ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration is an important study material for students from an exam perspective. Students can refer to the solutions designed by the subject matter experts at BYJU’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 grasping power.
ML Aggarwal Solutions for Class 9 Maths Chapter 16: Mensuration
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 = ½ × base × height
Substituting the values
= ½ × 6 × 4
By further calculation
= 6 × 2
= 12 cm2
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 cm2
(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 × 5 × 3 × 2
= 210 cm2
(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 × 2.4 × 6
= 34.56 cm2
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 × 6 × 4
= 336 cm2
Here the shortest side of the triangle is 20 cm
Consider h cm as the corresponding altitude
Area of triangle = ½ × base × height
Substituting the values
336 = ½ × 20 × h
By further calculation
h = (336 × 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 m2)
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 × 450 × 2
It is given that
1 hectare = 10000 m2
= (525 × 900)/ 10000
By further calculation
= (525 × 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 × 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
AB2 = AC2 + BC2
Substituting the values
132 = AC2 + 122
By further calculation
AC2 = 132 – 122
So we get
AC2 = 169 – 144 = 25
AC = √25 = 5 cm
We know that
Area of triangle ABC = ½ × base × height
Substituting the values
= ½ × 12 × 5
= 30 cm2
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 = √3/4 (side)2
Substituting the values
= √3/4 × 8 × 8
By further calculation
= √3 × 2 × 8
= 1.73 × 16
= 27.71 m2
7. If the area of an equilateral triangle is 81√3 cm2 find its perimeter.
Solution:
We know that
Area of equilateral triangle = √3/4 (side)2
Substituting the values
81 √3 = √3/4 (side)2
By further calculation
Side2 = (81 √3 × 4)/ √3
So we get
(side)2 = 81 × 4
side = √(81 × 4)
side = 9 × 2 = 18 cm
So the perimeter of equilateral triangle = 3 × side
= 3 × 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 × side
Substituting the values
36 = 3 × side
By further calculation
side = 36/3 = 12 cm
So AB = BC = CA = 12 cm
Here
Area of equilateral triangle = √3/4 (side)2
Substituting the values
= √3/4 (12)2
By further calculation
= √3/4 × 12 × 12
So we get
= √3 × 3 × 12
= 1.73 × 36
= 62.4 cm2
In triangle ABD
Using Pythagoras Theorem
AB2 = AD2 + BD2
Here BD = 12/2 = 6 cm
Substituting the values
122 = AD2 + 62
By further calculation
144 =AD2 + 36
AD2 = 144 – 36 = 108
So we get
AD = √108 = 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 × 4 = 12 cm
b = 4x = 4 × 4 = 16 cm
c = 5x = 5 × 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 × 4 × 2
= 96 cm2
(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 × 3)/ sum of ration
Substituting the values
= (300 × 3)/ (3 + 5 + 7)
By further calculation
= (300 × 3)/ 15
= 60 m
Second side = (300 × 5)/ 15 = 100 m
Third side = (300 × 7)/ 15 = 140 m
Here
S = perimeter/2 = 300/2 = 150 m
So we get
We get
= 1500 × 1.732
= 2598 m2
10. ABC is a triangle in which AB = AC = 4 cm and ∠A = 900. Calculate the area of △ABC. Also find the length of perpendicular from A to BC.
Solution:
It is given that
AB = AC = 4 cm
Using the Pythagoras theorem
BC2 = AB2 + AC2
Substituting the values
BC2 = 42 + 42
By further calculation
BC2 = 16 + 16 = 32
BC = √32 = 4√2 cm
We know that
Area of △ABC = ½ × BC × h
Substituting the values
8 = ½ × 4√2 × h
By further calculation
h = (8 × 2)/ 4√2
We can write it as
h = (2 × 2)/ √2 × √2/√2
So we get
h = 4√2/2 = 2 × √2
h = 2 × 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 – (12 + 12) = 30 – 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 × 3.873
= 34.857
= 34.86 cm2
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 △ABC = AB + BC + AC
Substituting the values
16 = x + 6 + x
By further calculation
16 = 2x + 6
16 – 6 = 2x
10 = 2x
So we get
x = 10/2 = 5
Here AB = AC = 5 cm
BC = ½ × 6 = 3 cm
In △ABD
AB2 = AD2 + BD2
Substituting the values
52 = AD2 + 32
25 = AD2 + 9
By further calculation
AD2 = 25 – 9 = 16
So we get
AD = 4 cm
Here
Area of △ABC = ½ × base × height
Substituting the values
= ½ × 6 × 4
= 3 × 4
= 12 cm2
13. The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm2.
Solution:
Consider ABC as a right angled triangle
AB = 5x cm and BC = (3x – 1) cm
We know that
Area of △ABC = ½ × AB × BC
Substituting the values
60 = ½ × 5x (3x – 1)
By further calculation
120 = 5x (3x – 1)
120 = 15x2 – 5x
It can be written as
15x2 – 5x – 120 = 0
Taking out the common terms
5 (3x2 – x – 24) = 0
3x2 – x – 24 = 0
3x2 – 9x + 8x – 24 = 0
Taking out the common terms
3x (x – 3) + 8 (x – 3) = 0
(3x + 8) (x – 3) = 0
Here
3x + 8 = 0 or x – 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 × 3 = 15 cm
BC = (3 × 3 – 1) = 9 – 1 = 8 cm
In right angled △ABC
Using Pythagoras theorem
AC2 = AB2 + BC2
Substituting the values
AC2 = 152 + 82
By further calculation
AC2 = 225 + 64 = 289
AC2 = 172
So AC = 17 cm
Therefore, the hypotenuse of the right angled triangle is 17 cm.
14. In △ABC, ∠B = 900, AB = (2A + 1) cm and BC = (A + 1) cm. If the area of the △ABC is 60 cm2, find its perimeter.
Solution:
It is given that
AB = (2x + 1) cm
BC = (x + 1) cm
We know that
Area of △ABC = ½ × AB × BC
Substituting the values
60 = ½ × (2x + 1) (x + 1)
By cross multiplication
60 × 2 = (2x + 1) (x + 1)
By further calculation
120 = 2x2 + 3x + 1
We can write it as
0 = 2x2 + 3x + 1 – 120
0 = 2x2 + 3x – 119
So we get
2x2 + 3x – 119 = 0
2x2 + 17x – 14x – 119 = 0
Taking out the common terms
x (2x + 17) – 7 (2x + 17) = 0
(x – 7) (2x + 17) = 0
Here
x – 7 = 0 or 2x + 17 = 0
x = 7 or 2x = – 17
x = 7 or x = -17/2
AB = (2x + 1) = 2 × 7 + 1
AB = 14 + 1 = 15 cm
BC = (x + 1) = 7 + 1 = 8 cm
In right angled △ABC
Using Pythagoras Theorem
AC2 = AB2 + BC2
Substituting the values
AC2 = 152 + 82
AC2 = 225 + 64
AC2 = 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 – 25 = 35 cm
Consider base = x cm
Altitude = (35 – x) cm
Using the Pythagoras theorem
x2 + (35 – x)2 = 252
By further calculation
x2 + 1225 + x2 – 70x = 625
2x2 – 70x + 1225 – 625 = 0
2x2 – 70x + 600 = 0
Dividing by 2
x2 – 35x + 300 = 0
x2 – 15x – 20x + 300 = 0
Taking out the common terms
x (x – 15) – 20 (x – 15) = 0
(x – 15) (x – 20) = 0
Here
x – 15 = 0
So we get
x = 15
Similarly
x – 20 = 0
So we get
x = 20 cm
So 15 cm and 20 cm are the sides of the triangle
Area = ½ × base × altitude
Substituting the values
= ½ × 15 × 20
= 150 cm2
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 cm2 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 cm2
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 △ABD
AB2 = BD2 + AD2
Substituting the values
132 = (x/2)2 + AD2
By further calculation
169 = x2/4 + AD2
AD2 = 169 – x2/4 …… (1)
We know that
Area = 60 cm2
AD = (area × 2)/ base
Substituting the values
AD = (60 × 2)/ x = 120/x …… (2)
Using both the equations
169 – x2 /4 = (120/x)2
By further calculation
(676 – x2)/ 4 = 14400/x2
By cross multiplication
676x2 – x4 = 57600
We can write it as
x4 – 676x2 + 57600 = 0
x4 – 576x2 – 100x2 + 57600 = 0
Taking out the common terms
x2 (x2 – 576) – 100 (x2 – 576) = 0
(x2 – 576) (x2 – 100) = 0
Here
x2 – 576 = 0 where x2 = 576
So x = 24
Similarly
x2 – 100 = 0 where x2 = 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 m2 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 m2 = Rs 60000
Here the cost of cultivating the field of Rs 25 for 100 m2
So the cost of cultivating the field of Rs 1 = 100/25 m2
Cost of cultivating the field of Rs 60000 = 100/25 × 60000
= 4 × 60000
= 240000 m2
So the area of the field = 240000 m2
½ × base × height = 240000 ….. (1)
Consider
Height of triangular field = h m2
Base of triangular field = 3h m2
Substituting the values in equation (1)
½ × 3h × h = 240000
By further calculation
½ × 3h2 = 240000
h2 = (240000 × 2)/ 3
h2 = 80000 × 2
h2 = 160000
So we get
h = √160000 = 400
Here the height of triangular field = 400 m
Base of triangular field = 3 × 400 = 1200 m2
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 built = 3m
Remaining perimeter = 250 – 3 = 247 m
So the length of the fence around it = 247 m
Rate of fencing = Rs 20 per m
Total cost of fencing = 20 × 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 × 200 √6 = 1000 √6 cm2
Area of triangular piece of second color = 1000 √6 cm2
21. (a) In the figure (1) given below, ABC is an equilateral triangle with each side of length 10 cm. In △BCD, ∠D = 900 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 = √3/4 × (side)2
Substituting the values
= √3/4 × 102
= √3/4 × 100
So we get
= √3× 25
= 1.73 × 25
= 43.3 cm2
In the right angled triangle BDC
∠D = 900
BC = 10 cm
CD = 6 cm
Using Pythagoras Theorem
BD2 + DC2 = BC2
Substituting the values
BD2 + 62 = 102
BD2 + 36 = 100
So we get
BD2 = 100 – 36 = 64
BD = √64 = 8 cm
We know that
Area of triangle BDC = ½ × base × height
So we get
= ½ × BD × DC
Substituting the values
= ½ × 8 × 6
= 4 × 6
= 24 cm2
Here the area of shaded portion = Area of triangle ABC – Area of triangle BDC
Substituting the values
= 43.3 – 24
= 19.3 cm2
(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
∠A = 900
We know that
Area of right triangle ADE = ½ × AD × AE
Substituting the values
= ½ × 3 × 3
= 9/2 cm2
So triangle BDG is an isosceles right triangle
Similarly
DG2 + BG2 = BD2
DG2 + DG2 = 42
By further calculation
2DG2 = 16
DG2 = 16/2 = 8
DG = √8 cm
Area of triangle BDG = ½ × BG × DG
We can write it as
= ½ × DG × DG
Substituting the values
= ½ (√8)2
= ½ × 8
= 4 cm2
Area of isosceles right triangle EFC = 4 cm2
So the area of shaded portion = 9/2 + 4 + 4
Taking LCM
= (9 + 8 + 8)/ 2
= 25/2
= 12.5 cm2
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
= ½ × base × height + ½ × base × height
So we get
= ½ × AC × BX + ½ × AC × DY
Substituting the values
= (½ × 20 × 9) + (½ × 20 × 15)
By further calculation
= (10 × 9 + 10 × 15)
= 90 + 150
= 240 cm2
(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 = ½ × diagonal AC × diagonal BD
Substituting the values
= ½ × 18 × 12
By further calculation
= 9 × 12
= 108 cm2
2. Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 900.
