cm^{2}

(b) We know that

Side of square ABCD = 4 cm

Radius of each quadrant circle = 1 cm

Radius of circle in the square = 2/2 = 1 cm

(i) Here

Perimeter of shaded region = circumference of 4 quadrants + circumference of circle + 4 Ã— Â½ side of square

It can be written as

= 4 Ã— Â¼ (2 Ï€r) + 2 Ï€r + 4 Ã— 2

By further calculation

= 2 Ï€r + 2 Ï€r + 8

= 4 Ï€r + 8

So we get

= 4 Ã— 3.14 Ã— 1 + 8

= 12.56 + 8

= 20.56 cm

(ii) Area of shaded region = area of square â€“ area of 4 quadrants â€“ area of circle

It can be written as

= side^{2} â€“ 4 Ã— Â¼ Ï€r^{2} â€“ Ï€r^{2}

= 4^{2} â€“ Ï€r^{2}– Ï€r^{2}

By further calculation

= 16 – 2Ï€r^{2}

So we get

= 16 â€“ 2 Ã— 3.14 Ã— 1^{2}

= 16 â€“ 6.28

= 9.72 cm^{2}

**33. (a) In the figure given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle. (Take Ï€ = 22/7)**

**(b) The figure (ii) given below shows a kite, in which BCD is in the shape of a quadrant of circle of radius 42 cm. ABCD is a square and â–³CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.**

**Solution:**

(a) Here

Area of remaining piece = area of rectangle ABCD â€“ area of semicircle DGE â€“ area of quarter BFEC

Substituting the values

= 14 Ã— 7 â€“ Â½ Ã— Ï€ (7/2)^{2} â€“ Â¼ Ï€ Ã— 7^{2}

By further calculation

= 14 Ã— 7 â€“ Â½ Ã— 22/7 Ã— 7/2 Ã— 7/2 â€“ Â¼ Ã— 22/7 Ã— 7 Ã— 7

So we get

= 98 â€“ 77/4 â€“ 154/4

= 98 â€“ 19.25 â€“ 38.5

= 98 â€“ 57.75

= 40.25 cm^{2}

(b) In the given figure

ABCD is a square of side = radius of quadrant = 42 cm

â–³CEF is an isosceles right triangle with each side = 6 cm

Area of shaded portion = area of quadrant + area of isosceles right triangle

It can be written as

= Â¼ Ï€r^{2} + Â½ EC Ã— FC

Substituting the values

= Â¼ Ã— 22/7 Ã— 42 Ã— 42 + Â½ Ã— 6 Ã— 6

By further calculation

= 1386 + 18

= 1404 cm^{2}

**34. (a) In the figure (i) given below, the boundary of the shaded region in the given diagram consists of four semicircular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, calculate**

**(i) the length of the boundary.**

**(ii) the area of the shaded region.**

**(b) In the figure (ii) given below, a piece of cardboard, in the shape of a trapezium ABCD, AB || CD and âˆ BOD = 90 ^{0}, quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE = 2 cm. Calculate the area of the remaining piece of the cardboard.**

**Solution:**

(a) (i) We know that

Length of boundary = Circumference of bigger semi-circle + Circumference of small semi-circle + 2 Ã— circumference of the smaller semi-circles

It can be written as

= Ï€R + Ï€r_{1} + 2 Ã— Ï€r_{2}

= Ï€ (R + r_{1}) + 2Ï€r_{2}

Substituting the values

= 22/7 (7 + 3.5) + 2 Ã— 22/7 Ã— 3.5/2

By further calculation

= 22/7 Ã— 10.5 + 11

So we get

= 33 + 11

= 44 cm

(ii) We know that

Area of shaded region = Area of bigger semicircle + area of small semicircle â€“ 2 Ã— area of smaller semicircles

It can be written as

= Â½ Ï€ (7)^{2} + Â½ Ï€ (3.5)^{2} â€“ 2 Ã— Â½ Ï€ (1.75)^{2}

By further calculation

= Â½ Ã— 22/7 Ã— 7 Ã— 7 + Â½ Ã— 22/7 Ã— 3.5 Ã— 3.5 â€“ 22/7 Ã— 1.75 Ã— 1.75

So we get

= 77 + 19.25 â€“ 9.625

= 86.625 cm^{2}

(b) Here

ABCD is a trapezium in which AB || DC and âˆ C = 90^{0}

AB = BC = 3.5 cm and DE = 2 cm

Radius of quadrant = 3.5 cm

We know that

Area of trapezium = Â½ (AB + DC) Ã— BC

Substituting the values

= Â½ (3.5 + 3.5 + 2) Ã— 3.5

By further calculation

= Â½ (9 Ã— 3.5)

= 4.5 Ã— 3.5

= 15.75 cm^{2}

So the area of quadrant = Â¼ Ï€r^{2}

Substituting the values

= Â¼ Ã— 22/7 Ã— 3.5 Ã— 3.5

= 9.625 cm^{2}

Area of shaded portions = 15.75 â€“ 9.625 = 6.125 cm^{2}

**35. (a) In the figure (i) given below, ABC is a right angled triangle, âˆ B = 90 ^{0}, AB = 28 cm and BC = 21 cm. With AC as diameter a semi-circle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.**

**(b) In the figure (ii) given below, ABC is an equilateral triangle of side 8 c. A, B and C are the centers of circular arcs of equal radius. Find the area of the shaded region correct upto 2 decimal places. (Take Ï€ = 3.142 and âˆš3 = 1.732)**

**Solution:**

(a) In right â–³ABC

âˆ B = 90^{0}

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

Substituting the values

= 28^{2} + 21^{2}

= 784 + 441

= 1225

So we get

AC = âˆš1225 = 35 cm

Here

Radius of semi-circle (R) = 35/2

Radius of quadrant (r) = 21 cm

So the area of shaded region = area of â–³ABC + area of semi-circle â€“ area of quadrant

= Â½ Ã— 28 Ã— 21 + Â½ Ï€R^{2} â€“ Â¼ r^{2}

Substituting the values

= 294 + Â½ Ã— 22/7 Ã— 35/2 Ã— 35/2 â€“ Â¼ Ã— 22/7 Ã— 21 Ã— 21

By further calculation

= 294 + 1925/4 â€“ 693/2

= 294 + 481.25 â€“ 346.5

So we get

= 775.25 â€“ 346.50

= 428.75 cm^{2}

(b) We know that

â–³ABC is an equilateral triangle of side 8 cm

A, B, C are the centres of three circular arcs of equal radius

Radius = 8/2 = 4 cm

Here

Area of â–³ABC = âˆš3/4a^{2}

Substituting the values

= âˆš3/4 Ã— 8 Ã— 8

= âˆš3/4 Ã— 64

So we get

= 16âˆš3cm^{2}

Substituting the value of âˆš3

= 16 Ã— 1.732

= 27.712 cm^{2}

So the area of 3 equal sectors of 60^{0} whose radius is 4 cm = 3 Ã— Ï€r^{2} Ã— 60/360

By further calculation

= 3 Ã— 3.142 Ã— 4 Ã— 4 Ã— 1/6

So we get

= 3.142 Ã— 8

= 25.136 cm^{2}

Area of shaded region = 27.712 â€“ 25.136 = 2.576 = 2.58 cm^{2}

**36. A circle is inscribed in a regular hexagon of side 2âˆš3 cm. Find**

**(i) the circumference of the inscribed circle**

**(ii) the area of the inscribed circle**

**Solution:**

It is given that

ABCDEF is a regular hexagon of side 2âˆš3 cm and a circle is inscribed in it with O as the centre

