In exercise 20.4 of Chapter 20 of RD Sharma Solutions for Class 7 Maths we shall study about the area of the triangle. Students can refer to RD Sharma Solutions for Class 7 which is designed by our expert tutors for the purpose of strengthening the concepts and problem-solving abilities. On regular practice, students can speed up the method of solving problems by using short-cut tips to secure high marks in their examination.

## Download the PDF of RD Sharma Solutions For Class 7 Chapter 20 – Mensuration – I (Perimeter and Area of rectilinear figures) Exercise 20.4

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 20 – Mensuration – I (Perimeter and Area of rectilinear figures) Exercise 20.4

**1. Find the area in square centimetres of a triangle whose base and altitude are as under:**

**(i) Base =18 cm, altitude = 3.5 cm**

**(ii) Base = 8 dm, altitude =15 cm**

**Solution:**

(i) Given base = 18 cm and height = 3.5 cm

We know that the area of a triangle = Â½ (Base x Height)

Therefore area of the triangle = Â½ x 18 x 3.5

= 31.5 cm^{2}

(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]

And height = 3.5 cm

We know that the area of a triangle = Â½ (Base x Height)

Therefore area of the triangle = Â½ x 80 x 15

= 600 cm^{2}

Â

**2. Find the altitude of a triangle whose area is 42 cm ^{2}Â and base is 12 cm.**

**Solution:**

Given base = 12 cm and area = 42 cm^{2}

We know that the area of a triangle = Â½ (Base x Height)

Therefore altitude of a triangle = (2 x Area)/Base

Altitude = (2 x 42)/12

= 7 cm

**3. The area of a triangle is 50 cm ^{2}. If the altitude is 8 cm, what is its base?**

**Solution:**

Given, altitude = 8 cm and area = 50 cm^{2}

We know that the area of a triangle = Â½ (Base x Height)

Therefore base of a triangle = (2 x Area)/ Altitude

Altitude = (2 x 50)/ 8

= 12.5 cm

**4. Find the area of a right angled triangle whose sides containing the right angle are of lengths 20.8 m and 14.7 m.**

**Solution:**

In a right-angled triangle,

The sides containing the right angles are of lengths 20.8 m and 14.7 m.

Let the base be 20.8 m and the height be 141 m.

Then,

Area of a triangle = 1/2 (Base x Height)

= 1/2 (20.8 Ã— 14.7)

= 152.88 m^{2}

**5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.**

**Solution:**

For the first triangle, given that

Base = 15 cm and altitude = 7 cm

We know that area of a triangle = Â½ (Base x Altitude)

= Â½ (15 x 7)

= 52.5 cm^{2}

It is also given that the area of the first triangle and the second triangle are equal.

Area of the second triangle = 52.5 c m^{2}

One side of the second triangle = 10.5 cm

Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle

= (2x 52.5)/10.5

=10 cm

Hence, the other side of the second triangle will be 10 cm.

**6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this field?**

**Solution:**

Given length of the rectangular field = 48 m

Breadth of the rectangular field = 20 m

Area of the rectangular field = Length x Breadth

= 48 m x 20 m

= 960 m^{2}

Area of one right triangular flower bed = Â½ (12 x 5) = 30 m^{2}

Therefore, required number of right triangular flower beds = area of the rectangular field/ area of one right triangular flower bed.

= 960/30

Number of right triangular flower bed = 32

**7. In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DLÂ âŠ¥AC, BMÂ âŠ¥Â AC, DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.**

**Solution:**

Given AC = 84 cm, DL = 16.5 cm and BM = 12 cm

We know that area of triangle = Â½ x base x height

Area of triangle ADC = Â½ (AC x DL)

= Â½ (84 x 16.5)

= 693 cm^{2}

Area of triangle ABC = Â½ (AC x BM)

= Â½ (84 x 12) = 504 cm^{2}

Hence, Area of quadrilateral ABCD = Area of ADC + Area of ABC

= (693 + 504) cm^{2}Â

= 1197 cm^{2}

**8. Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD measure 48 m and 32 m respectively and are perpendicular to each other.**