Solution:
It is given that
ABCD is a quadrilateral field
AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 900
In triangle BAD
∠A = 900
Using the Pythagoras Theorem
BD2 = BA2 + AD2
Substituting the values
BD2 = 402 + 92
By further calculation
BD2 = 1600 + 81 = 1681
So we get
BD = 41
We know that
Area of quadrilateral ABCD = Area of △BAD + Area of △BDC
It can be written as
= ½ × base × height + Area of △BDC
Substituting the values
= ½ × 40 × 9 + Area of △BDC
By further calculation
= 180 m2 + Area of △BDC
Determining the area of △BDC
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 × 7 × 3 × 3
= 126 m2
So the area of quadrilateral ABCD = 180 m2 + Area of △BDC
Substituting the values
= 180 + 126
= 306 m2
3. Find the area of the quadrilateral ABCD in which ∠BCA = 900, AB = 13 cm and ACD is an equilateral triangle of side 12 cm.
Solution:
It is given that
ABCD is a quadrilateral in which ∠BCA = 900 and AB = 13 cm
ABCD is an equilateral triangle in which AC = CD = AD = 12 cm
In right-angled △ABC
Using Pythagoras theorem,
AB2 = AC2 + BC2
Substituting the values
132 = 122 + BC2
By further calculation
BC2 = 132 – 122
BC2 = 169 – 144 = 25
So we get
BC = √25 = 5 cm
We know that
Area of quadrilateral ABCD = Area of △ABC + Area of △ACD
It can be written as
= ½ × base × height + √3/4 × side2
= ½ × AC × BC + √3/4 × 122
Substituting the values
= ½ × 12 × 5 + √3/4 × 12 × 12
So we get
= 6 × 5 + √3 × 3 × 12
= 30 + 36√3
Substituting the value of √3
= 30 + 36 × 1.732
= 30 + 62.28
= 92.28 cm2
4. Find the area of quadrilateral ABCD in which ∠B = 900, AB = 6 cm, BC = 8 cm and CD = AD = 13 cm.
Solution:
It is given that
ABCD is a quadrilateral in which ∠B = 900, AB = 6 cm, BC = 8 cm and CD = AD = 13 cm
In △ABC
Using Pythagoras theorem
AC2 = AB2 + BC2
Substituting the values
AC2 = 62 + 82
By further calculation
AC2 = 36 + 64 = 100
So we get
AC2 = 102
AC = 10 cm
We know that
Area of quadrilateral ABCD = Area of △ABC + Area of △ACD
It can be written as
= ½ × base × height + Area of △ACD
= ½ × AB × BC + Area of △ACD
Substituting the values
= ½ × 6 × 8 + Area of △ACD
By further calculation
= 24 cm2 + Area of △ACD …… (1)
Finding the area of △ACD
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 × 2 × 2 × 5
= 60 cm2
Using equation (1)
Area of quadrilateral ABCD = 24 cm2 + Area of △ACD
Substituting the values
= 24 + 60
= 84 cm2
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 × (l + b) = 96 cm
Substituting the values
2 × (l + 18) = 96 cm
By further calculation
(l + 18) = 96/2
l + 18 = 48
So we get
l = 48 – 18 = 30 cm
Here
Area of rectangular cardboard = l × b
Substituting the values
= 30 × 18
= 540 cm2
6. The length of a rectangular hall is 5 cm more than its breadth, if the area of the hall is 594 m2, 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 × b
Substituting the values
594 = x (x + 5)
By further calculation
594 = x2 + 5x
0 = x2 + 5x – 594
x2 + 5x – 594 = 0
It can be written as
x2 + 27x – 22x – 594 = 0
Taking out the common terms
x (x + 27) – 22 (x + 27) = 0
So we get
(x – 22) (x + 27) = 0
Here
x – 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 × 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 – Area of square LMNO
Substituting the values
= 50 × 5 + 5 × 35 – 5 × 5
By further calculation
= 250 + 175 – 25
So we get
= 250 + 150
= 400 m2
(b) We know that
Area of shaded portion = Area of ABCD – 5 × Area of any small square
It can be written as
= l × b – 5 × side × side
Substituting the values
= 8 × 6 – 5 × 2 × 2
By further calculation
= 48 – 20
= 28 cm2
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 – 2 × 2
By further calculation
= 20 – 4
= 16 m
Breadth of grassy lawn = 14 – 2 × 2
By further calculation
= 14 – 4
= 10 m
Area of grassy lawn = length × breadth
Substituting the values
= 16 × 10
= 160 m2
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 – beds.
Solution:
Consider BCDE as the lawn
(i) We know that
Length of lawn BCDE = BC
It can be written as
= AD – AB – CD
Substituting the values
= 30 – 2 – 2
By further calculation
= 30 – 4
= 26 m
Breadth of lawn BDCE = BE
It can be written as
= AG – GH
Substituting the values
= 12 – 2
= 10 m
(ii) We know that
Area of flower beds = Area of rectangle ADFG – Area of lawn BCDE
It can be written as
= AD × AG – BC × BE
Substituting the values
= 30 × 12 – 26 × 10
By further calculation
= 360 – 260
= 100 m2
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 m2, 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 × b
Substituting the values
= 50 × 38
= 1900 m2
Let x m as the width of foot path all around the inside of a rectangular field
Length of rectangular field PQRS = (50 – x – x) = (50 – 2x) m
Breadth of rectangular field PQRS = (38 – x – x) = (38 – 2x) m
Here
Area of foot path = Area of rectangular field ABCD – Area of rectangular field PQRS
Substituting the values
492 = 1900 – (50 – 2x) (38 – 2x)
It can be written as
492 = 1900 – [50 (38 – 2x) – 2x (38 – 2x)]
By further calculation
492 = 1900 – (1900 – 100x – 76x + 4x2)
492 = 1900 – 1900 + 100x + 76x – 4x2
On further simplification
492 = 176x – 4x2
Taking out 4 as common
492 = 4 (44x – x2)
44x – x2 = 492/4 = 123
We get
x2 – 44x + 123 = 0
It can be written as
x2 – 41x – 3x + 123 = 0
Taking out the common terms
x (x – 41) – 3 (x – 41) = 0
(x – 3) (x – 41) = 0
Here
x – 3 = 0 or x – 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 – 3000
x = 2400/30 = 80 m
Breadth of garden = 80 m
So the area of rectangular field = l × b
Substituting the values
= 100 × 80
= 8000 m2
12. A rectangular floor which measures 15 m × 8 m is to be laid with tiles measuring 50 cm × 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 × 8m
Length = 15 m
Breadth = 8 m
Here the area = l × b = 15 × 8 = 120 m2
Measurement of tiles = 50 cm × 25 cm
Length = 50 cm = 50/100 = ½ m
Breadth = 25 cm = 25/100 = ¼ m
So the area of one tile = ½ × ¼ = 1/8 m2
No. of required tiles = Area of rectangular field/Area of one tile
Substituting the values
= 120/ (1/8)
By further calculation
= (120 × 8)/ 1
= 960 tiles
Length of carpet = 15 – 1 – 1
= 15 – 2
= 13 m
Breadth of carpet = 8 – 1 – 1
= 8 – 2
= 6 m
Area of carpet = l × b
= 13 × 6
= 78 m2
We know that
Area of floor which is uncovered by carpet = Area of floor – Area of carpet
Substituting the values
= 120 – 78
= 42 m2
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 × 8x/5
y = 16x/5
We get
5y = 16x
16x = 5y ….. (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 × 32
By further calculation
x = (5 × 32)/ 16
x = (5 × 2)/ 1
x = 10 m
Breadth = 3/5 × x
Substituting the value of x
= 3/5 × 10
= 3 × 2
= 6 m
Here the floor area of the room = l × b
Substituting the values
= 10 × 6
= 60 m2
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 × b
Substituting the values
= 10 × 16
= 160 m2
Consider x m as the width of the walk
Length of rectangular garden PQRS = 10 – x – x = (10 – 2x) m
Breadth of rectangular garden PQRS = 16 – x – x = (16 – 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) × h
Substituting the values
= 2 (6 + 4.8) × 3.5
By further calculation
= 2 × 10.8 × 3.5
= 21.6 × 3.5
= 75.6 m2
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) × h
Substituting the values
= 2 (44 + 25.5) × 2
By further calculation
= 2 × 69.5 × 2
= 2 × 139
= 278 m2
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 × breadth + length × breadth
Substituting the values
= 2 × 5.5 + 4 × 1.5
= 11 + 6
= 17 m2
(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 × 3 + 8 × 2 + 7 × 2
By further calculation
= 21 + 16 + 14
= 51 m2
(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 × 2 + 2 × 3 + (2 + 3) × 2 + 2 × 3 + 7 × 2
By further calculation
= 14 + 6 + 10 + 6 + 14
= 50 cm2
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 × b
Substituting the values
= 12 × 9
= 108 cm2
Using the condition
Area of triangle ABC = 1/3 × area of rectangle
Substituting the values
= 1/3 × 108
= 36 cm2
Consider h cm as the height of triangle ABC
Area of triangle ABC = ½ × base × height
Substituting the values
36 = ½ × 9 × h
By further calculation
36 × 2 = 9 × h
h = (36 × 2)/ 9
So we get
h = 4 × 2
h = 8 cm
Therefore, height of triangle ABC is 8 cm.