Radius of inscribed circle = âˆš3/2 Ã— side of regular hexagon

Substituting the values

= âˆš3/2 Ã— 2âˆš3

= 3 cm

(i) We know that

Circumference of the circle = 2Ï€r

Substituting the values

= 2Ï€ Ã— 3

By further calculation

= (6 Ã— 22)/ 7

= 132/7 cm

(ii) Area of the circle = Ï€r^{2}

Substituting the values

= Ï€ Ã— 3 Ã— 3

By further calculation

= (9 Ã— 22)/ 7

= 198/7 cm^{2}

**37. In the figure (i) given below, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Take Ï€ = 3.14.**

**Solution:**

It is given that

Radius of the circle = 10 cm

Angle at the centre subtended by a chord AB = 90^{0}

We know that

Area of sector OACB = Ï€r^{2} Ã— 90/360

Substituting the values

= 3.14 Ã— 10 Ã— 10 Ã— 90/360

So we get

= 314 Ã— Â¼

= 78.5 cm^{2}

Here

Area of â–³OAB = Â½ Ã— 10 Ã— 10 = 50 cm^{2}

Area of minor segment = Area of sector â–³ACB â€“ Area of â–³OAB

Substituting the values

= 78.5 â€“ 50

= 28.5 cm^{2}

Area of circle = Ï€r^{2}

Substituting the values

= 3.14 Ã— 10 Ã— 10

= 314 cm^{2}

Area of major segment = area of circle â€“ area of minor segment

Substituting the values

= 314 â€“ 28.5

= 285.5 cm^{2}

Exercise 16.4

**1. Find the surface area and volume of a cube whose one edge is 7 cm.**

**Solution:**

It is given that

One edge of cube a = 7 cm

We know that

Surface area of cube = 6a^{2}

Substituting the values

= 6 (7)^{2}

= 6 Ã— 7 Ã— 7

= 294 cm^{2}

Volume of cube = a^{3}

Substituting the values

= 7^{3}

= 7 Ã— 7 Ã— 7

= 343 cm^{3}

**2. Find the surface area and the volume of a rectangular solid measuring 5 m by 4 m by 3 m. Also find the length of a diagonal.**

**Solution:**

In a rectangular solid

l = 5 m, b = 4 m and h = 3 m

Here

Surface area of rectangular solid = 2 (lb + bh + lh)

Substituting the values

= 2 (5 Ã— 4 + 4 Ã— 3 + 5 Ã— 3)

By further calculation

= 2 (20 + 12 + 15)

= 2 Ã— 47

= 94 sq. m

Volume of rectangular solid = l Ã— b Ã— h

Substituting the values

= 5 Ã— 4 Ã— 3

= 60 m^{3}

We know that

It can be written as

= 5 âˆš2 m

Substituting the value of âˆš2

= 5 Ã— 1.414

= 7.07 m

Therefore, the length of diagonal is 7.07 m.

**3. The length and breadth of a rectangular solid are respectively 25 cm and 20 cm. If the volume is 7000 cm ^{3}, find its height.**

**Solution:**

It is given that

Length of rectangular solid = 25 cm

Breadth of rectangular solid = 20 cm

Volume of rectangular solid = 7000 cm^{3}

Consider height of rectangular solid = h cm

Here

Volume = l Ã— b Ã— h

Substituting the values

7000 = 25 Ã— 20 Ã— h

By further calculation

25 Ã— 20 Ã— h = 7000

h = 7000/ (25 Ã— 20)

So we get

h = 700/ (25 Ã— 2)

h = 350/25

By division

h = 70/5

h = 14 cm

Therefore, height of rectangular solid is 14 cm.

**4. A class room is 10 m long, 6 m broad and 4 m high. How many students can it accommodate if one student needs 1.5 m ^{2} of floor area? How many cubic metres of air will each student have?**

**Solution:**

The given dimensions of class room are

Length (l) = 10 m

Breadth (b) = 6 m

Height (h) = 4 m

We know that

Floor area of class room = l Ã— b

Substituting the values

= 10 Ã— 6

= 60 m^{2}

Here

One student needs 1.5 m^{2} floor area

So the number of students = 60/1.5

Multiply and divide by 10

= (60 Ã— 10)/ 15

= 600/15

= 40 students

Volume of class room = l Ã— b Ã— h

Substituting the values

= 10 Ã— 6 Ã— 4

= 240 m^{3}

So the cubic metres of air for each student = volume of classroom/ number of students

Substituting the values

= 240/40

= 6 m^{3}

**5. (a) The volume of a cuboid is 1440 cm ^{3}. Its height is 10 cm and the cross-section is a square. Find the side of the square.**

**(b) The perimeter of one face of a cube is 20 cm. Find the surface area and the volume of the cube.**

**Solution:**

(a) It is given that

Volume of cuboid = 1440 cm^{3}

Height of cuboid = 10 cm

We know that

Volume of cuboid = area of square Ã— height

Substituting the values

1440 = area of square Ã— 10

By further calculation

Area of square = 1440/10 = 144 cm^{2}

Here

Side Ã— side = 144

So we get

Side = âˆš144 = 12 cm

Therefore, the side of square is 12 cm.

(b) It is given that

Perimeter of one face of a cube = 20 cm

Perimeter of one face of a cube = 4 Ã— side

We can write it as

20 = 4 Ã— side

By further calculation

Side = 20/4 = 5 cm

Here

Area of one face = side Ã— side

Substituting the values

= 5 Ã— 5

= 25 cm^{2}

Area of 6 faces = 6 Ã— 25 = 150 cm^{2}

So the volume of cube = side Ã— side Ã— side

Substituting the values

= 5 Ã— 5 Ã— 5

= 125 cm^{3}

**6. Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured papers with pictures of Santa Claus. She must know the exact quantity of paper to buy for this purpose. If the box has length 80 cm, breadth 40 cm and height 20 cm respectively, then how many square sheets of paper of side 40 cm would she require?**

**Solution:**

It is given that

Length of box (l) = 80 cm

Breadth (b) = 40 cm

Height (h) = 20 cm

We know that

Surface area of the box = 2 (lh + bh + hl)

Substituting the values

= 2 (80 Ã— 40 + 40 Ã— 20 + 20 Ã— 80)

By further calculation

= 2 (320 + 800 + 1600)

= 2 Ã— 5600

= 11200 cm^{2}

So the area of square sheet = side^{2}Â

= 40^{2}

= 1600 cm^{2}

Here

Number of sheets = area of box/ area of one sheet

Substituting the values

= 11200/1600

= 7

**7. The volume of a cuboid is 3600 cm ^{3} and its height is 12 cm. The cross-section is a rectangle whose length and breadth are in the ratio 4: 3. Find the perimeter of the cross-section.**

**Solution:**

It is given that

Volume of a cuboid = 3600 cm^{3}

Height of cuboid = 12 cm

We know that

Volume of cuboid = Area of rectangle Ã— height

Substituting the values

3600 = area of rectangle Ã— 12

By further calculation

Area of rectangle = 3600/ 12

Area of rectangle = 300 cm^{2} â€¦.. (1)

Here

Ratio of length and breadth of rectangle = 4: 3

Consider

Length of rectangle = 4x

Breadth of rectangle = 3x

Area of rectangle = length Ã— breadth

Substituting the values

Area of rectangle = 4x Ã— 3x

So we get

Area of rectangle = 12x^{2} cm^{2} â€¦.. (2)

