In this chapter, we provide RD Sharma Class 7 ex 20.4 Solutions Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures) for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 20.4 Solutions Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures) Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 20 |

Chapter Name | Mensuration I (Perimeter and Area of Rectilinear Figures) |

Exercise | Ex 20.4 |

Table of Contents

**RD Sharma Solutions for Class 7 Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures)** **Ex 20.4 Download PDF**

**Mensuration I (Perimeter and Area of Rectilinear Figures) Exercise 20.4**

**Question: 1**

Find the area in square centimeters 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:**

We know that the area of a triangle = 1/2 (Base x Height)

(i) Here, base = 18 cm and height = 3.5 cm

Area of the triangle = 1/2 x 18 x 3.5

= 31.5 cm^{2}

(ii) Here, base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm] and height = 3.5 cm

Area of the triangle = 1/2 x 80 x 15

= 600 c m^{2}

**Question: 2**

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

**Solution:**

We have,

Attitude of a triangle = (2 x Area)/Base

Here, base = 12 cm and area = 42 cm^{2}

Attitude = (2 x 42)/12 = 7 cm

**Question: 3**

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

**Solution:**

We have,

Base of a triangle = (2 x Area)/ Altitude

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

Altitude = (2 x 50)/ 8 = 12.5 cm

**Question: 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 × 14.7)

= 152.88 m^{2}

**Question: 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, we have,

Base = 15 cm and altitude = 7 cm

Thus, area of a triangle = 1/2 (Base x Altitude)

= 1/2 (15 x 7)

= 52.5 cm^{2}

It is 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.

**Question: 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:**

We have,

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 = ½ (12 m x 5m) = 30 m^{2}

Therefore,

Required number of right triangular flower beds = 960 m^{2}/30 m^{2}= 32

**Question: 7**

In Figure, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ⊥AC, BM ⊥ AC, DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.

**Solution:**

We have,

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

Area of triangle ADC = 1/2 (AC x DL) = 1/2 (84 cm x 16.5 cm) = 693 cm^{2}

Area of triangle ABC = 1/2 (AC x BM) = 1/2 (84 cm x 12 cm) = 504 cm^{2}

Hence, Area of quadrilateral ABCD = Area of ADC + Area of ABC = (693 + 504) cm^{2} = 1197 cm^{2}

**Question: 8**

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

**Solution:**

We have,

Diagonal AC = 48 cm and diagonal BD = 32 m

Area of a quadrilateral = 1/2 (Product of diagonals)

= 1/2(AC x BD) = 1/2 (48 x 32) m^{2}

= (24 x 32) m^{2} = 768 m^{2}

**Question: 9**

In Fig below, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such that EF⊥ AD and EF= 14 cm. Calculate the area of the shaded region.

**Solution:**

We have,

Area of the rectangle = AB x BC = 32 m x 18 m = 576 m^{2}

Area of the triangle = 1/2 (AD x FE)

= 1/2 (BC x FE) [Since AD = BC]

= 1/2 (18 m x 14 m)

= 9 m x 14 m

= 126 m^{2}

Area of the shaded region = Area of the rectangle – Area of the triangle

= (576 – 126) m^{2}

= 450 m^{2}

**Question: 10**

In Fig. below, 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:**

Join points PR and SQ. These two lines bisect each other at point 0.

Here, AB = DC = SQ = 40 cm and AD = BC = RP = 25 cm

Also OP = OR = RP/2 = 25/2 = 12.5 cm

From the figure we observed 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 cm x 12.5 cm))

= 500 cm^{2}

**Question: 11**

Calculate the area of the quadrilateral ABCD as shown in Figure, given that BD = 42 cm, AC = 28 cm, OD = 12 cm and AC ⊥ BO.

**Solution:**

We have,

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

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

= 1/2 (AC x (BD – OD))

= 1/2 (28 cm x (42 cm – 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}

**Question: 12**

Find the area of a 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 – 8) cm = 10 cm

x = 5 cm

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

**Question: 13**

Find the area of Figure, in the following ways: (i) Sum of the areas of three triangles (ii) Area of a rectangle — sum of the areas of five triangles

**Solution:**

We have,

(i) 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 = 1/2 (AB x AP ) = 1/2 (20 cm x 25 cm) = 250 cm^{2}

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

Area of Triangle RPQ = 1/2 (Base x Height) = 1/2 (25 cm x 10 cm) = 125 cm^{2}

Hence, Sum of the three triangles = (250 + 250 + 125) cm^{2} = 625 cm^{2}

(ii) Area of the rectangle ABCD = 50 cm x 20 cm = 1000 cm^{2}

Thus, Area of the rectangle – Sum of the areas of three triangles

= (1000 – 625 ) cm^{2} = 375 cm^{2}

**Question: 14**

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

**Solution:**

We have,

Join CE , which intersect AD at point E.

Here, AE = ED = BC = 25 m and EC = AB = 30 m

Area of the rectangle ABCE = AB x BC = 30 m x 25 m = 750 m^{2}

Area of Triangle CED = 1/2 (EC x ED) = 1/2 ( 30 m x 25 m) = 375 m^{2}

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

**Question: 15**

Calculate the area of the pentagon ABCDE, where AB = AE and with dimensions as shown in Figure.

**Solution:**

Join BE.

Area of the rectangle BCDE = CD x DE

= 10 cm x 12 cm

= 120 c m^{2}

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

= 1/2 (10 cm x (20 – 12) cm)

= 1/2 (10 cm x 8 cm)

= 40 cm^{2}

Hence, Area of the pentagon ABCDE = (120 + 40) cm^{2} = 160 cm^{2}

**Question: 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.

Area of the triangular field = 1/2 (h x 3h )=3h^{2}/2 m^{2} —–(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} [Since 1 hectare = 10000 m^{2}] —–(ii)

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

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

3h^{2} = 135000 x 2 = 270000 m^{2}

h^{2} = 270000/3 m^{2}= 90000 m^{2} = (300 m)^{2}

h = 300 m

Hence, Height of the triangular field = 300 m and base of the triangular field = 3 x 300 m = 900 m

**Question: 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 Figure. below. Find the cost of painting the wall (leaving windows) at the rate of Rs 15 per m^{2}.

**Solution:**

We have,

Length of a wall = 4.5 m

Breadth of the wall = 3 m

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} + 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 – 0.96) m^{2}

= 12.54 m^{2}

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

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

= Rs. 188.1

(In the book, the answer is given for one window, but we have 2 windows.)

**All Chapter RD Sharma Solutions For Class 7 Maths**

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