19. The area of a square plot is 484 m2. 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 m2
Sides of square are AB, BC, CD and AD
We know that
Area of square = side × side
Substituting the values
484 = (side)2
So we get
Side = √484 = 22 m
AB = BC = 22 m
In triangle ABC
Using Pythagoras Theorem
AC2 = AB2 + BC2
Substituting the values
AC2 = 222 + 222
AC2 = 484 + 484 = 968
By further calculation
AC = √968 = √ (484 × 2)
AC = 22 × √2
So we get
AC = 22 × 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 × side
Substituting the values
56 = 4x
By further calculation
x = 56/4 = 14 m
In triangle ABC
Using Pythagoras theorem
AC2 = AB2 + BC2
Substituting the values
AC2 = 142 + 142
By further calculation
AC2 = 196 + 196 = 392
So we get
AC = √392
AC = √ (196 × 2)
AC = 14 √2
Substituting the value of √2
AC = 14 × 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 √3 cm2. 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 √3 cm2
Consider x cm as the side of equilateral triangle
We know that
Area = √3/4 (side)2
Substituting the values
36 √3 = √3/4 × (x)2
By further calculation
x2 = (36 √3 × 4)/ √3
x2 = 36 × 4
So we get
x = √ (36 × 4)
x = 6 × 2
x = 12 cm
Here
Perimeter of equilateral triangle = 3 × side
= 3 × 12
= 36 cm
(i) We know that
Perimeter of equilateral triangle = Perimeter of square
It can be written as
36 = 4 × side
So we get
Side = 36/4 = 9 cm
Area of square = side × side
= 9 × 9
= 81 cm2
(ii) We know that
Perimeter of triangle = Perimeter of rectangle …. (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 – 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 × breadth
= 10 × 8
= 80 cm2
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 × height
We can write it as
= AB × DM
Substituting the values
= 15 × 8
= 120 cm2
If base is AD
Area of parallelogram = AD × DN
Substituting the values
120 = 10 × DN
So we get
DN = 120/10 = 12 cm
Therefore, the area of parallelogram is 120 cm2 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 √399 cm2
We know that
Area of parallelogram = 2 × Area of triangle ABC
Substituting the values
= 2 × 3 √399
= 6 √399
So we get
= 6 × 19.96
= 119.8 cm2
Consider DM as the distance between the shorter lines
Base = AD = BC = 10 cm
Area of parallelogram = AD × DM
Substituting the values
119.8 = 10 × 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 × DM
Substituting the values
= 12 × 6
= 72 cm2
Similarly
Area of triangle AOD = ¼ × Area of parallelogram
Substituting the values
= ¼ × 72
= 18 cm2
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 × 2
= 24 cm2
We know that
Area of parallelogram ABCD = 4 × Area of triangle AOB
Substituting the values
= 4 × 24
= 96 cm2
26. The area of a parallelogram is p cm2 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 cm2
Height of first parallelogram = q cm
We know that
Area of parallelogram = base × height
Substituting the values
p = base × 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 cm2
It can be written as
Base × height = p cm2
Substituting the values
[(p + qr)/q] × 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 = ½ × AC × BD
Substituting the values
= ½ × 16 × 12
By further calculation
= 8 × 12
= 96 cm2
28. The area of a rhombus is 98 cm2. 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 cm2
One of its diagonal = 14 cm
We know that
Area of rhombus = ½ × product of diagonals
Substituting the values
98 = ½ × one diagonal × other diagonal
98 = ½ × 14 × other diagonal
By further calculation
Other diagonal = (98 × 2)/ 14
= 7 × 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 × height
Substituting the values
= 45/4 × 8
= 45 × 2
= 90 cm2
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 × height
Substituting the values
= 3 × 2.5
= 7.5 cm2
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
AB2 = AO2 + BO 2
Substituting the values
AB2 = 42 + 32
By further calculation
AB2 = 16 + 9 = 25
So we get
AB = √25 = 5 cm
Side of rhombus ABCD = 5 cm
Here
Perimeter of rhombus = 4 × side
Substituting the values
= 4 × 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
AB2 = AO2 + BO2
Substituting the values
52 = 42 + BO2
By further calculation
25 = 16 + BO2
BO2 = 25 – 16 = 9
So we get
BO = √9 = 3 cm
BD = 2 × BO = 2 × 3 = 6 cm
Length of other diagonal = 6 cm
Area of rhombus = ½ × product of diagonals
Substituting the values
= ½ × 8 × 6
By further calculation
= 4 × 6
= 24 cm2
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 m2, 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 – ND
CN = CD – AO
CN = 8 – 5 = 3 cm
In triangle BCN
Using Pythagoras theorem
BC2 = BN2 + CN2
Substituting the values
BC2 = 42 + 32
By further calculation
BC2 = 16 + 9 = 25
BC = √25 = 5 cm
Length of BC = 5 cm
Area of trapezium = ½ × sum of parallel sides × height
It can be written as
= ½ × (AB + CD) × AD
Substituting the values
= ½ × (5 + 8) × 4
By further calculation
= ½ × 13 × 4
So we get
= 13 × 2
= 26 cm2
Area of trapezium = 26 cm2
(b) From figure (ii)
AD = 8 units
BC = 2 units
CD = 10 units
Construct CN perpendicular to AD
AN = 2 units
We know that
DN = AD – DN
Substituting the values
= 8 – 6
= 2 units
In triangle CDN
Using Pythagoras theorem
CD2 = DN2 + NC2
Substituting the values
102 = 62 + NC2
By further calculation
NC2 = 102 – 62
NC2 = 100 – 36 = 64
So we get
NC = √64 = 8 units
From the figure NC = AB = 8 units
We know that
Area of trapezium = ½ × sum of parallel sides × height
It can be written as
= ½ × (BC + AD) × AB
Substituting the values
= ½ × (2 + 8) × 8
By further calculation
= ½ × 10 × 8
= 5 × 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 m2
It can be written as
½ × sum of parallel sides × depth = 16.8
½ × (AB + DC) × AL = 16.8
Substituting the values
½ × (6 + 8) × AL = 16.8
By further calculation
½ × 14 × AL = 16.8
AL = (16.8 × 2)/ 14
So we get
AL = (16.8 × 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 = ½ (AB + DC) = 18 cm
Here
Area of trapezium ABCD = ½ (AB + DC) × height
Substituting the values
= 18 × 12
= 216 cm2
35. The area of a trapezium is 540 cm2. 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 cm2
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 = ½ × sum of parallel sides × height
Substituting the values
540 = ½ × (7x + 5x) × 18
By further calculation
540 = ½ × 12x × 18
540 = 6x × 18
540 = 108x
x = 540/108 = 5
Here
First parallel side = 7x = 7 × 5 = 35 cm
Second parallel side = 5x = 5 × 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 cn2, 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 = ½ (sum of parallel sides) × height
Substituting the values
60 = ½ × (2x + 3x) × 4
By further calculation
60 = ½ × 5x × 4
60 = 5x × 2
60 = 10x
So we get
x = 60/10 = 6
Here
CD = 2x = 2 × 6 = 12 cm
AB = 3x = 3 × 6 = 18 cm
AN = BM
We can write it as
AN = AB – BN
AN = AB – (MN + BM)
We know that
MN = CD
AN = AB – (CD + BM)
Similarly BM = AN
AN = AB – (CD + AN)
Substituting the values
AN = 18 – (12 + AN)
AN = 18 – 12 – AN
AN + AN = 6
2AN = 6
By division
AN = 6/2 = 3
In triangle AND
Using Pythagoras theorem
AD2 = DN2 + AN2
Here DN = 4 cm
AD2 = 42 + 32
By further calculation
AD2 = 16 + 9 = 25
So we get
AD= √25 = 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 cm2. 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 cm2
Condition – If one altitude is half the corresponding base
Take base = x cm
Corresponding altitude = x/2 cm
We know that
Area of parallelogram = base × altitude
Substituting the values
98 = x × x/2
98 = x2/2
By cross multiplication
x2 = 98 × 2 = 196
So we get
x = √196 = 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 × b
Substituting the values
Area = (x + 12) × x
Area = (x2 + 12x) m2
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 × numerical value of perimeter
x 2 + 12x = 4 × (4x + 24)
By further calculation
x2 + 12x = 16x + 96
x2 + 12x – 16x – 96 = 0
x2 – 4x – 96 = 0
It can be written as
x2 – 12x + 8x – 96 = 0
Taking out the common terms
x (x – 12) + 8 (x – 12) = 0
(x + 8) (x – 12) = 0
Here
x + 8 = 0 or x – 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 – y …… (1)
In triangle ABC
Using Pythagoras theorem
AC2 = AB2 + BC2
Substituting the values
262 = x2 + y2
x2 + y2 = 676
Now substituting the value of x in equation (1)
(34 – y)2 + y2 = 676
1156 + y2 – 68y + y2 = 676
By further calculation
2y2 – 68y + 1156 – 676 = 0
2y2 – 68y – 480 = 0
Taking 2 as common
2 (y2 – 34y – 240) = 0
y2 – 34y – 240 = 0
It can be written as
y2 – 24y – 10y – 240 = 0
Taking out the common terms
y (y – 24) – 10 (y – 24) = 0
(y – 10) (y – 24) = 0
Here
y – 10 = 0 or y – 24 = 0
y = 10 m or y = 24 m
Now substituting the value of y in equation (1)
y = 10 m, x = 34 – 10 = 24 m
y = 24 m, x = 34 – 24 = 10 m
Area in both cases = xy
= 24 × 10 or 10 × 24
= 240 m2
Therefore, the area of the rectangular block is 240 m2.
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 × side = x × x = x2 cm2
Area of rectangle = l × b
Substituting the values
= (x + 12) (x + 8) cm2
Based on the question
Area of rectangle = 2 × area of square
Substituting the values
(x + 12) (x + 8) = 2 × x2
It can be written as
x (x + 8) + 12 (x + 8) = 2x2
x2 + 8x + 12x + 96 = 2x2
x2 – 2x2 + 8x + 12x + 96 = 0
By further calculation
-x2 + 20x + 96 = 0
– (x2 – 20x – 96) = 0
x2 – 20x – 96 = 0
We can write it as
x2 – 24x + 4x – 96 = 0
x (x – 24) + 4 (x – 24) = 0
(x + 4) (x – 24) = 0
Here
x + 4 = 0 or x – 24 = 0
x = -4 or x = 24 cm
Side of square = 24 cm
Perimeter of square = 4 × side
Substituting the values
= 4 × 24
= 96 cm
41. The perimeter of a square is 48 cm. The area of a rectangle is 4 cm2 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 = side2 = 122 = 144 cm2
Area of rectangle = 144 – 4 = 140 cm2
Take breadth of rectangle = x cm
Length of rectangle = x + 4 cm
So the area = (x + 4) × x cm2
Substituting the values
(x + 4) x = 140
By further calculation
x2 + 4x – 140 = 0
x2 + 14x – 10x – 140 = 0
Taking out the common terms
x (x + 14) – 10 (x + 14) = 0
(x + 14) (x – 10) = 0
Here
x + 14 = 0 where x = – 14
x – 10 = 0 where x = 10
Breadth = 10 cm
Length = 10 + 4 = 14 cm
Perimeter = 2 (l + b)
Substituting the values
= 2 (14 + 10)
= 2 × 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
AE2 + EB2 = AB2
Substituting the values
x2 + x2 = 102
2x2 = 100
By further calculation
x2 = 100/2 = 50
x = √50 = √ (25 × 2) = 5√2 cm
We know that
Area of triangle AEB = Area of triangle GCD
It can be written as
= ½ × x × x
= ½ x2 cm2
Substituting the value of x
= ½ × 50
= 25 cm2
Area of triangle HAD = Area of BFC
It can be written as
= √3/4 × 82
= √3/4 × 64
= 16√3 cm2
Area of shaded portion = Area of rectangle ABCD + 2 area of triangle AEB + 2 area of triangle BFC
Substituting the values
= (10 × 8 + 2 × 25 + 2 × 16√3)
By further calculation
= 80 + 50 + 32√3
So we get
= (130 + 32√3) cm2
Here
Perimeter of the figure = AE + EB + BF + FC + CD + GD + DH + HA
It can be written as
= 4AE + 4BF
Substituting the values
= 4 × 5√2 + 4 × 8
So we get
= 20√2 + 32
= (32 + 20√2) 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 × 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
CE2 = EF2 – CF2
Substituting the values
= 52 – 32
= 25 – 9
= 16
So we get
CE = √16 = 4 cm
Area of triangle ABC = √3/4 × 62
By further calculation
= √3/4 × 36
= 9√3 cm2
Area of trapezium DEFG = ½ (DE + GF) × CE
Substituting the values
= ½ × (6 + 12) × 4
By further calculation
= ½ × 18 × 4
= 36 cm2
So the area of figure = 9√3 + 36
Substituting the values
= 9 × 1.732 + 36
= 15.59 + 36
= 51.59 cm2
(b) We know that
Length of rectangle = 2 + 2 + 2 + 2 = 8 cm
Width of rectangle = 2 cm
Area of rectangle = l × b
Substituting the values
= 8 × 2
= 16 cm2
Here
Area of each trap = ½ (2 + 2) × (6 – 2)