Using equations (1) and (2)

12x^{2} = 300

x^{2} = 300/12

So we get

x^{2} = 25

x = âˆš25 = 5

Here

Length of rectangle = 4 Ã— 5 = 20 cm

Breadth of rectangle = 3 Ã— 5 = 15 cm

Perimeter of the cross section = 2 (l + b)

Substituting the values

= 2 (20 + 15)

= 2 Ã— 35

= 70 cm

**8. The volume of a cube is 729 cm ^{3}. Find its surface area and the length of a diagonal.**

**Solution:**

It is given that

Volume of a cube = 729 cm^{3}

We can write it as

side Ã— side Ã— side = 729

(side)^{3} = 729

So we get

Side = 9 cm

We know that

Surface area of cube = 6 (side)^{2}

Substituting the values

= 6 Ã— (9)^{2}

= 6 Ã— 9 Ã— 9

= 486 cm^{2}

So the length of a diagonal = âˆš3 Ã— side

Substituting the value

= âˆš3 Ã— 9

= 1.73 Ã— 9

= 15.57 cm

**9. The length of the longest rod which can be kept inside a rectangular box is 17 cm. If the inner length and breadth of the box are 12 cm and 8 cm respectively, find its inner height.**

**Solution:**

Consider h m as the inner height

It is given that

Length of longest rod inside a rectangular box = 17 cm which is same as diagonal of rectangular box

By squaring on both sides

17^{2} = 12^{2} + 8^{2} + h^{2}

By further calculation

289 = 144 + 64 + h^{2}

289 = 208 + h^{2}

So we get

h^{2} + 208 = 289

h^{2} = 289 â€“ 208 = 81

h = âˆš81 = 9 cm

Therefore, the inner height of rectangular box is 9 cm.

**10. A closed rectangular box has inner dimensions 90 cm by 80 cm by 70 cm. Calculate its capacity and the area of tin-foil needed to line its inner surface.**

**Solution:**

It is given that

Inner length of rectangular box = 90 cm

Inner breadth of rectangular box = 80 cm

Inner height of rectangular box = 70 cm

We know that

Capacity of rectangular box = volume of rectangular box = l Ã— b Ã— h

Substituting the values

= 90 Ã— 80 Ã— 70

= 504000 cm^{3}

Here

Required area of tin foil = 2 (lb + bh + lh)

Substituting the values

= 2 (90 Ã— 80 + 80 Ã— 70 + 90 Ã— 70)

By further calculation

= 2 (7200 + 5600 + 6300)

So we get

= 2 Ã— 19100

= 38200 cm^{2}

**11. The internal measurements of a box are 20 cm long, 16 cm wide and 24 cm high. How many 4 cm cubes could be put into the box?**

**Solution:**

It is given that

Volume of box = 20 cm Ã— 16 cm Ã— 14 cm

Volume of cubes = 4 cm Ã— 4 cm Ã— 4 cm

We know that

Number of cubes put into the box = volume of box/ volume of cubes

Substituting the values

= (20 Ã— 16 Ã— 24)/ (4 Ã— 4 Ã— 4)

= 5 Ã— 4 Ã— 6

= 120

Therefore, 120 cubes can be put into the box.

**12. The internal measurements of a box are 10 cm long, 8 cm wide and 7 cm high. How many cubes of side 2 cm can be put into the box?**

**Solution:**

It is given that

Length of box = 10 cm

Breadth of box = 8 cm

Height of box = 7 cm

We know that

3 number of cubes of side 2 cm can be put in box

i.e. height of box is 7 cm so only 3 cubes can be put height wise

**13. A certain quantity of wood costs Rs 250 per m ^{3}. A solid cubical block of such wood is bought for Rs 182.25. Calculate the volume of the block and use the method of factors to find the length of one edge of the block.**

**Solution:**

It is given that

Cost of Rs 250 for 1 m^{3} wood

Cost of Rs 1 for 1/250 m^{3} wood

Cost of Rs 182.25 for 182.25/250 m^{3} wood

Here

Quantity of wood = 182.25/250 m^{3}

Multiply and divide by 100

= 18225/ (250 Ã— 100)

So we get

= 18225/ (25 Ã— 1000)

= 729/ 1000

= 0.729 m^{3}

We know that

Volume of given block = 0.729 m^{3}

Consider the length of one edge of block = x m

So we get

x^{3} = 0.729 m^{3}

Now taking cube root on both sides

So we get

= (3 Ã— 3)/ (2 Ã— 5)

= 9/10

= 0.9 m

Therefore, the length of one edge of the block is 0.9 m.

**14. A cube of 11 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base of the vessel are 15 cm Ã— 12 cm, find the rise in the water level in centimetres correct to 2 decimal places, assuming that no water over flows.**

**Solution:**

It is given that

Edge of cube = 11 cm

Volume of cube = edge^{3}

Substituting the values

= 11^{3}

= 11 Ã— 11 Ã— 11

= 1331 cm^{3}

We know that

The dimensions of the base of the vessel are 15 cm Ã— 12 cm

Consider the rise in the water level = h cm

So the volume of cube = volume of vessel

Substituting the values

1331 = 15 Ã— 12 Ã— h

By further calculation

h = 1331/ (15 Ã— 12)

So we get

h = 1331/ 180 = 7.39 cm

Therefore, the rise in the water level is 7.39 cm.

**15. A rectangular container, whose base is a square of side 6 cm, stands on a horizontal table and holds water upto 1 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 2 cm ^{3} of water over flows, calculate the volume of the cube.**

**Solution:**

If the base of rectangular container is a square

l = 6 cm and b = 6 cm

When a cube is placed in it, water rises to top i.e. through height 1 cm and 2 cm^{3} of water overflows

We know that

Volume of cube = Volume of water displaced

Substituting the values

= 6 Ã— 6 Ã— 1 + 2

= 36 + 2

= 38 cm^{3}

**16. (a) Two cubes, each with 12 cm edge, are joined end to end. Find the surface area of the resulting cuboid.**

**(b) A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between the surface area of the original cube and the sum of the surface areas of the new cubes.**

**Solution:**

(a) We know that

By joining two cubes end to end a cuboid is formed whose dimensions are

l = 12 + 12 = 24 cm

b = 12 cm

h = 12 cm

Here

Total surface area of cuboid = 2 (lb + bh + hl)

Substituting the values

= 2 (24 Ã— 12 + 12 Ã— 12 + 12 Ã— 24)

By further calculation

= 2 (288 + 144 + 288)

So we get

= 2 Ã— 720

= 1440 cm^{2}

(b) We know that

Side of a cube = 12 cm

Here

Volume = side^{3} = 12^{3} = 1728 cm^{3}

If cut into 8 equal cubes

Volume of each cube = 1728/8 = 216 cm^{3}

Surface area of original cube = 6 Ã— side^{2}

Substituting the values

= 6 Ã— 12^{2}

= 6 Ã— 144

= 864 cm^{2}

Surface area of one smaller cube = 6 Ã— 6^{2}

So we get

= 6 Ã— 36

= 216 cm^{2}

Surface area of 8 cube = 216 Ã— 8

= 1728 cm^{2}

So the ratio between their areas = 864: 1728 = 1: 2

**17. A cube of a metal of 6 cm edge is melted and cast into a cuboid whose base is 9 cm Ã— 8 cm. Find the height of the cuboid.**

**Solution:**

It is given that

Edge of melted cube = 6 cm

Volume of melted cube = 6 cm Ã— 6 cm Ã— 6 cm = 216 cm^{3}

Dimensions of cuboid are

Length = 9 cm

Breadth = 8 cm

h cm is the height

We know that

Volume of cuboid = l Ã— b Ã— h

Substituting the values

= 9 Ã— 8 Ã— h

= 72 h cm^{3}

Here

Volume of cuboid = Volume of melted metal cube

Substituting the values

72h = 216

h = 216/72 = 3 cm

Therefore, the height of cuboid is 3 cm.