**Solution:**

Given diagonal AC = 48 cm and diagonal BD = 32 m

Area of a quadrilateral = Â½ (Product of diagonals)

= Â½ (AC x BD)

= Â½ (48 x 32) m^{2}

= 768 m^{2}

**9. Â In Fig 31, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such that EFâŠ¥Â AD and EF= 14 cm. Calculate the area of the shaded region.**

**Solution:**

Given length of rectangle = 32m and breadth = 18m

We know that area of rectangle = length x breadth

Therefore area of the rectangle = AB x BC

= 32 m x 18 m

= 576 m^{2}

Also given that base of triangle = 18m and height = 14m and EFâŠ¥Â AD

We know that area of triangle = Â½ x base x height

Area of the triangle = Â½ (AD x FE)

= Â½ (BC x FE) [Since AD = BC]

= Â½ (18 m x 14 m)

= 126 m^{2}

Area of the shaded region = Area of the rectangle â€“ Area of the triangle

= (576 â€“ 126) m^{2}

= 450 m^{2}

**10. In Fig. 32, ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm. If P, Q, R, S be the mid-points of the sides AB, BC, CD and DA respectively, find the area of the shaded region.**

**Solution:**

Given ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm.

Join PR and SQ so that these two lines bisect each other at point O

Also OP = OR = RP/2

= 25/2

= 12.5 cm

From the given figure it is clear that,

Area of Triangle SPQ = Area of Triangle SRQ

Hence, area of the shaded region = 2 x (Area of SPQ)

= 2 x (1/2 (SQ x OP))

= 2 x (1/2 (40 x 12.5))

= 500 cm^{2}

**11. Â Calculate the area of the quadrilateral ABCD as shown in Fig.33, given that BD = 42 cm, AC = 28 cm, OD = 12 cm and ACÂ âŠ¥Â BO.**

**Solution:**

BD = 42 cm, AC = 28 cm, OD= 12 cm

Area of Triangle ABC = 1/2 (AC x OB)

= 1/2 (AC x (BD â€“ OD))

= 1/2 (28 cm x (42 cm â€“ 12 cm))

= 1/2 (28 cm x 30 cm)

= 14 cm x 30 cm

= 420 cm^{2}

Area of Triangle ADC = 1/2 (AC x OD) = 1/2 (28 cm x 12 cm)

= 14 cm x 12 cm

= 168 cm^{2}

Hence, Area of the quadrilateral ABCD = Area of ABC + Area of ADC

= (420 + 168) cm^{2}

= 588 cm^{2}

**12. Find the area of figure formed by a square of side 8 cm and an isosceles triangle with base as one side of the square and perimeter as 18 cm.**

**Solution:**

Let x cm be one of the equal sides of an isosceles triangle.

Given that the perimeter of the isosceles triangle = 18 cm

Then, x + x + 8 =18

2x = (18 â€“ 8) = 10 cm

2x = 10

x = 5 cm

Area of the figure formed = Area of the square + Area of the isosceles triangle

= (side of square)^{2} + Â½ (base x âˆš[(equal side)^{2} â€“ Â¼ x (base)^{2}]

= 8^{2} + Â½ (8) x âˆš [5^{2} â€“ Â¼ x 8^{2}]

= 64 + 4 x âˆš [25 â€“ Â¼ x 64]

= 64 + 4 x âˆš (25 â€“ 16)

= 64 + 4 x âˆš9

= 64 + 4 x 3

= 64 + 12

= 76 cm^{2}

**13. Find the area of Fig. 34, in the following ways: (i) Sum of the areas of three triangles (ii) Area of a rectangle â€” sum of the areas of five triangles**

**Solution:**

(i) From the figure, P is the midpoint of AD.