By further calculation
= ½ × 4 × 4
= 8 cm2
So the total area = area of rectangle + area of 2 trapezium
Substituting the values
= 16 + 8 + 8
= 32 cm2
(c) We know that
Length of each rectangle = 24 cm
Width of each rectangle = 6 cm
Area of each rectangle = l × b
Substituting the values
= 24 × 6
= 144 cm2
Base of each parallelogram = 8 cm
Height of each parallelogram = 6 cm
So the area of each parallelogram = 8 × 6 = 48 cm2
Here
Area of the M-shaped figure = 2 × 144 + 2 × 48
So we get
= 288 + 96
= 384 cm2
Area of the square cardboard = 24 × 24 = 576 cm2
Area of the removing cardboard = 576 – 384 = 192 cm2
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 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
∠FEG = ∠GAF = 900
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 × b + ½ (sum of parallel sides × height)
= AF × AG + ½ (GB + CD) × DG
Substituting the values
= 2.4 × 0.3 + ½ [(AB – AG) + CD] × (DE + EG)
Here AG = FE and using EG = AF
= 0.72 + ½ [(1.8 – 0.3) + 0.6] × (1.2 + 2.4)
By further calculation
= 0.72 + ½ [1.5 + 0.6] × 3.6
= 0.72 + ½ × 2.1 × 3.6
So we get
= 0.72 + 2.1 × 1.8
= 0.72 + 3.78
= 4.5 m2
(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
= ½ × base × height + ½ (sum of parallel sides) × height + ½ × base × height + ½ × base × height
= ½ × BX × AX + ½ (BX + CZ) × XZ + ½ × CZ × DZ + ½ × AD × EY
Substituting the values
= ½ × 30 × 12 + ½ (30 + 25) × 15 + ½ × 25 × 10 + ½ × 37 × 20
By further calculation
= 15 × 12 + 7.5 × 55 + 25 × 5 + 37 × 10
So we get
= 180 + 412.5 + 125 + 370
= 1087.5 m2
(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
DE2 = DY2 + EY2
Substituting the values
52 = DY2 + 42
25 = DY2 + 16
DY2 = 25 – 16 = 9
So we get
DY = √9 = 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
= ½ × base × height + ½ × base × height + ½ × (sum of parallel sides) × height + ½ × (sum of parallel sides) × height
= ½ × EY × DY + ½ × CY × DY + ½ × (EY + AX) × XY + ½ × (CY + BX) × XY
Substituting the values
= ½ × 4 × 3 + ½ × 4 × 3 + ½ × (4 + 6) × (DX – DY) + ½ (4 + 6) × (DX – DY)
By further calculation
= 2 × 3 + 2 × 2 + ½ × 10 × (9 – 3) + ½ × 10 × (9 – 3)
So we get
= 6 + 6 + 5 × 6 + 5 × 6
= 6 + 6 + 30 + 30
= 72 cm2
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 × b = xy m2
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 × new breadth
Substituting the values
= (x + 1) (y + 1) m2
Based on the question
xy = (x + 1) (y + 1) – 21
By further calculation
xy = x (y + 1) + 1 (y + 1) – 21
xy = xy + x + y + 1 – 21
So we get
0 = x + y + 1 – 21
0 = x + y – 20
x + y – 20 = 0
x + y = 20 ….. (1)
Similarly
Length is increased by 1 metre then new length = (x + 1) metre
Breadth is decreased by 1 metre than new breadth = (y – 1) metre
So the new area = new length × new breadth
= (x + 1) (y – 1) m2
Based on the question
xy = (x + 1) (y – 1) + 5
By further calculation
xy = x (y – 1) + 1 (y – 1) + 5
xy = xy – x + y – 1 + 5
0 = -x + y + 4
So we get
x – y = 4 …… (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 – 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 × 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 × 4 × 6 × 7
= 336 cm2
We know that
Base = 28 cm
Height = Area/base
Substituting the values
= 336/28
= 12 cm
47. A rectangle of area 105 cm2 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 cm2
Length of rectangle = x cm
We know that
Area = length × breadth
Substituting the values
105 = x × 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
(x2 + 105)/ x = 22
By cross multiplication
x2 + 105 = 22x
x2 – 22x + 105 = 0
We can write it as
x2 – 15x – 7x + 105 = 0
x (x – 15) – 7 (x – 15) = 0
(x – 7) (x – 15) = 0
Here
x – 7 = 0 or x – 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 m2. 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 m2
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 – x m
We know that
Area of rectangle = l × b
Substituting the values
1800 = x × (90 – x)
90x – x2 = 1800
It can be written as
– (x2 – 90x) = 1800
x2 – 90x = – 1800
By further calculation
x2 – 90x + 1800 = 0
x2 – 60x – 30x + 1800 = 0
x (x – 60) – 30 (x – 60) = 0
(x – 30) (x – 60) = 0
Here
x – 30 = 0 or x – 60 = 0
x = 30 or x = 60
If x = 30 m
Breadth = 90 – 30 = 60 m
If x = 60 m
Breadth = 90 – 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 πr
We know that
2 πr = 44
So we get
(2 × 22)/ 7 r = 44
By further calculation
r = (44 × 7)/ (2 × 22) = 7 cm
Diameter = 2r = 2 × 7 = 14 cm
2. Find the radius and area of a circle if its circumference is 18π cm.
Solution:
Consider the radius of the circle = r
Circumference = 2 πr
We know that
2 πr = 18π
So we get
2r = 18
r = 18/2 = 9 cm
Here
Area = πr2
Substituting the value of r
= π × 9 × 9
= 81π cm2
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 = πr
Perimeter = πr + 2r = r (π + 2)
Substituting the values
= 3.85 (22/7 + 2)
By further calculation
= 3.85 × 36/7
= 0.55 × 36
= 19.8 cm
4. Find the radius and circumference of a circle whose area is 144 π cm2.
Solution:
It is given that
Area of a circle = 144 π cm2
Consider radius = r
π r2 = 144 π
r2 = 144
So we get
r = √144 = 12 cm
Here
Circumference = 2 πr
So we get
= 2 × 12 × π
= 24π 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 × 2/0.5
Multiply and divide by 10
= (11 × 10)/5 × (2 × 10)/5
By further calculation
= 22 × 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 cm2, find its perimeter.
Solution:
It is given that
Area of a semicircular region = 77 cm2
Consider r as the radius of the region
We know that
Area = ½ πr2
½ πr2 = 77
By further calculation
½ × 22/7 r2 = 77
So we get
r2 = (77 × 2 × 7)/ 22 = 49 = 72
r = 7 cm
Here
Perimeter of the region = πr + 2r
By further calculation
= 22/7 × 7 + 2 × 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 cm2, 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 π = 3.14)
Solution:
(a) It is given that
Area of shaded portion = Area of semicircle = 308 cm2
Consider r as the radius of circle
½ πr2 = 308
By further calculation
½ × 22/7 r2 = 308
r2 = (308 × 2 × 7)/ 22
r2 = 196 = 142
So r = 14 cm
(i) AC = 2r = 2 × 14 = 28 cm
(ii) We know that
Circumference of the circle = 2πr
Substituting the values
= 28 × 22/7
So we get
= 4 × 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 × area of one quadrant
So we get
= ½ πr2
Substituting the values
= ½ × 3.14 × 8 × 8
= 100.48 cm2
We know that
Perimeter of shaded part = ½ of circumference + 4r
It can be written as
= ½ × 2πr + 4r
= πr + 4r
Taking r as common
= r (π + 4)
Substituting the values
= 8 (3.14 + 4)
= 8 × 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 = 2πr
Substituting the values
= 2 × 22/7 × 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 = 2πr
Substituting the values
= 2 × 22/7 × 42
= 264 cm
So the distance covered in 5 reductions = 264 × 5 = 1320 cm
Time = 1 second
We know that
Speed of wheel = 1320/1 × (60 × 60)/ (100 × 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 cm2, correct to the nearest cm2.
(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
2πr = 123.2
By further calculation
(2 × 22)/ 7 r = 1232/10
So we get
r = (1232 × 7)/ (10 × 2 × 22)
r = 19.6 cm
Therefore, radius of the circle is 19.6 cm
(ii) Here
Area of the circle = πr2
Substituting the values
= 22/7 × 19.6 × 19.6
= 1207.36
= 1207 cm2
(iii) We know that
If radius is doubled = 19.6 × 2 = 39.2 cm
So the area of circle = πr2
Substituting the values
= 22/7 × 39.2 × 39.2
= 4829.44 cm2
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 cm2. 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 cm2. 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 = π (R2 – r2)
Substituting the values
= 22/7 (212 – r2)
= 22/7 (441 – r2)
Area of the ring = 770 cm2
So we get
22/7 (441 – r2) = 770
By further calculation
441 – r2 = (770 × 7)/ 22 = 245
r2 = 441 – 245 = 196
r = √196 = 14
Hence, the radius of inner circle is 14 cm.
(b) It is given that
Area of ring = 346.5 cm2
Circumference of inner circle = 88 cm
We know that
Radius = (88 × 7)/ (2 × 22) = 14 cm
Consider R cm as the radius of outer circle
Area of ring = π (R2 – r2)
Substituting the values
= 22/7 (R2 – 142)
= 22/7 (R2 – 196) cm2
Area of ring = 346.5 cm2
By equating we get
22/7 (R2 – 196) = 346.5
By further calculation
R2 – 196 = (346.5 × 7)/ 22 = 110.25
R2 = 110.25 + 196 = 306.25
So we get
R = √306.25 = 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 m2.
Solution:
It is given that
Circumference of circular plot = 44 m
Radius of circular plot = (44 × 7)/ (22 × 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 = π (R2 – r2)
Substituting the values
= 22/7 (10.52 – 72)
We can write it as
= 22/7 (10.5 + 7) (10 – 7)
= 22/7 × 17.5 × 3.5
= 192.5 m2
Here
Rate of paving the road = Rs 50 per m2
Total cost = 192.5 × 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 …… (1)
We know that
Difference of their circumferences = 8 cm
2 πR – 2 πr = 8
Taking out the common terms
2 π (R – r) = 8
(2 × 22)/ 7 (R – r) = 8
By further calculation
R – r = (8 × 7)/ (2 × 22) = 14/11 ….. (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 × 2) = 91/22
From equation (1)
R + r = 7
Substituting the value of R
91/22 + r = 7
r = 7 – 91/22
Taking LCM
r = (154 – 91)/ 22 = 63/22
We know that
Circumference of first circle = 2 πr
Substituting the values
= 2 × 22/7 × 91/22
= 26 cm
Circumference of second circle = 2 π R
Substituting the values
= 2 × 22/7 × 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 = πr2
= π (2)2
= 4 π cm2
Radius of second circle = 3 cm
Area of first circle = πr2
= π (3)2
= 9 π cm2
Radius of second circle = 6 cm
Area of first circle = πr2
= π (6)2
= 36 π cm2
So the total area of the three circles = 4 π + 9 π + 36 π = 49 π cm2
Area of the given circle = 49 π cm2
We know that
Radius = √ (49 π/ π) = √49 = 7 cm
Circumference = 2 πr = 2 × 22/7 × 7 = 44 cm
15. A copper wire when bent in the form of a square encloses an area of 121 cm2. 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 cm2
So side = √121 = 11 cm
Perimeter = 4a = 4 × 11 = 44 cm
Circumference of circle = 44 cm
Radius of circle = (44 × 7)/ (2 × 22) = 7 cm
We know that
Area of the circle = πr2
Substituting the values
= 22/7 (7)2
So we get
= 22/7 × 7× 7
= 154 cm2
16. A copper wire when bent into an equilateral triangle has area 121 √3 cm2. 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 √3 cm2
Consider a as the side of triangle
Area = √3/4 a2
It can be written as
√3/4 a2 = 121√3
By further calculation
a2 = (121 × √3 × 4)/ √3
a2 = 484
So we get
a = √484 = 22 cm
Here
Length of the wire = 66 cm
Radius of the circle = 66/2π
By further calculation
= (66 × 7)/ (2 × 22)
= 21/2 cm
We know that
Area of the circle = πr2
By further calculation
= 22/7 × (21/2)2
= 22/7 × 21/2 × 21/2
So we get
= 693/2
= 346.5 cm2
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 π = 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 = πr2
Substituting the values
= 22/7 × 7/2 × 7/2
= 77/2 cm2
Here
Area of bigger circle = 77/2 × 16 = 616 cm2
Consider r as the radius
πr2 = 616
We can write it as
22/7 r2 = 616
By further calculation
r2 = (616 × 7)/ 22
r2 = 196 cm2
So we get
r = √196 = 14 cm
Circumference = 2πr
Substituting the values
= 2 × 22/7 × 14
= 88 cm
(b) It is given that
Radius of each circle = 3 cm
Diameter of each circle = 2 × 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 × b
Substituting the values
= 15 × 6
= 90 cm2
We know that
Area of 2 ½ circles = 5/2 πr2
Substituting the values
= 5/2 × 3.14 × 3 × 3
By further calculation
= 5 × 1.57 × 9
= 70.65 cm2
So the area of unshaded portion = 90 – 70.65
= 19.35 cm2
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 π = 22/7.