**18. The area of a playground is 4800 m ^{2}. Find the cost of covering it with gravel 1 cm deep, if the gravel costs Rs 260 per cubic metre.**

**Solution:**

It is given that

Area of playground = 4800 m^{2}

We can write it as

l Ã— b = 4800

We know that

Depth of level = 1 cm

h = 1 cm = 1/100 m

Here

Volume of gravel = l Ã— b Ã— h

Substituting the values

= 4800 Ã— 1/100

= 48 m^{3}

Cost of gravel = Rs 260 per cubic metre

So the total cost = 260 Ã— 48 = Rs 12480

**19. A field is 30 m long and 18 m broad. A pit 6 m long, 4 m wide and 3 m deep is dug out from the middle of the field and the earth removed is evenly spread over the remaining area of the field. Find the rise in the level of the remaining part of the field in centimetres correct to two decimal places.**

**Solution:**

Consider ABCD is a field

Let ABCD be a part of the field where a pit is dug

Here

Volume of the earth dug out = 6 Ã— 4 Ã— 3 = 72 m^{3}

h m is the level raised over the field uniformly

Now divide the raised level of the field into parts I and II

Volume of part I = 14 Ã— 6 Ã— h = 84h m^{3}

Volume of part II = 24 Ã— 18 Ã— h = 432h m^{3}

Total volume of part I and II = [84h + 432h] = 516h m^{3}

We know that

516 = volume of earth dug out

Substituting the values

516 = 72

So we get

h = 72/516 = 0.1395 m

Multiply by 100

h = 0.1395 Ã— 100

h = 13.95 cm

Therefore, the level has been raised by 13.95 cm.

**20. A rectangular plot is 24 m long and 20 m wide. A cubical pit of edge 4 m is dug at each of the four corners of the field and the soil removed is evenly spread over the remaining part of the plot. By what height does the remaining plot get raised?**

**Solution:**

It is given that

Length of plot (l) = 24 m

Width of plot (b) = 20 m

So the area of plot = l Ã— b = 24 Ã— 20 = 480 m^{2}

We know that

Side of cubical pit = 4 m

Volume of each pit = 4^{3} = 64 m^{3}

Here

Volume of 4 pits at the corners = 4 Ã— 64 = 256 m^{3}

Area of surface of 4 pits = 4a^{2}

= 4 Ã— 4^{2}

= 64 m^{2}

So the area of remaining plot = 480 â€“ 64 = 416 m^{2}

Height of the soil spread over the remaining plot = 256/416 = 8/13 m

**21. The inner dimensions of a closed wooden box are 2 m, 1.2 m and 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1 m ^{3} of wood costs Rs 5400.**

**Solution:**

It is given that

Inner dimensions of wooden box are 2 m, 1.2 m and 0.75 m

Thickness of the wood = 2.5 cm

So we get

= 25/10 cm

It can be written as

= 25/10 Ã— 1/100

= 1/10 Ã— Â¼

So we get

= 1/40

= 0.025 m

So the external dimensions of wooden box are

(2 + 2 Ã— 0.025), (1.2 + 2 Ã— 0.025), (0.75 + 2 Ã— 0.025)

By further calculation

= (2 + 0.05), (1.2 + 0.05), (0.75 + 0.5)

= 2.05, 1.25, 0.80

Here

Volume of solid = External volume of box â€“ Internal volume of box

Substituting the values

= 2.05 Ã— 1.25 Ã— 0.80 â€“ 2 Ã— 1.2 Ã— 0.75

By further calculation

= 2.05 â€“ 1.80

= 0.25 m^{3}

Cost = Rs 5400 for 1 m^{3}

So the total cost = 5400 Ã— 0.25

Multiply and divide by 100

= 5400 Ã— 25/100

= 54 Ã— 25

= Rs 1350

**22. A cubical wooden box of internal edge 1 m is made of 5 cm thick wood. The box is open at the top. If the wood costs Rs 9600 per cubic metre, find the cost of the wood required to make the box.**

**Solution:**

It is given that

Internal edge of cubical wooden box = 1 m

Thickness of wood = 5 cm

We know that

External length = 1 m + 10 cm = 1.1 m

Breadth = 1 m + 10 m = 1.1 m

Height = 1 m + 5 cm = 1.05 m

Here

Volume of the wood used = Outer volume â€“ Inner volume

Substituting the values

= 1.1 Ã— 1.1 Ã— 1.05 â€“ 1 Ã— 1 Ã— 1

By further calculation

= 1.205 â€“ 1

= 0.2705 m^{3}

Cost of 1 m^{3} = Rs 9600

So the cost of 0.2705 m^{3} = 9600 Ã— 0.2705

= Rs 2596.80

**23. A square brass plate of side x cm is 1 mm thick and weighs 4725 g. If one cc of brass weights 8.4 gm, find the value of x.**

**Solution:**

It is given that

Side of square brass plate = x cm

Here l = x cm and b = x cm

Thickness of plate = 1 mm = 1/10 cm

We know that

Volume of the plate = l Ã— b Ã— h

Substituting the values

= x Ã— x Ã— 1/10

= x^{2}/ 10 cm^{3} â€¦.. (1)

Here

8.4 gm weight brass having volume = 1 cc

1 gm weight brass having volume = 1/8.4 cc

So the 4725 gm weight brass having volume = 4725 Ã— 1/8.4 = 562.5 cc

Volume of plate = 562.5 cc = 562.5 cm^{3} â€¦. (2)

Using both the equations

x^{2}/10 = 562.5

By cross multiplication

x^{2} = 562.5 Ã— 10 = 5625

Here

x = âˆš5625 = 75 cm

Therefore, the value of x is 75 cm.

**24. Three cubes whose edges are x cm, 8 cm and 10 cm respectively are melted and recast into a single cube of edge 12 cm. Find x.**

**Solution:**

It is given that

Edges of three cubes are x cm, 8 cm and 10 cm

So the volumes of these cubes are x^{3}, 8^{3} and 10^{3} i.e. x^{3}, 512 cm^{3} and 1000 cm^{3}

We know that

Edges of new cube formed = 12 cm

Volume of new cube = 12^{3} = 1728 cm^{3}

Based on the question

x^{3 }+ 512 + 1000 = 1728

By further calculation

x^{3} + 1512 = 1728

x^{3} = 216

It can be written as

x^{3} = 6 Ã— 6 Ã— 6

So we get

x = 6 cm

**25. The area of cross-section of a pipe is 3.5 cm ^{2} and water is flowing out of pipe at the rate of 40 cm/s. How much water is delivered by the pipe in one minute?**

**Solution:**

It is given that

Area of cross-section of pipe = 3.5 cm^{2}

Speed of water = 40 cm/sec

Length of water column in 1 sec = 40 cm

We know that

Volume of water flowing in 1 second = Area of cross section Ã— length

Substituting the values

= 3.5 Ã— 40

= 35 Ã— 4

= 140 cm^{3}

So the volume of water flowing in 1 minute i.e. 60 sec = 140 Ã— 60 cm^{3}

Here

1 litre = 1000 cm^{3}

So we get

Volume = (140 Ã— 60)/ 100

= (14 Ã— 6)/ 10

= 84/10

= 8.4 litres

**26. (a) The figure (i) given below shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in cm and all angles in the figure are right angles.**