Thus AP = PD = 25 cm and AB = CD = 20 cm

From the figure, we observed that,

Area of Triangle APB = Area of Triangle PDC

Area of Triangle APB = Â½ (AB x AP)

= Â½ (20 x 25)

= 250 cm^{2}

Area of Triangle PDC = Area of Triangle APB = 250 c m^{2}

Area of Triangle RPQ = Â½ (Base x Height)

= Â½ (25 cm x 10 cm)

= 125 cm^{2}

Hence, Sum of the three triangles = (250 + 250 + 125) cm^{2}Â

= 625 cm^{2}

(ii) From the figure, area of the rectangle ABCD = 50 cm x 20 cm

= 1000 cm^{2}

Thus, Area of the rectangle â€“ Sum of the areas of three triangles

= (1000 â€“ 625) cm^{2}Â

= 375 cm^{2}

**14. Calculate the area of quadrilateral field ABCD as shown in Fig.35, by dividing it into a rectangle and a triangle.**

**Solution:**

Join CE, so that which intersect AD at point E.

Given AE = ED = BC = 25 m and EC = AB = 30 m

Area of rectangle = length x breadth

Area of the rectangle ABCE = AB x BC

= 30 m x 25 m

= 750 m^{2}

Area of triangle = Â½ x base x height

Area of Triangle CED = Â½ (EC x ED)

= Â½ (30 m x 25 m)

= 375 m^{2}

Hence, Area of the quadrilateral ABCD = (750 + 375) m^{2}Â = 1125 m^{2}

**15. Calculate the area of the pentagon ABCDE, where AB = AE and with dimensions as shown in Fig. 36.**

**Solution:**

Join BE so that we can get rectangle and triangle.

We know that area of rectangle = length x breadth

Area of the rectangle BCDE = CD x DE

= 10 cm x 12 cm

= 120 c m^{2}

Area of triangle = Â½ x base x height

Area of Triangle ABE = 1/2 (BE x height of the triangle)

= Â½ (10 cm x (20 â€“ 12) cm)

= Â½ (10 cm x 8 cm)

= 40 cm^{2}

Hence, Area of the pentagon ABCDE = area of rectangle + area of triangle

= (120 + 40) cm^{2}Â

= 160 cm^{2}

**16. The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10, find its base and height.**

**Solution:**

Let altitude of the triangular field be h m

Then base of the triangular field is 3h m.

We know that area of triangle = Â½ x b x h

Area of the triangular field = Â½ (h x 3h) = 3h^{2}/2 m^{2} â€¦â€¦. (i)

The rate of cultivating the field is Rs 24.60 per hectare.

Therefore,

Area of the triangular field = 332.10 /24.60

= 13.5 hectare

= 135000 m^{2}Â [Since 1 hectare = 10000 m^{2}] â€¦â€¦ (ii)

From equation (i) and (ii) we have,

3h^{2}/2 = 135000 m^{2}

3h^{2}Â = 135000 x 2 = 270000 m^{2}

h^{2}Â = 270000/3

= 90000 m^{2}Â

= (300)^{2}

h = 300 m

Hence, Height of the triangular field = 300 m and

Base of the triangular field = 3 x 300 m = 900 m

**17. A wall is 4.5 m long and 3 m high. It has two equal windows, each having form and dimensions as shown in Fig. 37. Find the cost of painting the wall (leaving windows) at the rate of Rs 15 per m ^{2}.**

**Solution:**

Given length of a wall = 4.5 m

Breadth of the wall = 3 m

We know that the area of triangle = length x Breadth

Area of the wall = Length x Breadth

= 4.5 m x 3 m = 13.5 m^{2}

From the figure we observed that,

Area of the window = Area of the rectangle + Area of the triangle

= (0.8 m x 0.5 m) + (12 x 0.8 m x 0.2 m) [Since 1 m = 100 cm]

= 0.4 m^{2}Â + 0.08 m^{2}

= 0.48 m^{2}

Area of two windows = 2 x 0.48 = 0.96 m^{2}

Area of the remaining wall (leaving windows) = (13.5 â€“ 0.96) m^{2}

= 12.54 m^{2}

Cost of painting the wall per m^{2}Â = Rs. 15

Hence, the cost of painting on the wall = Rs. (15 x 12.54)

= Rs. 188.1