Solution:
It is given that
Side of square = 21 cm
So the area of square = side2 = 212 = 441 cm2
We know that
∠AOD + ∠COD + ∠AOB + ∠BOC = 441
Substituting the values
x + x + x + x = 441
4x = 441
So we get
x = 441/4 = 110.25 cm2
Based on the question
We should find the area of shaded portion in square ABCD which is ∠AOD and ∠BOC
∠AOD + ∠BOC = 110.25 + 110.25 = 220.5 cm2
Here
Area of two semicircle = πr2
Substituting the values
= 22/7 × 10.5 × 10.5
= 346.50 cm2
So the area of shaded portion = 220.5 + 346.5 = 567 cm2
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 π = 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 = a2 = 142 = 196 cm2
Area of two semicircles = 2 × ½ πr2
= πr2
Substituting the values
= 22/7 × 7 × 7
= 154 cm2
So the area of shaded portion = 196 – 154 = 42 cm2
(ii) Here
Length of arcs of two semicircles = 2πr
Substituting the values
= 2 × 22/7 × 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 = a2 = 142 = 196 cm2
Area of 4 circles = 4 × πr2
Substituting the values
= 4 × 22/7 × 7/2 × 7/2
= 154 cm2
So the area of shaded portion = 196 – 154 = 42 cm2
(ii) Here
Perimeter of 4 circles = 4 × 2 πr
Substituting the values
= 4 × 2 × 22/7 × 7/2
= 88 cm
So the perimeter of shaded portion = perimeter of 4 circles + perimeter of square
Substituting the values
= 88 + 4 × 14
So we get
= 88 + 56
= 144 cm
(c) We know that
Area of rectangle ACDE = ED × AE
Substituting the values
= 14 × 7
= 98 cm2
Here
Area of semicircle DEF = πr2/2
Substituting the values
= (22 × 7 × 7)/ (7 × 2)
= 77 cm2
So the area of shaded region = 77 + (98 – 2 × ¼ × 22/7 × 7 × 7)
= 77 + 21
= 98 cm2
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 △ABC = 35 cm2. 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 – Area of smaller circle
= ½ πR2 – ½ πr2
We can write it as
= ½ π (R2 – r2)
Substituting the values
= ½ × 22/7 (142 – 72)
By further calculation
= 11/7 (14 + 7) (14 – 7)
So we get
= 11/7 × 21 × 7
= 231 cm2
(ii) Here
Perimeter of shaded portion = circumference of larger semicircle + circumference of smaller semicircle + radius of larger semicircle
= πR + πr + R
Substituting the values
= 22/7 × 14 + 22/7 × 7 + 14
By further calculation
= 44 + 22 + 14
= 80 cm
(b) We know that
Area of △ABC formed in a semicircle = 3.5 cm
Altitude CD = 5 cm
So the base AB = (area × 2)/ altitude
Substituting the values
= (35 × 2)/ 5
= 14 cm
Here
Diameter of semicircle = 14 cm
Radius of semicircle (R) = 14/2 = 7 cm
So the area of semicircle = ½ πR2
Substituting the values
= ½ × 22/7 × 7 × 7
= 77 cm2
Area of shaded portion = Area of semicircle – Area of triangle
= 77 – 35
= 42 cm2
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 π = 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 – △AOB
Radius of the quadrant = 10 m
Here
Area of quadrant = ¼ πr2
Substituting the values
= ¼ × 3.14 × 10 × 10
By further calculation
= (3.14 × 100)/ 4
= 314/4
= 78.5 m2
We know that
Area of △AOB = ½ × AO × OB
Substituting the values
= ½ × 10 × 10
= 50 m2
So the area of shaded portion = 78.5 – 50 = 28.5 m2
(b) In the figure
Radius of quadrant = 3.5 cm
(i) We know that
Area of quadrant = ¼ πr2
Substituting the values
= ¼ × 22/7 × 3.5 × 3.5
= 9.625 cm2
(ii) Here
Area of △AOD = ½ × AO × OD
Substituting the values
= ½ × 3.5 × 2
= 3.5 cm2
So the area of shaded portion = Area of quadrant – Area of △AOD
Substituting the values
= 9.625 – 3.6
= 6.125 cm2
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 π = 22/7)
Solution:
It is given that
Length of rectangle = 30 cm
Width of rectangle = 21 cm
We know that
Area of rectangle = l × b
= 30 × 21
= 630 cm2
So the radius of the biggest circle = 21/2 cm
Here
Area of the circle = πr2
Substituting the values
= 22/7 × 21/2 × 21/2
So we get
= 693/2
= 346.5 cm2
So the area of remaining part = 630 – 346.5 = 283.5 cm2
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 × 2 = 5 cm
Here
So we get
= 3 cm
We know that
Area of rectangle = AB × BC
Substituting the values
= 4 × 3
= 12 cm2
24. (a) In the figure (i) given below, calculate the area of the shaded region correct to two decimal places. (Take π = 3.142)
(b) In the figure (ii) given below, ABC is an isosceles right angled triangle with ∠ABC = 900. A semicircle is drawn with AC as diameter. If AB = BC = 7 cm, find the area of the shaded region. Take π = 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 = πr2
Substituting the values
= 3.142 × (6.5)2
By further calculation
= 3.142 × 42.25
= 132.75 cm2
Area of rectangle = l × b
Substituting the values
= 12 × 5
= 60 cm2
So the area of the shaded portion = 132.75 – 60 = 72.75 cm2
(b) We know that
Area of △ABC = ½ × AB × BC
Substituting the values
= ½ × 7 × 7
= 49/2 cm2
Here
AC2 = AB2 + BC2
Substituting
= 49 + 49
So we get
AC = 7√2
Radius of semi-circle = 7√2/ 2 cm
Area of semi-circle = π/2 × (7√2/ 2)2
By further calculation
= ½ × 22/7 × 98/4
= 77/2 cm2
Area of the shaded region = Area of the semi-circle – Area of △ABC
Substituting the values
= 77/2 – 49/2
= 28/2
= 14 cm2
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 π
Substituting the values
= (660 × 7)/ (2 × 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 = √2 a
a = AC/√2
Substituting the values
a = (105 × 2)/ √2
Multiply and divide by √2
a = (105 × 2 × √2)/ (√2 × √2)
By further calculation
a = (105 × 2 × √2)/ 2
a = 105 √2 m
We know that
Area of the square = a2
It can be written as
= (105√2)2
= 105√2 × 105√2
= 22050 m2
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 = 4a2
Diagonal of AC = √2AB
(i) We know that
Radius of the circumcircle = ½ AC
It can be written as
= ½ (√2 × AB)
Substituting the values
= √2/2 × 2a
= √2a
Circumference = 2 πr = 2 × π × √2a = 2√2 πa
Radius of incircle = AB = ½ × 2a = a
Circumference = 2 πr = 2 πa
Here
Ratio between the circumference incircle and circumcircle = 2 πa: 2√2 πa
= 1: √2
(ii) We know that
Area of incircle = πr2 = πa2
Area of circumcircle = πR2
= π (√2a)2
= π2a2
= 2 πa2
So the ratio = πa2: 2 πa2 = 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 m2.