**(b) The figure (ii) given below shows the cross section of a concrete wall to be constructed. It is 2 m wide at the top, 3.5 m wide at the bottom and its height is 6 m and its length is 400 m. Calculate **

**(i) the cross sectional area and **

**(ii) volume of concrete in the wall.**

**(c) The figure (iii) given below show the cross section of a swimming pool 10 m broad, 2 m deep at one end and 3 m deep at the other end. Calculate the volume of water it will hold when full, given that its length is 40 m.**

**Solution:**

(a) We know that

The given figure can be divided into two cuboids of dimensions 4 cm, 4 cm, 2 cm, 4 cm, 2 cm and 6 cm

Volume of solid = 4 Ã— 4 Ã— 2 + 4 Ã— 2 Ã— 6

= 32 + 48

= 80 cm^{3}

(b) We know that

Figure (ii) is a trapezium with parallel sides 2 m and 3.5 m

(i) Area of cross section = Â½ (sum of parallel sides) Ã— height

Substituting the values

= Â½ (2 + 3.5) Ã— 6

By further calculation

= Â½ Ã— 5.5 Ã— 6

So we get

= 5.5 Ã— 3

= 16.5 m^{2}

(ii) Volume of concrete in the wall = Area of cross section Ã— length

Substituting the values

= 16.5 Ã— 400

= 165 Ã— 40

= 6600 m^{3}

(c) We know that

From figure (iii) we know that it is trapezium with parallel sides 2 m and 3m

Here

Area of cross section = Â½ (sum of parallel sides) Ã— height

Substituting the values

= Â½ (2 + 3) Ã— 10

By further calculation

= Â½ Ã— 5 Ã— 10

= 5 Ã— 5

= 25 m^{2}

So the volume of water it will hold when full = area of cross section Ã— height

= 25 Ã— 40

= 1000 m^{3}

**27. A swimming pool is 50 metres long and 15 metres wide. Its shallow and deep ends are 1 Â½ metres and 14 Â½ metres deep respectively. If the bottom of the pool slopes uniformly, find the amount of water required to fill the pool.**

**Solution:**

It is given that

Length of swimming pool = 50 m

Width of swimming pool = 15 m

Its shallow and deep ends are 1 Â½ m and 5 Â½ m deep

We know that

Area of cross section of swimming pool = Â½ (sum of parallel sides) Ã— width

Substituting the values

= Â½ (1 Â½ + 4 Â½) Ã— 15

By further calculation

= Â½ (3/2 + 9/2) Ã— 15

So we get

= Â½ [(3 + 9)/ 2] Ã— 15

= Â½ Ã— 12/2 Ã— 15

= Â½ Ã— 6 Ã— 15

= 3 Ã— 15

= 45 m^{2}

Here

Amount of water required to fill pool = Area of cross section Ã— length

= 45 Ã— 50

= 2250 m^{3}

Chapter Test

**Take Ï€ = 22/7, unless stated otherwise.**

**1. (a) Calculate the area of the shaded region.**

**(b) If the sides of a square are lengthened by 3 cm, the area becomes 121 cm ^{2}. Find the perimeter of the original square.**

**Solution:**

(a) From the figure,

OA is perpendicular to BC

It is given that

AC = 15 cm, AO = 12 cm, BO = 5 cm, BC = 14 cm

OC = BC â€“ BO = 14 â€“ 5 = 9 cm

Here

Area of right â–³AOC = Â½ Ã— base Ã— altitude

Substituting the values

= Â½ Ã— 9 Ã— 12

= 54 cm^{2}

(b) Consider the side of original square = x cm

So the length of given square = (x + 3) cm

We know that

Area = side Ã— side

Substituting the values

121 = (x + 3) (x + 3)

It can be written as

11^{2} = (x + 3)^{2}

By further calculation

11 = x + 3

x = 11 â€“ 3

x = 8 cm

**2. (a) Find the area enclosed by the figure (i) given below. All measurements are in centimetres.**

**(b) Find the area of the quadrilateral ABCD shown in figure (ii) given below. All measurements are in centimetres.**

**(c) Calculate the area of the shaded region shown in figure (iii) given below. All measurements are in metres.**

**Solution:**

(a) We know that

Area of figure (i) = Area of ABCD â€“ Area of both triangles

Substituting the values

= (9 Ã— 9) â€“ (Â½ Ã— 5 Ã— 6 ) Ã— 2

By further calculation

= 81 â€“ (15 Ã— 2)

So we get

= 81 â€“ 30

= 51 cm^{2}

(b) In â–³ABD

Using Pythagoras theorem

BD^{2} = AB^{2} + AD^{2}

Substituting the values

BD^{2} = 6^{2} + 8^{2}

BD^{2} = 36 + 64

So we get

BD^{2} = 100

BD = 10 cm

In â–³BCD

Using Pythagoras theorem

BC^{2} = BD^{2} + CD^{2}

Substituting the values

26^{2} = 10^{2} + CD^{2}

676 = 100 + CD^{2}

By further calculation

CD^{2} = 676 â€“ 100 = 576

CD = âˆš576

CD = 24 cm

Here

Area of the given figure = Area of â–³ABD + Area of â–³BCD

We can write it as

Area of the given figure = Â½ Ã— base Ã— height + Â½ Ã— base Ã— height

Area of the given figure = Â½ Ã— AB Ã— AD + Â½ Ã— CD Ã— BD

Substituting the values

Area of the given figure = Â½ Ã— 6 Ã— 8 + Â½ Ã— 24 Ã— 10

So we get

Area of the given figure = 3 Ã— 8 + 12 Ã— 10

= 24 + 120

= 144 cm^{2}

(c) We know that

Area of the figure (iii) = Area of ABCD â€“ (Area of 1^{st} part + Area of 2^{nd} part + Area of 3^{rd} part)

It can be written as

= (AB Ã— BC) â€“ [(1/2 Ã— base Ã— height) + (1/2 Ã— base Ã— height) + Â½ (sum of parallel side Ã— height)]

Substituting the values

= (12 Ã— 12) â€“ [1/2 Ã— 5 Ã— 5 + Â½ Ã— 5 Ã— 7 + Â½ (7 + 3) Ã— 12]

By further calculation

= 144 â€“ [25/2 + 35/2 + 10 Ã— 6]

= 144 â€“ (60/2 + 60)

= 144 â€“ (30 + 60)

So we get

= 144 â€“ 90

= 54 m^{2}

Therefore, the required area of given figure = 54 m^{2}.