(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 × b
= 200 × 70
= 14000 m2
Radius of each semi-circular part on either side of rectangle = 70/2 = 35 m
Area of both semi-circular parts = 2 × ½ πr2
Substituting the values
= 22/7 × 35 × 35
= 3850 m2
So the total area of grassed enclosure = 1400 + 3850 = 17850 m2
(ii) We know that
Width of track around the enclosure = 7 m
Outer length = 200 m
So the width = 70 + 7 × 2
= 70 + 14
= 84 m
Outer radius = 84/2 = 42 m
Here
Circumference of both semi-circular part = 2πr
Substituting the values
= 2 × 22/7 × 42
= 264 m
Outer perimeter = 264 + 200 × 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 – 180 = 132 m
Here
Radius of each semi-circle = 132/2π
So we get
= 66/3.14
= 21.02 m
Diameter of each semi-circle = 66/π × 2
So we get
= 132/ π
= 132/3.14
Multiply and divide by 100
= (132 × 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 × ½ × πR2
= πR2
Substituting the values
= 3.14 × (23.02)2
= 3.14 × 23.02 × 23.02
= 1663.95 m2
Area of rectangle = 90 × 46.04
= 4143.6 m2
Total area = 1663.95 + 4143.60 = 5807.55 m2
Area of two inner circles = 2 × ½ πr2
Substituting the values
= 3.14 × 21.02 × 21.02
= 1387.38 m2
Area of inner rectangle = 90 × 42.04
= 3783.6 m2
Total inner area = 3783.60 + 1387.38
= 5170.98 m2
Here
Area of path = 5807.55 – 5170.98
= 636.57 m2
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 – AB = 8 – 3 = 5 cm
We know that
Area of big circle of radius AC = πR2
Substituting the values
= 22/7 × 8 × 8
By further calculation
= 64 × 22/7 cm2
Area of smaller circle = πr2
Substituting the values
= 22/7 × 5 × 5
By further calculation
= (25 × 22)/ 7 cm2
Here
Area of shaded portion = (64 × 22)/ 7 – (25 × 22)/ 7
Taking out the common terms
= 22/7 (64 – 25)
By further calculation
= 22/7 × 39
= 122.57 cm2
(b) We know that
Radius of each quadrant = 7 cm
Here
Area of shaded region = Area of square – 4 area of the quadrant
It can be written as
= (side)2 – 4 × ¼ πr2
Substituting the values
= 142 – 22/7 × 7 × 7
By further calculation
= 196 – 154
= 42 cm2
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. (π = 3.14)
Solution:
(a) It is given that
Side of square lawn ABCD (a) = 56 cm
Area = a2 = 52 = 3136 cm2
We know that
Length of the diagonal of the square = √2 a
= √2 × 56 cm
Radius of each quadrant = (√2 × 56)/ 2
= 28 √2 cm
So the area of each segment = ¼ πr2 – area of △OBC
Substituting the values
= ¼ × 22/7 × 28√2 × 28√2 – ½ × 28√2 × 28√2
Taking out the common terms
= 28√2 × 28√2 (1/4 × 22/7 – 1/2)
By further calculation
= 784 × 2 (11/14 – ½)
So we get
= 784 × 2 × 4/14
= 448 cm2
Here
Area of two segments = 448 × 2 = 896 cm2
So the total area of the lawn and beds = 3136 + 896 = 4032 cm2
(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 = √2 a = √2 ×20 cm
Radius of quadrant = OB = 20√2cm
We know that
Area of quadrant = ¼ πr2
Substituting the values
= ¼ × 3.14 × (20√2)2
By further calculation
= ¼ × 3.14 × 800
So we get
= 314 × 2
= 628 cm2
Area of square = a2 = 202 = 400 cm2
So the area of shaded portion = 628 – 400 = 228 cm2
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 ∠BOD = 900. Find the area of the shaded region. (Use π = 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 × b
Substituting the values
= 14 × 7
= 98 cm2
Radius of each outer semicircles = 7/2 cm
So the area = 2 × ½ πr2
Substituting the values
= 22/7 × 7/2 × 7/2
= 77/2
= 38.5 cm2
Here
Area of semicircle drawn on CD as diameter = ½ πR2
Substituting the values
= ½ × 22/7 × 72
By further calculation
= 11/7 × 7 × 7
= 77 cm2
So the area of shaded region = 98 + 38.5 – 77
= 59.5 cm2
(b) In the figure
AC = 24 cm
AB = 7 cm
∠BOD = 900
In △ABC
Using Pythagoras theorem
BC2 = AC2 + AB2
Substituting the values
BC2 = 242 + 72
By further calculation
BC = √ (576 + 49) = √625 = 25 cm
So the radius of circle = 25/2 cm
We know that
Area of △ABC = ½ × AB × AC
Substituting the values
= ½ × 7 × 24
= 84 cm2
Area of quadrant COD = ¼ πr2
Substituting the values
= ¼ × 3.14 × 25/2 × 25/2
By further calculation
= 1962.5/16
= 122.66 cm2
Area of circle = πr2
Substituting the values
= 3.14 × 25/2 × 25/2
By further calculation
= 1962.5/4
= 490.63 cm2
Area of shaded portion = Area of circle – (Area of △ABC + Area of quadrilateral COD)
Substituting the values
= 490.63 – (84 + 122.66)
By further calculation
= 490.63 – 206.66
= 283.97 cm2
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 π 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 = a2
Substituting the values
= 14 × 14
= 196 cm2
Area of 4 sectors of 900 each = 4 × π × r2
Substituting the values
= 4 × 22/7 × 7 × 7 × ¼
= 154 cm
Area of each sector of 2700 angle = 3/4 πr2
Substituting the values
= ¾ × 22/7 × 7 × 7
= 231/2
= 115.5 cm2
Here
Area of 4 sectors = 115.5 × 4 = 462 cm2
So the area of shaded portion = area of square + area of 4 bigger sector – area of 4 smaller sector
Substituting the values
= 196 + 462 – 154
= 658 – 154
= 504 cm2
(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
= πR + πr + πr
= 3.14 (R + 2r)
Substituting the values
= 3.14 (5 + 2 × 5/2)
By further calculation
= 3.14 × 10
= 31.4 cm
(ii) Here
Area of shaded region = area of bigger semi-circle + area of one smaller semi-circle – area of other smaller semi-circle
We know that
Area of bigger semi-circle = ½ πR2
Substituting the values
= 3.14/2 × 5 × 5
= 1.57 × 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 π = 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 π = 3.14
Solution:
(a) It is given that
Radius of bigger circle = 5 cm
Radius of small circle (r1) = 3 cm
Radius of smaller circle (r2) = 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
= πR + π r1 + π r2
= π (R + r1 + r2)
Substituting the values
= π (5 + 3 + 2)
= 3.14 × 10
= 31.4 cm2
(ii) We know that
Area of shaded region = area of bigger semi-circle + area of smaller semi-circle – area of small semicircle
It can be written as
= ½ πR2 + ½ πr22 – ½ πr12
= ½ π (R2 + r22 – r12)
Substituting the values
= ½ π (5 2 + 22 – 32)
By further calculation
= ½ π (25 + 4 – 9)
= ½ π × 20
So we get
= 10 × 3.14
= 31.4 cm2
(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 × ½ side of square
It can be written as
= 4 × ¼ (2 πr) + 2 πr + 4 × 2
By further calculation
= 2 πr + 2 πr + 8
= 4 πr + 8
So we get
= 4 × 3.14 × 1 + 8
= 12.56 + 8
= 20.56 cm
(ii) Area of shaded region = area of square – area of 4 quadrants – area of circle
It can be written as
= side2 – 4 × ¼ πr2 – πr2
= 42 – πr2– πr2
By further calculation
= 16 – 2πr2
So we get
= 16 – 2 × 3.14 × 12
= 16 – 6.28
= 9.72 cm2
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 π = 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 △CEF 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 – area of semicircle DGE – area of quarter BFEC
Substituting the values
= 14 × 7 – ½ × π (7/2)2 – ¼ π × 72
By further calculation
= 14 × 7 – ½ × 22/7 × 7/2 × 7/2 – ¼ × 22/7 × 7 × 7
So we get
= 98 – 77/4 – 154/4
= 98 – 19.25 – 38.5
= 98 – 57.75
= 40.25 cm2
(b) In the given figure
ABCD is a square of side = radius of quadrant = 42 cm
△CEF 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
= ¼ πr2 + ½ EC × FC
Substituting the values
= ¼ × 22/7 × 42 × 42 + ½ × 6 × 6
By further calculation
= 1386 + 18
= 1404 cm2
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 ∠BOD = 900, 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 × circumference of the smaller semi-circles
It can be written as
= πR + πr1 + 2 × πr2
= π (R + r1) + 2πr2
Substituting the values
= 22/7 (7 + 3.5) + 2 × 22/7 × 3.5/2
By further calculation
= 22/7 × 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 – 2 × area of smaller semicircles
It can be written as
= ½ π (7)2 + ½ π (3.5)2 – 2 × ½ π (1.75)2
By further calculation
= ½ × 22/7 × 7 × 7 + ½ × 22/7 × 3.5 × 3.5 – 22/7 × 1.75 × 1.75
So we get
= 77 + 19.25 – 9.625
= 86.625 cm2
(b) Here
ABCD is a trapezium in which AB || DC and ∠C = 900
AB = BC = 3.5 cm and DE = 2 cm
Radius of quadrant = 3.5 cm
We know that
Area of trapezium = ½ (AB + DC) × BC
Substituting the values
= ½ (3.5 + 3.5 + 2) × 3.5
By further calculation
= ½ (9 × 3.5)
= 4.5 × 3.5
= 15.75 cm2
So the area of quadrant = ¼ πr2
Substituting the values
= ¼ × 22/7 × 3.5 × 3.5
= 9.625 cm2
Area of shaded portions = 15.75 – 9.625 = 6.125 cm2
35. (a) In the figure (i) given below, ABC is a right angled triangle, ∠B = 900, 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 π = 3.142 and √3 = 1.732)
Solution:
(a) In right △ABC
∠B = 900
Using Pythagoras theorem
AC2 = AB2 + BC2
Substituting the values
= 282 + 212
= 784 + 441
= 1225
So we get
AC = √1225 = 35 cm
Here
Radius of semi-circle (R) = 35/2
Radius of quadrant (r) = 21 cm
So the area of shaded region = area of △ABC + area of semi-circle – area of quadrant
= ½ × 28 × 21 + ½ πR2 – ¼ r2
Substituting the values
= 294 + ½ × 22/7 × 35/2 × 35/2 – ¼ × 22/7 × 21 × 21
By further calculation
= 294 + 1925/4 – 693/2
= 294 + 481.25 – 346.5
So we get
= 775.25 – 346.50
= 428.75 cm2
(b) We know that
△ABC 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 △ABC = √3/4a2
Substituting the values
= √3/4 × 8 × 8
= √3/4 × 64
So we get
= 16√3cm2
Substituting the value of √3
= 16 × 1.732
= 27.712 cm2
So the area of 3 equal sectors of 600 whose radius is 4 cm = 3 × πr2 × 60/360
By further calculation
= 3 × 3.142 × 4 × 4 × 1/6
So we get
= 3.142 × 8
= 25.136 cm2
Area of shaded region = 27.712 – 25.136 = 2.576 = 2.58 cm2
36. A circle is inscribed in a regular hexagon of side 2√3 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 2√3 cm and a circle is inscribed in it with O as the centre
Radius of inscribed circle = √3/2 × side of regular hexagon
Substituting the values
= √3/2 × 2√3
= 3 cm
(i) We know that
Circumference of the circle = 2πr
Substituting the values
= 2π × 3
By further calculation
= (6 × 22)/ 7
= 132/7 cm
(ii) Area of the circle = πr2
Substituting the values
= π × 3 × 3
By further calculation
= (9 × 22)/ 7
= 198/7 cm2
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 π = 3.14.
Solution:
It is given that
Radius of the circle = 10 cm
Angle at the centre subtended by a chord AB = 900
We know that
Area of sector OACB = πr2 × 90/360
Substituting the values
= 3.14 × 10 × 10 × 90/360
So we get
= 314 × ¼
= 78.5 cm2
Here
Area of △OAB = ½ × 10 × 10 = 50 cm2
Area of minor segment = Area of sector △ACB – Area of △OAB
Substituting the values
= 78.5 – 50
= 28.5 cm2
Area of circle = πr2
Substituting the values
= 3.14 × 10 × 10
= 314 cm2
Area of major segment = area of circle – area of minor segment
Substituting the values
= 314 – 28.5
= 285.5 cm2
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 = 6a2
Substituting the values
= 6 (7)2
= 6 × 7 × 7
= 294 cm2
Volume of cube = a3
Substituting the values
= 73
= 7 × 7 × 7
= 343 cm3
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 × 4 + 4 × 3 + 5 × 3)
By further calculation
= 2 (20 + 12 + 15)
= 2 × 47
= 94 sq. m
Volume of rectangular solid = l × b × h
Substituting the values
= 5 × 4 × 3
= 60 m3
We know that
It can be written as
= 5 √2 m
Substituting the value of √2
= 5 × 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 cm3, 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 cm3
Consider height of rectangular solid = h cm
Here
Volume = l × b × h
Substituting the values
7000 = 25 × 20 × h
By further calculation
25 × 20 × h = 7000
h = 7000/ (25 × 20)
So we get
h = 700/ (25 × 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 m2 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 × b
Substituting the values
= 10 × 6
= 60 m2
Here
One student needs 1.5 m2 floor area
So the number of students = 60/1.5
Multiply and divide by 10
= (60 × 10)/ 15
= 600/15
= 40 students
Volume of class room = l × b × h
Substituting the values
= 10 × 6 × 4
= 240 m3
So the cubic metres of air for each student = volume of classroom/ number of students
Substituting the values
= 240/40
= 6 m3
5. (a) The volume of a cuboid is 1440 cm3. 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 cm3
Height of cuboid = 10 cm
We know that
Volume of cuboid = area of square × height
Substituting the values
1440 = area of square × 10
By further calculation
Area of square = 1440/10 = 144 cm2
Here
Side × side = 144
So we get
Side = √144 = 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 × side
We can write it as
20 = 4 × side
By further calculation
Side = 20/4 = 5 cm
Here
Area of one face = side × side
Substituting the values
= 5 × 5
= 25 cm2
Area of 6 faces = 6 × 25 = 150 cm2
So the volume of cube = side × side × side
Substituting the values
= 5 × 5 × 5
= 125 cm3
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 × 40 + 40 × 20 + 20 × 80)
By further calculation
= 2 (320 + 800 + 1600)
= 2 × 5600
= 11200 cm2
So the area of square sheet = side2
= 402
= 1600 cm2
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 cm3 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 cm3
Height of cuboid = 12 cm
We know that
Volume of cuboid = Area of rectangle × height
Substituting the values
3600 = area of rectangle × 12
By further calculation
Area of rectangle = 3600/ 12
Area of rectangle = 300 cm2 ….. (1)
Here
Ratio of length and breadth of rectangle = 4: 3
Consider
Length of rectangle = 4x
Breadth of rectangle = 3x
Area of rectangle = length × breadth
Substituting the values
Area of rectangle = 4x × 3x
So we get
Area of rectangle = 12x2 cm2 ….. (2)
Using equations (1) and (2)
12x2 = 300
x2 = 300/12
So we get
x2 = 25
x = √25 = 5
Here
Length of rectangle = 4 × 5 = 20 cm
Breadth of rectangle = 3 × 5 = 15 cm
Perimeter of the cross section = 2 (l + b)
Substituting the values
= 2 (20 + 15)