**3. Asifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.**

**Solution:**

Here the whole figure of an aeroplane has 5 figures i.e. three triangles, one rectangle and one trapezium

Consider M as the midpoint of AB

AM = MB = 1 cm

Now join MN and CN

â–³AMN, â–³NCB and â–³MNC are equilateral triangles having 1 cm side each

Area of â–³GHF

Here

= Â¾ Ã— 3.316

= 3 Ã— 0.829

= 2.487

= 2.48 cm^{2}

Area of rectangle II (MCFH) = l Ã— b

Substituting the values

= 6.5 Ã— 1

= 6.5 cm^{2}

Area of â–³ III + IV = 2 Ã— Â½ Ã— 6 Ã— 1.5 = 9 cm^{2}

Area of three equilateral triangles formed trapezium III = 3 Ã— âˆš3/4 Ã— 1^{2}

= Â¾ Ã— 1.732

= 3 Ã— 0.433

= 1.299

= 1.3 cm^{2}

So we get

Total area = 2.48 + 6.50 + 9 + 1.30

= 19.28

= 19.3 cm^{2}

**4. If the area of a circle is 78.5 cm ^{2}, find its circumference. (Take Ï€ = 3.14)**

**Solution:**

It is given that

Area of a circle = 78.5 cm^{2}

Consider r as the radius

r^{2} = Area/Ï€

Substituting the values

r^{2} = 78.50/3.14

r^{2} = 25 = 5^{2}

So we get

r = 5 cm

Here

Circumference = 2 Ï€r

Substituting the values

= 2 Ã— 3.14 Ã— 5

= 31.4 cm

**5. From a square cardboard, a circle of biggest area was cut out. If the area of the circle is 154 cm ^{2}, calculate the original area of the cardboard.**

**Solution:**

It is given that

Area of circle cut out from the square board = 154 cm^{2}

Consider r as the radius

Ï€r^{2} = 154

22/7 r^{2} = 154

By further calculation

r^{2} = (154 Ã— 7)/ 22 = 49 = 7^{2}

r = 7 cm

We know that

Side of square = 7 Ã— 2 = 14 cm

So the area of the original cardboard = a^{2}

= 14^{2}

= 196 cm^{2}

**6. (a) From a sheet of paper of dimensions 2 m Ã— 1.5 m, how many circles of radius 5 cm can be cut? Also find the area of the paper wasted. Take Ï€ = 3.14.**

**(b) If the diameter of a semi-circular protractor is 14 cm, then find its perimeter.**

**Solution:**

(a) It is given that

Length of sheet of paper = 2 m = 200 cm

Breadth of sheet = 1.5 m = 150 cm

Area = l Ã— b

Substituting the values

= 200 Ã— 150

= 30000 cm^{2}

We know that

Radius of circle = 5 cm

Number of circles in lengthwise = 200/ (5 Ã— 2) = 20

Number of circles in widthwise = 150/10 = 15

So the number of circles = 20 Ã— 15 = 300

Here

Area of one circle = Ï€r^{2}

Substituting the values

= 3.14 Ã— 5 Ã— 5 cm^{2}

Area of 300 circles = 300 Ã— 314/100 Ã— 25 = 23550 cm^{2}

So the area of remaining portion = area of square â€“ area of 300 circles

= 30000 â€“ 23550

= 6450 cm^{2}

(b) It is given that

Diameter of semicircular protractor = 14 cm

We know that

Perimeter = Â½ Ï€d + d

Substituting the values

= Â½ Ã— 22/7 Ã— 14 + 14

= 22 + 14

= 36 cm

**7. A road 3.5 m wide surrounds a circular park whose circumference is 88 m. Find the cost of paving the road at the rate of â‚¹ 60 per square metre.**

**Solution:**

It is given that

Width of the road = 3.5 m

Circumference of the circular park = 88 m

Consider r as the radius of the park

2 Ï€r = 88

Substituting the values

2 Ã— 22/7 r = 88

By further calculation

r = (88 Ã— 7)/ (2 Ã— 22) = 14 m

Here

Outer radius (R) = 14 + 3.5 = 17.5 m

Area of the path = 22/7 Ã— (17.5 + 14) (17.5 â€“ 14)

We know that

Ï€ (R^{2} â€“ r^{2}) = 22/7 [(17.5)^{2} â€“ (14)^{2}]

By further calculation

= 22/7 (17.5 + 14) (17.5 â€“ 14)

= 22/7 Ã— 31.5 Ã— 3.5

= 246.5 m^{2}

It is given that

Rate of paving the road = â‚¹ 60 per m^{2}

So the total cost = 60 Ã— 346.5 = â‚¹ 20790

**8. The adjoining sketch shows a running track 3.5 m wide all around which consists of two straight paths and two semicircular rings. Find the area of the track.**

**Solution:**

It is given that

Width of track = 3.5 m

Inner length of rectangular base = 140 m

Width = 42 m

Outer length of rectangular base = 140 + 2 Ã— 3.5

= 140 + 7

= 147 m

Width = 42 + 2 Ã— 3.5

= 42 + 7

= 49 m

We know that

Radius of inner semicircle (r) = 42/2 = 21 m

Outer radius (R) = 21 + 3.5 = 24.5 m

Here

Area of track = 2 (140 Ã— 3.5) + 2 Ã— Â½ Ï€ (R^{2} â€“ r^{2})

Substituting the values

= 2 (490) + 22/7 [(24.5)^{2} â€“ (21)^{2}]

By further calculation

= 980 + 22/7 (24.5 + 21) (24.5 â€“ 21)

= 980 + 22/7 Ã— 45.5 Ã— 3.5

So we get

= 980 + 500.5

= 1480.5 m^{2}

**9. In the adjoining figure, O is the centre of a circular arc and AOB is a line segment. Find the perimeter and the area of the shaded region correct to one decimal place. (Take Ï€ = 3.142)**

**Hint. Angle in a semicircle is a right angle.**

**Solution:**

We know that

In a semicircle âˆ ACB = 90^{0}

â–³ ABC is a right-angled triangle

Using Pythagoras theorem

AB^{2} = AC^{2}Â + BC^{2}

Substituting the values

= 12^{2} + 16^{2}

= 144 + 256

= 400

AB^{2} = (20)^{2}

AB = 20 cm

Radius of semicircle = 20/2 = 10 cm

(i) We know that

Area of shaded portion = Area of semicircle â€“ Area of â–³ ABC

It can be written as

= Â½ Ï€r^{2} â€“ (AC Ã— BC)/ 2

Substituting the values

= Â½ Ã— 3.142 (10)^{2} â€“ (12 Ã— 16)/2

By further calculation

= 314.2/2 â€“ 96

= 157.1 â€“ 96

= 61.1 cm^{2}

(ii) Here

Perimeter of shaded portion = circumference of semicircle + AC + BC

It can be written as

= Ï€r + 12 + 16

= 3.142 Ã— 10 + 28

So we get

= 31.42 + 28

= 59.42 cm

= 59.4 cm

**10. (a) In the figure (i) given below, the radius is 3.5 cm. Find the perimeter of the quarter of the circle.**

**(b) In the figure (ii) given below, there are five squares each of side 2 cm.**

**(i) Find the radius of the circle.**

**(ii) Find the area of the shaded region. (Take Ï€ = 3.14).**

**Solution:**

(a) It is given that

Radius of quadrant = 3.5 cm

Perimeter = 2r + Â¼ Ã— 2 Ï€r

= 2r + Â½ Ã— Ï€r

Substituting the values

= 2 Ã— 3.5 + Â½ Ã— 22/7 Ã— 3.5

So we get

= 7 + 5.5

= 12.5 cm

(b) From the figure

OB = 2 + 1 = 3 cm

AB = 1 cm

Using Pythagoras theorem

OA = âˆšOB^{2} + AB^{2}

Substituting the values

OA = âˆš(3)^{2} + (1)^{2}

= âˆš9 + 1

= âˆš10

So the radius of the circle = âˆš10 cm

We know that

Area of the circle = Ï€r^{2}

Substituting the values

= 3.14 Ã— (âˆš10)^{2}

= 3.14 Ã— 10

= 31.4 cm^{2}

Area of 5 square of side 2 cm each = 2^{2} Ã— 5

= 4 Ã— 5

= 20 cm^{2}

So the area of shaded portion = 31.4 â€“ 20 = 11.4 cm^{2}

**11. (a) In the figure (i) given below, a piece of cardboard in the shape of a quadrant of a circle of radius 7 cm is bounded by the perpendicular radii OX and OY. Points A and B lie on OX and OY respectively such that OA = 3 cm and OB = 4 cm. The triangular part OAB is removed. Calculate the area and the perimeter of the remaining piece.**