= 2 × 35
= 70 cm
8. The volume of a cube is 729 cm3. Find its surface area and the length of a diagonal.
Solution:
It is given that
Volume of a cube = 729 cm3
We can write it as
side × side × 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 × (9)2
= 6 × 9 × 9
= 486 cm2
So the length of a diagonal = √3 × side
Substituting the value
= √3 × 9
= 1.73 × 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
172 = 122 + 82 + h2
By further calculation
289 = 144 + 64 + h2
289 = 208 + h2
So we get
h2 + 208 = 289
h2 = 289 – 208 = 81
h = √81 = 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 × b × h
Substituting the values
= 90 × 80 × 70
= 504000 cm3
Here
Required area of tin foil = 2 (lb + bh + lh)
Substituting the values
= 2 (90 × 80 + 80 × 70 + 90 × 70)
By further calculation
= 2 (7200 + 5600 + 6300)
So we get
= 2 × 19100
= 38200 cm2
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 × 16 cm × 14 cm
Volume of cubes = 4 cm × 4 cm × 4 cm
We know that
Number of cubes put into the box = volume of box/ volume of cubes
Substituting the values
= (20 × 16 × 24)/ (4 × 4 × 4)
= 5 × 4 × 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 m3. 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 m3 wood
Cost of Rs 1 for 1/250 m3 wood
Cost of Rs 182.25 for 182.25/250 m3 wood
Here
Quantity of wood = 182.25/250 m3
Multiply and divide by 100
= 18225/ (250 × 100)
So we get
= 18225/ (25 × 1000)
= 729/ 1000
= 0.729 m3
We know that
Volume of given block = 0.729 m3
Consider the length of one edge of block = x m
So we get
x3 = 0.729 m3
Now taking cube root on both sides
So we get
= (3 × 3)/ (2 × 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 × 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 = edge3
Substituting the values
= 113
= 11 × 11 × 11
= 1331 cm3
We know that
The dimensions of the base of the vessel are 15 cm × 12 cm
Consider the rise in the water level = h cm
So the volume of cube = volume of vessel
Substituting the values
1331 = 15 × 12 × h
By further calculation
h = 1331/ (15 × 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 cm3 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 cm3 of water overflows
We know that
Volume of cube = Volume of water displaced
Substituting the values
= 6 × 6 × 1 + 2
= 36 + 2
= 38 cm3
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 × 12 + 12 × 12 + 12 × 24)
By further calculation
= 2 (288 + 144 + 288)
So we get
= 2 × 720
= 1440 cm2
(b) We know that
Side of a cube = 12 cm
Here
Volume = side3 = 123 = 1728 cm3
If cut into 8 equal cubes
Volume of each cube = 1728/8 = 216 cm3
Surface area of original cube = 6 × side2
Substituting the values
= 6 × 122
= 6 × 144
= 864 cm2
Surface area of one smaller cube = 6 × 62
So we get
= 6 × 36
= 216 cm2
Surface area of 8 cube = 216 × 8
= 1728 cm2
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 × 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 × 6 cm × 6 cm = 216 cm3
Dimensions of cuboid are
Length = 9 cm
Breadth = 8 cm
h cm is the height
We know that
Volume of cuboid = l × b × h
Substituting the values
= 9 × 8 × h
= 72 h cm3
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 m2. 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 m2
We can write it as
l × b = 4800
We know that
Depth of level = 1 cm
h = 1 cm = 1/100 m
Here
Volume of gravel = l × b × h
Substituting the values
= 4800 × 1/100
= 48 m3
Cost of gravel = Rs 260 per cubic metre
So the total cost = 260 × 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 × 4 × 3 = 72 m3
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 × 6 × h = 84h m3
Volume of part II = 24 × 18 × h = 432h m3
Total volume of part I and II = [84h + 432h] = 516h m3
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 × 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 × b = 24 × 20 = 480 m2
We know that
Side of cubical pit = 4 m
Volume of each pit = 43 = 64 m3
Here
Volume of 4 pits at the corners = 4 × 64 = 256 m3
Area of surface of 4 pits = 4a2
= 4 × 42
= 64 m2
So the area of remaining plot = 480 – 64 = 416 m2
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 m3 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 × 1/100
= 1/10 × ¼
So we get
= 1/40
= 0.025 m
So the external dimensions of wooden box are
(2 + 2 × 0.025), (1.2 + 2 × 0.025), (0.75 + 2 × 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 – Internal volume of box
Substituting the values
= 2.05 × 1.25 × 0.80 – 2 × 1.2 × 0.75
By further calculation
= 2.05 – 1.80
= 0.25 m3
Cost = Rs 5400 for 1 m3
So the total cost = 5400 × 0.25
Multiply and divide by 100
= 5400 × 25/100
= 54 × 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 – Inner volume
Substituting the values
= 1.1 × 1.1 × 1.05 – 1 × 1 × 1
By further calculation
= 1.205 – 1
= 0.2705 m3
Cost of 1 m3 = Rs 9600
So the cost of 0.2705 m3 = 9600 × 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 × b × h
Substituting the values
= x × x × 1/10
= x2/ 10 cm3 ….. (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 × 1/8.4 = 562.5 cc
Volume of plate = 562.5 cc = 562.5 cm3 …. (2)
Using both the equations
x2/10 = 562.5
By cross multiplication
x2 = 562.5 × 10 = 5625
Here
x = √5625 = 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 x3, 83 and 103 i.e. x3, 512 cm3 and 1000 cm3
We know that
Edges of new cube formed = 12 cm
Volume of new cube = 123 = 1728 cm3
Based on the question
x3 + 512 + 1000 = 1728
By further calculation
x3 + 1512 = 1728
x3 = 216
It can be written as
x3 = 6 × 6 × 6
So we get
x = 6 cm
25. The area of cross-section of a pipe is 3.5 cm2 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 cm2
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 × length
Substituting the values
= 3.5 × 40
= 35 × 4
= 140 cm3
So the volume of water flowing in 1 minute i.e. 60 sec = 140 × 60 cm3
Here
1 litre = 1000 cm3
So we get
Volume = (140 × 60)/ 100
= (14 × 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 × 4 × 2 + 4 × 2 × 6
= 32 + 48
= 80 cm3
(b) We know that
Figure (ii) is a trapezium with parallel sides 2 m and 3.5 m
(i) Area of cross section = ½ (sum of parallel sides) × height
Substituting the values
= ½ (2 + 3.5) × 6
By further calculation
= ½ × 5.5 × 6
So we get
= 5.5 × 3
= 16.5 m2
(ii) Volume of concrete in the wall = Area of cross section × length
Substituting the values
= 16.5 × 400
= 165 × 40
= 6600 m3
(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 = ½ (sum of parallel sides) × height
Substituting the values
= ½ (2 + 3) × 10
By further calculation
= ½ × 5 × 10
= 5 × 5
= 25 m2
So the volume of water it will hold when full = area of cross section × height
= 25 × 40
= 1000 m3
27. A swimming pool is 50 metres long and 15 metres wide. Its shallow and deep ends are 1 ½ metres and 14 ½ 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 ½ m and 5 ½ m deep
We know that
Area of cross section of swimming pool = ½ (sum of parallel sides) × width
Substituting the values
= ½ (1 ½ + 4 ½) × 15
By further calculation
= ½ (3/2 + 9/2) × 15
So we get
= ½ [(3 + 9)/ 2] × 15
= ½ × 12/2 × 15
= ½ × 6 × 15
= 3 × 15
= 45 m2
Here
Amount of water required to fill pool = Area of cross section × length
= 45 × 50
= 2250 m3
Chapter Test
Take π = 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 cm2. 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 – BO = 14 – 5 = 9 cm
Here
Area of right △AOC = ½ × base × altitude
Substituting the values
= ½ × 9 × 12
= 54 cm2
(b) Consider the side of original square = x cm
So the length of given square = (x + 3) cm
We know that
Area = side × side
Substituting the values
121 = (x + 3) (x + 3)
It can be written as
112 = (x + 3)2
By further calculation
11 = x + 3
x = 11 – 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 – Area of both triangles
Substituting the values
= (9 × 9) – (½ × 5 × 6 ) × 2
By further calculation
= 81 – (15 × 2)
So we get
= 81 – 30
= 51 cm2
(b) In △ABD
Using Pythagoras theorem
BD2 = AB2 + AD2
Substituting the values
BD2 = 62 + 82
BD2 = 36 + 64
So we get
BD2 = 100
BD = 10 cm
In △BCD
Using Pythagoras theorem
BC2 = BD2 + CD2
Substituting the values
262 = 102 + CD2
676 = 100 + CD2
By further calculation
CD2 = 676 – 100 = 576
CD = √576
CD = 24 cm
Here
Area of the given figure = Area of △ABD + Area of △BCD
We can write it as
Area of the given figure = ½ × base × height + ½ × base × height
Area of the given figure = ½ × AB × AD + ½ × CD × BD
Substituting the values
Area of the given figure = ½ × 6 × 8 + ½ × 24 × 10
So we get
Area of the given figure = 3 × 8 + 12 × 10
= 24 + 120
= 144 cm2
(c) We know that
Area of the figure (iii) = Area of ABCD – (Area of 1st part + Area of 2nd part + Area of 3rd part)
It can be written as
= (AB × BC) – [(1/2 × base × height) + (1/2 × base × height) + ½ (sum of parallel side × height)]
Substituting the values
= (12 × 12) – [1/2 × 5 × 5 + ½ × 5 × 7 + ½ (7 + 3) × 12]
By further calculation
= 144 – [25/2 + 35/2 + 10 × 6]
= 144 – (60/2 + 60)
= 144 – (30 + 60)
So we get
= 144 – 90
= 54 m2
Therefore, the required area of given figure = 54 m2.
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
△AMN, △NCB and △MNC are equilateral triangles having 1 cm side each
Area of △GHF
Here
= ¾ × 3.316
= 3 × 0.829
= 2.487
= 2.48 cm2
Area of rectangle II (MCFH) = l × b
Substituting the values
= 6.5 × 1
= 6.5 cm2
Area of △ III + IV = 2 × ½ × 6 × 1.5 = 9 cm2
Area of three equilateral triangles formed trapezium III = 3 × √3/4 × 12
= ¾ × 1.732
= 3 × 0.433
= 1.299
= 1.3 cm2
So we get
Total area = 2.48 + 6.50 + 9 + 1.30
= 19.28
= 19.3 cm2
4. If the area of a circle is 78.5 cm2, find its circumference. (Take π = 3.14)
Solution:
It is given that
Area of a circle = 78.5 cm2
Consider r as the radius
r2 = Area/π
Substituting the values
r2 = 78.50/3.14
r2 = 25 = 52
So we get
r = 5 cm
Here
Circumference = 2 πr
Substituting the values
= 2 × 3.14 × 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 cm2, calculate the original area of the cardboard.