**(b) In the figure (ii) given below, ABCD is a square. Points A, B, C and D are centres of quadrants of circles of the same radius. If the area of the shaded portion is 21 3/7 cm ^{2}, find the radius of the quadrants.**

**Solution:**

(a) It is given that

Radius of quadrant = 7 cm

OA = 3 cm, OB = 4 cm

AX = 7 â€“ 3 = 4 cm

BY = 7 â€“ 4 = 3 cm

We know that

AB^{2} = OA^{2} + OB^{2}

Substituting the values

= 3^{2} + 4^{2}

= 9 + 16

= 25

So we get

AB = âˆš25 = 5 cm

(i) Area of shaded portion = Â¼ Ï€r^{2} â€“ Â½ OA Ã— OB

Substituting the values

= Â¼ Ã— 22/7 Ã— 7^{2} â€“ Â½ Ã— 3 Ã— 4

= Â¼ Ã— 22/7 Ã— 49 â€“ 6

By further calculation

= 77/2 â€“ 6

= 65/2

= 32.5 cm^{2}

(ii) Perimeter of shaded portion = Â¼ Ã— 2 Ï€r + AX + BY + AB

Substituting the values

= Â½ Ã— 22/7 Ã— 7 + 4 + 3 + 5

= 11 + 12

= 23 cm

(b) We know that

ABCD is a square with centres A, B, C and D quadrants drawn.

Consider a as the side of the square

Radius of each quadrant = a/2

Here

Area of shaded portion = a^{2} â€“ 4 Ã— [Â¼ Ï€(a/2)^{2}]

We can write it as

= a^{2} â€“ 4 Ã— Â¼ Ï€ a^{2}/4

Substituting the values

= a^{2} â€“ 22/7 Ã— a^{2}/4

So we get

= a^{2} â€“ 11a^{2}/14

= 3a^{2}/14

Here

Area of shaded portion = 21 3/7 = 150/7 cm^{2}

By equating both we get

3a^{2}/14 = 150/7

On further calculation

a^{2} = 150/7 Ã— 14/3

a^{2} = 100 = 10^{2}

a = 10 cm

So the radius of each quadrant = a/2 = 10/2 = 5 cm

**12. In the adjoining figure, ABC is a right angled triangle right angled at B. Semicircles are drawn on AB, BC and CA as diameter. Show that the sum of areas of semicircles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.**

**Solution:**

It is given that

ABC is a right angled triangle right angled at B

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2} â€¦â€¦(i)

Area of semicircle on AC as diameter = Â½ Ï€ (AC/2)^{2}

So we get

= Â½ Ï€ Ã— AC^{2}/4

= Ï€AC^{2}/8

Area of semicircle on AB as diameter = Â½ Ï€ (AB/2)^{2}

So we get

= Â½ Ï€ Ã— AB^{2}/4

= Ï€AB^{2}/8

Area of semicircle on BC as diameter = Â½ Ï€ (BC/2)^{2}

So we get

= Â½ Ï€ Ã— BC^{2}/4

= Ï€BC^{2}/8

We know that

Ï€AB^{2}/8 + Ï€BC^{2}/8 = Ï€/8 (AB^{2} + BC^{2})

From equation (i)

= Ï€/8 (AC^{2})

= Ï€AC^{2}/8

Therefore, it is proved.

**13. The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 15 minutes.**

**Solution:**

It is given that

Radius of hand = 14 cm

So the area swept in 15 minutes = Ï€r^{2} Ã— 15/60

By further calculation

= 22/7 Ã— 14 Ã— 14 Ã— Â¼

= 154 cm^{2}

**14. Find the radius of a circle if a 90 ^{0} arc has a length of 3.5 Ï€ cm. Hence, find the area of the sector formed by this arc.**

**Solution:**

It is given that

Length of arc of the sector of a circle = 3.5 Ï€ cm

Angle at the centre = 90^{0}

We know that

Radius of the arc = 3.5 Ï€/2 Ï€ Ã— 360/90

So we get

= (3.5 Ã— 4)/2

= 7 cm

Area of the sector = Ï€r^{2} Ã— 90^{0}/360^{0}

By further calculation

= 22/7 Ã— 7 Ã— 7 Ã— Â¼

= 77/2

= 38.5 cm^{2}

**15. A cube whose each edge is 28 cm long has a circle of maximum radius on each of its face painted red. Find the total area of the unpainted surface of the cube.**

**Solution:**

It is given that

Edge of cube = 28 cm

Surface area = 6 a^{2}

Substituting the value

= 6 Ã— (28)^{2}

= 6 Ã— 28 Ã— 28

= 4704 cm^{2}

Diameter of each circle = 28 cm

So the radius = 28/2 = 14 cm

We know that

Area of each circle = Ï€r^{2}

Substituting the values

= 22/7 Ã— 14 Ã— 14

= 616 cm^{2}

Area of such 6 circles drawn on 6 faces of cube = 616 Ã— 6 = 3696 cm^{2}

Here

Area of remaining portion of the cube = 4704 â€“ 3696 = 1008 cm^{2}

**16. Can a pole 6.5 m long fit into the body of a truck with internal dimensions of 3.5 m, 3 m and 4 m?**

**Solution:**

No, a pole 6.5 m long cannot fit into the body of a truck with internal dimensions of 3.5 m, 3 m and 4 m

Length of pole = 6.5 m

Internal dimensions of truck are 3.5 m, 3 m and 4 m

The internal dimensions are less than the length of pole. So the pole cannot fit into the body of truck with given dimensions.

**17. A car has a petrol tank 40 cm long, 28 cm wide and 25 cm deep. If the fuel consumption of the car averages 13.5 km per litre, how far can the car travel with a full tank of petrol?**

**Solution:**

It is given that

Capacity of car tank = 40 cm Ã— 28 cm Ã— 25 cm = (40 Ã— 28 Ã— 25) cm^{3}

Here 1000 cm^{3} = 1 litre

So we get

= (40 Ã— 28 Ã— 25)/ 1000 litre

Average of car = 13.5 km per litre

We know that

Distance travelled by car = (40 Ã— 28 Ã— 25)/ 1000 Ã— 13.5

Multiply and divide by 10

= (40 Ã— 25) Ã— 28/ 1000 Ã— 135/10

So we get

= (1 Ã— 28)/ 1 Ã— 135/10

= (14 Ã— 135)/5

= 14 Ã— 27

= 378 km

Therefore, the car can travel 378 km with a full tank of petrol.

**18. An aquarium took 96 minutes to completely fill with water. Water was filling the aquarium at a rate of 25 litres every 2 minutes. Given that the aquarium was 2 m long and 80 cm wide, compute the height of the aquarium.**

**Solution:**

We know that

Water filled in 2 minutes = 25 litres

Water filled in 1 minute = 25/2 litres

Water filled in 96 minutes = 25/2 Ã— 96

= 25 Ã— 48

= 1200 litres

So the capacity of aquarium = 1200 litres â€¦.. (1)

Here

Length of aquarium = 2m = 2 Ã— 100 = 200 cm

Breadth of aquarium = 80 cm

Consider h cm as the height of aquarium

So the capacity of aquarium = 200 Ã— 80 Ã— h cm^{3}

We can write it as

= (200 Ã— 80 Ã— h)/ 1000 litre

= 1/5 Ã— 80 Ã— h litre

= 16 h litre â€¦.. (2)

Using equation (1) and (2)

16h = 1200

So we get

h = 1200 / 16

h = 75 cm

Therefore, height of aquarium = 75 cm.

**19. The lateral surface area of a cuboid is 224 cm ^{2}. Its height is 7 cm and the base is a square. Find**

**(i) a side of the square, and **

**(ii) the volume of the cuboid.**

**Solution:**

It is given that

Lateral surface area of a cuboid = 224 cm^{2}

Height of cuboid = 7 cm

Base is square

Consider x cm as the length of cuboid

x cm as the breadth of cuboid (Since the base is square both length and breadth are same)

Here

Lateral surface area = 2 (l + b) Ã— h

Substituting the values

224 = 2 (x + x) Ã— 7

By further calculation

224 = 2 Ã— 2x Ã— 7

224 = 28x

So we get

28x = 224

x = 224/28 = 8 cm

(i) Side of the square = 8 cm

(ii) We know that

Volume of the cuboid = l Ã— b Ã— h

Substituting the values

= 8 Ã— 8 Ã— 7

= 448 cm^{3}

**20. If the volume of a cube is V m ^{3}, its surface area is S m^{2} and the length of a diagonal is d metres, prove that 6âˆš3 V = Sd.**

**Solution:**

We know that

Volume of cube = (V) = (Side)^{3}

Consider a as the side of cube

V = a^{3} and S = 6a^{2}

Diagonal (d) = âˆš3. a

Sd = 6a^{2} Ã— âˆš3a = 6âˆš3a^{3}

Here V = a^{3}

So we get

Sd = 6âˆš3V

Therefore, 6âˆš3V = Sd.

**21. The adjoining figure shows a victory stand, each face is rectangular. All measurements are in centimetres. Find its volume and surface area (the bottom of the stand is open).**

**Solution:**

Three parts are indicated as 3, 1 and 2 in the figure

We know that

Volume of part (3) = 50 Ã— 40 Ã— 12 = 24000 cm^{3}

Volume of part (1) = 50 Ã— 40 Ã— (16 + 24)

By further calculation

= 50 Ã— 40 Ã— 40

= 80000 cm^{3}

Volume of part (2) = 50 Ã— 40 Ã— 24 = 48000 cm^{3}

So the total volume = 24000 + 8000 + 48000 = 153000 cm^{3}

We know that

Total surface area = Area of front and back + Area of vertical faces + Area of top faces

Substituting the values

= 2 (50 Ã— 12 + 50 Ã— 40 + 50 Ã— 24) cm^{2} + (12 Ã— 40 + 28 Ã— 40 + 16 Ã— 40 + 24 Ã— 40) cm^{2} + 3 (50 Ã— 40) cm^{2}

By further calculation

= 2 (600 + 2000 + 1200) cm^{2} + (480 + 1120 + 640 + 960) cm^{2} + (3 Ã— 2000) cm^{2}

= 2 (3800) + 3200 + 6000 cm^{2}

So we get

= 7600 + 3200 + 6000

= 16800 cm^{2}

**22. The external dimensions of an open rectangular wooden box are 98 cm by 84 cm by 77 cm. If the wood is 2 cm thick all around, find**

**(i) the capacity of the box**

**(ii) the volume of the wood used in making the box, and**

**(iii) the weight of the box in kilograms correct to one decimal place, given that 1 cm ^{3} of wood weighs 0.8 g.**

**Solution:**

It is given that

External dimensions of open rectangular wooden box = 98 cm, 84 cm and 77 cm

Thickness = 2 cm

So the internal dimensions of open rectangular wooden box = (98 â€“ 2 Ã— 2) cm, (84 â€“ 2 Ã— 2) cm and (77 â€“ 2) cm

= (98 â€“ 4) cm, (84 â€“ 4) cm, 75 cm

= 94 cm, 80 cm, 75 cm

(i) We know that

Capacity of the box = 94 cm Ã— 80 cm Ã— 75 cm

= 564000 cm^{3}

(ii) Internal volume of box = 564000 cm^{3}

External volume of box = 98 cm Ã— 84 cm Ã— 77 cm = 633864 cm^{3}

So the volume of wood used in making the box = 633864 â€“ 564000 = 69864 cm^{3}

(iii) Weight of 1 cm^{3} wood = 0.8 gm

So the weight of 69864 cm^{3} wood = 0.8 Ã— 69864 gm

By further calculation

= (0.8 Ã— 69864)/ 1000 kg

= 55891.2/1000 kg

= 55.9 kg (correct to one decimal place)

**23. A cuboidal block of metal has dimensions 36 cm by 32 cm by 0.25 m. It is melted and recast into cubes with an edge of 4 cm.**

**(i) How many such cubes can be made?**

**(ii) What is the cost of silver coating the surfaces of the cubes at the rate of â‚¹ 1.25 per square centimetre?**

**Solution:**

It is given that

Dimensions of cuboidal block = 36 cm, 32 cm and 0.25 m

We can write it as

= 36 cm Ã— 32 cm Ã— (0.25 Ã— 100) cm

= (36 Ã— 32 Ã— 25) cm^{3}

Volume of cube having edge 4 cm = 4 Ã— 4 Ã— 4 = 64 cm^{3}

(i) We know that

Number of cubes = Volume of cuboidal block/ Volume of one cube

Substituting the values

= (36 Ã— 32 Ã— 25)/ 64

= (36 Ã— 25)/ 2

So we get

= 18 Ã— 25

= 250

(ii) Here

Total surface area of one cube = 6(a)^{2}

Substituting the values

= 6 (4)^{2}

= 6 Ã— 4 Ã— 4

= 96 cm^{2}

So the total surface area of 450 cubes = 450 Ã— 96 = 43200 cm^{2}

Given

Cost of silver coating the surface for 1 cm^{2} = â‚¹ 1.25

Cost of silver coating the surface for 43200 cm^{2} = 43200 Ã— 1.25 = â‚¹ 54000

**24. Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube. Find the cost of coating the surface of the new cube with gold at the rate of â‚¹ 3.50 per square centimetre.**

**Solution:**

We know that

Volume of first cube = edge^{3}

Substituting the values

= (3 cm)^{3}

= 3 Ã— 3 Ã— 3

= 27 cm^{3}

Volume of second cube = edge^{3}

Substituting the values

= (4 cm)^{3}

= 4 Ã— 4 Ã— 4

= 64 cm^{3}

Volume of third cube = edge^{3}

Substituting the values

= (5 cm)^{3}

= 5 Ã— 5 Ã— 5

= 125 cm^{3}

So the total volume = 27 + 64 + 125 = 216 cm^{3}

Make new cube whose volume = 216 cm^{3}

So we get

Edge^{3} = 216 cm^{3}

Edge^{3} = (6 cm)^{3}

Edge = 6 cm

Surface area of new cube = 6 (edge)^{2}

Substituting the values

= 6 (6)^{2}

= 6 Ã— 6 Ã— 6

= 216 cm^{2}

Given

Cost of coating the surface for 1 cm^{2} = â‚¹ 3.50

So the cost of coating the surface for 216 cm^{2} = 3.50 Ã— 216 = â‚¹ 756