Solution:
It is given that
Area of circle cut out from the square board = 154 cm2
Consider r as the radius
πr2 = 154
22/7 r2 = 154
By further calculation
r2 = (154 × 7)/ 22 = 49 = 72
r = 7 cm
We know that
Side of square = 7 × 2 = 14 cm
So the area of the original cardboard = a2
= 142
= 196 cm2
6. (a) From a sheet of paper of dimensions 2 m × 1.5 m, how many circles of radius 5 cm can be cut? Also find the area of the paper wasted. Take π = 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 × b
Substituting the values
= 200 × 150
= 30000 cm2
We know that
Radius of circle = 5 cm
Number of circles in lengthwise = 200/ (5 × 2) = 20
Number of circles in widthwise = 150/10 = 15
So the number of circles = 20 × 15 = 300
Here
Area of one circle = πr2
Substituting the values
= 3.14 × 5 × 5 cm2
Area of 300 circles = 300 × 314/100 × 25 = 23550 cm2
So the area of remaining portion = area of square – area of 300 circles
= 30000 – 23550
= 6450 cm2
(b) It is given that
Diameter of semicircular protractor = 14 cm
We know that
Perimeter = ½ πd + d
Substituting the values
= ½ × 22/7 × 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 ₹ 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 πr = 88
Substituting the values
2 × 22/7 r = 88
By further calculation
r = (88 × 7)/ (2 × 22) = 14 m
Here
Outer radius (R) = 14 + 3.5 = 17.5 m
Area of the path = 22/7 × (17.5 + 14) (17.5 – 14)
We know that
π (R2 – r2) = 22/7 [(17.5)2 – (14)2]
By further calculation
= 22/7 (17.5 + 14) (17.5 – 14)
= 22/7 × 31.5 × 3.5
= 246.5 m2
It is given that
Rate of paving the road = ₹ 60 per m2
So the total cost = 60 × 346.5 = ₹ 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 × 3.5
= 140 + 7
= 147 m
Width = 42 + 2 × 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 × 3.5) + 2 × ½ π (R2 – r2)
Substituting the values
= 2 (490) + 22/7 [(24.5)2 – (21)2]
By further calculation
= 980 + 22/7 (24.5 + 21) (24.5 – 21)
= 980 + 22/7 × 45.5 × 3.5
So we get
= 980 + 500.5
= 1480.5 m2
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 π = 3.142)
Hint. Angle in a semicircle is a right angle.
Solution:
We know that
In a semicircle ∠ACB = 900
△ ABC is a right-angled triangle
Using Pythagoras theorem
AB2 = AC2 + BC2
Substituting the values
= 122 + 162
= 144 + 256
= 400
AB2 = (20)2
AB = 20 cm
Radius of semicircle = 20/2 = 10 cm
(i) We know that
Area of shaded portion = Area of semicircle – Area of △ ABC
It can be written as
= ½ πr2 – (AC × BC)/ 2
Substituting the values
= ½ × 3.142 (10)2 – (12 × 16)/2
By further calculation
= 314.2/2 – 96
= 157.1 – 96
= 61.1 cm2
(ii) Here
Perimeter of shaded portion = circumference of semicircle + AC + BC
It can be written as
= πr + 12 + 16
= 3.142 × 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 π = 3.14).
Solution:
(a) It is given that
Radius of quadrant = 3.5 cm
Perimeter = 2r + ¼ × 2 πr
= 2r + ½ × πr
Substituting the values
= 2 × 3.5 + ½ × 22/7 × 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 = √OB2 + AB2
Substituting the values
OA = √(3)2 + (1)2
= √9 + 1
= √10
So the radius of the circle = √10 cm
We know that
Area of the circle = πr2
Substituting the values
= 3.14 × (√10)2
= 3.14 × 10
= 31.4 cm2
Area of 5 square of side 2 cm each = 22 × 5
= 4 × 5
= 20 cm2
So the area of shaded portion = 31.4 – 20 = 11.4 cm2
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 cm2, 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 – 3 = 4 cm
BY = 7 – 4 = 3 cm
We know that
AB2 = OA2 + OB2
Substituting the values
= 32 + 42
= 9 + 16
= 25
So we get
AB = √25 = 5 cm
(i) Area of shaded portion = ¼ πr2 – ½ OA × OB
Substituting the values
= ¼ × 22/7 × 72 – ½ × 3 × 4
= ¼ × 22/7 × 49 – 6
By further calculation
= 77/2 – 6
= 65/2
= 32.5 cm2
(ii) Perimeter of shaded portion = ¼ × 2 πr + AX + BY + AB
Substituting the values
= ½ × 22/7 × 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 = a2 – 4 × [¼ π(a/2)2]
We can write it as
= a2 – 4 × ¼ π a2/4
Substituting the values
= a2 – 22/7 × a2/4
So we get
= a2 – 11a2/14
= 3a2/14
Here
Area of shaded portion = 21 3/7 = 150/7 cm2
By equating both we get
3a2/14 = 150/7
On further calculation
a2 = 150/7 × 14/3
a2 = 100 = 102
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
AC2 = AB2 + BC2 ……(i)
Area of semicircle on AC as diameter = ½ π (AC/2)2
So we get
= ½ π × AC2/4
= πAC2/8
Area of semicircle on AB as diameter = ½ π (AB/2)2
So we get
= ½ π × AB2/4
= πAB2/8
Area of semicircle on BC as diameter = ½ π (BC/2)2
So we get
= ½ π × BC2/4
= πBC2/8
We know that
πAB2/8 + πBC2/8 = π/8 (AB2 + BC2)
From equation (i)
= π/8 (AC2)
= πAC2/8
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 = πr2 × 15/60
By further calculation
= 22/7 × 14 × 14 × ¼
= 154 cm2
14. Find the radius of a circle if a 900 arc has a length of 3.5 π 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 π cm
Angle at the centre = 900
We know that
Radius of the arc = 3.5 π/2 π × 360/90
So we get
= (3.5 × 4)/2
= 7 cm
Area of the sector = πr2 × 900/3600
By further calculation
= 22/7 × 7 × 7 × ¼
= 77/2
= 38.5 cm2
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 a2
Substituting the value
= 6 × (28)2
= 6 × 28 × 28
= 4704 cm2
Diameter of each circle = 28 cm
So the radius = 28/2 = 14 cm
We know that
Area of each circle = πr2
Substituting the values
= 22/7 × 14 × 14
= 616 cm2
Area of such 6 circles drawn on 6 faces of cube = 616 × 6 = 3696 cm2
Here
Area of remaining portion of the cube = 4704 – 3696 = 1008 cm2
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 × 28 cm × 25 cm = (40 × 28 × 25) cm3
Here 1000 cm3 = 1 litre
So we get
= (40 × 28 × 25)/ 1000 litre
Average of car = 13.5 km per litre
We know that
Distance travelled by car = (40 × 28 × 25)/ 1000 × 13.5
Multiply and divide by 10
= (40 × 25) × 28/ 1000 × 135/10
So we get
= (1 × 28)/ 1 × 135/10
= (14 × 135)/5
= 14 × 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 × 96
= 25 × 48
= 1200 litres
So the capacity of aquarium = 1200 litres ….. (1)
Here
Length of aquarium = 2m = 2 × 100 = 200 cm
Breadth of aquarium = 80 cm
Consider h cm as the height of aquarium
So the capacity of aquarium = 200 × 80 × h cm3
We can write it as
= (200 × 80 × h)/ 1000 litre
= 1/5 × 80 × h litre
= 16 h litre ….. (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 cm2. 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 cm2
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) × h
Substituting the values
224 = 2 (x + x) × 7
By further calculation
224 = 2 × 2x × 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 × b × h
Substituting the values
= 8 × 8 × 7
= 448 cm3
20. If the volume of a cube is V m3, its surface area is S m2 and the length of a diagonal is d metres, prove that 6√3 V = Sd.
Solution:
We know that
Volume of cube = (V) = (Side)3
Consider a as the side of cube
V = a3 and S = 6a2
Diagonal (d) = √3. a
Sd = 6a2 × √3a = 6√3a3
Here V = a3
So we get
Sd = 6√3V
Therefore, 6√3V = 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 × 40 × 12 = 24000 cm3
Volume of part (1) = 50 × 40 × (16 + 24)
By further calculation
= 50 × 40 × 40
= 80000 cm3
Volume of part (2) = 50 × 40 × 24 = 48000 cm3
So the total volume = 24000 + 8000 + 48000 = 153000 cm3
We know that
Total surface area = Area of front and back + Area of vertical faces + Area of top faces
Substituting the values
= 2 (50 × 12 + 50 × 40 + 50 × 24) cm2 + (12 × 40 + 28 × 40 + 16 × 40 + 24 × 40) cm2 + 3 (50 × 40) cm2
By further calculation
= 2 (600 + 2000 + 1200) cm2 + (480 + 1120 + 640 + 960) cm2 + (3 × 2000) cm2
= 2 (3800) + 3200 + 6000 cm2
So we get
= 7600 + 3200 + 6000
= 16800 cm2
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 cm3 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 – 2 × 2) cm, (84 – 2 × 2) cm and (77 – 2) cm
= (98 – 4) cm, (84 – 4) cm, 75 cm
= 94 cm, 80 cm, 75 cm
(i) We know that
Capacity of the box = 94 cm × 80 cm × 75 cm
= 564000 cm3
(ii) Internal volume of box = 564000 cm3
External volume of box = 98 cm × 84 cm × 77 cm = 633864 cm3
So the volume of wood used in making the box = 633864 – 564000 = 69864 cm3
(iii) Weight of 1 cm3 wood = 0.8 gm
So the weight of 69864 cm3 wood = 0.8 × 69864 gm
By further calculation
= (0.8 × 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 ₹ 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 × 32 cm × (0.25 × 100) cm
= (36 × 32 × 25) cm3
Volume of cube having edge 4 cm = 4 × 4 × 4 = 64 cm3
(i) We know that
Number of cubes = Volume of cuboidal block/ Volume of one cube
Substituting the values
= (36 × 32 × 25)/ 64
= (36 × 25)/ 2
So we get
= 18 × 25
= 250
(ii) Here
Total surface area of one cube = 6(a)2
Substituting the values
= 6 (4)2
= 6 × 4 × 4
= 96 cm2
So the total surface area of 450 cubes = 450 × 96 = 43200 cm2
Given
Cost of silver coating the surface for 1 cm2 = ₹ 1.25
Cost of silver coating the surface for 43200 cm2 = 43200 × 1.25 = ₹ 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 ₹ 3.50 per square centimetre.
Solution:
We know that
Volume of first cube = edge3
Substituting the values
= (3 cm)3
= 3 × 3 × 3
= 27 cm3
Volume of second cube = edge3
Substituting the values
= (4 cm)3
= 4 × 4 × 4
= 64 cm3
Volume of third cube = edge3
Substituting the values
= (5 cm)3
= 5 × 5 × 5
= 125 cm3
So the total volume = 27 + 64 + 125 = 216 cm3
Make new cube whose volume = 216 cm3
So we get
Edge3 = 216 cm3
Edge3 = (6 cm)3
Edge = 6 cm
Surface area of new cube = 6 (edge)2
Substituting the values
= 6 (6)2
= 6 × 6 × 6
= 216 cm2
Given
Cost of coating the surface for 1 cm2 = ₹ 3.50
So the cost of coating the surface for 216 cm2 = 3.50 × 216 = ₹ 756
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