RD Sharma Solutions (Part - 2) - Ex-20.4, Mensuration - I, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

Question 11:

Calculate the area of the quadrilateral ABCD as shown in Fig. 33, given that BD = 42 cm, AC = 28 cm, OD= 12 cm and AC ⊥ BD.

Answer 11:

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

= 14 cm x 30 cm = 420 cm²

Hence,
Area of the quadrilateral ABCD = Area of Δ ABC + Area of Δ ADC
                                      
                                              = (420  + 168) cm² = 588 cm²

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.

Answer 12:

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 Fig. 34 in the following ways:
 (i) Sum of the areas of three triangles
 (ii) Area of a rectangle − sum of the areas of five triangles

Answer 13:

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 Δ APB = Area of Δ PDC

Area of Δ PDC = Area of Δ APB = 250 cm²
 

 Hence,
    Sum of the three triangles = (250  + 250 + 125) cm² 
                                         = 625 cm²

(ii) Area of the rectangle ABCD = 50 cm x 20 cm = 1000 cm²
      Thus,
      Area of the rectangle − Sum of the areas of three triangles  (There is a mistake in the question; it should be area of three triangles)
       = (1000 − 625 ) cm² = 375 cm²

Question 14:

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

Answer 14:

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 m²

Hence,
Area of the quadrilateral ABCD  = (750  + 375) m²
                                                     = 1125 m²

Question 15:

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

Answer 15:

Join BE.
Area of the rectangle BCDE = CD x DE
                                        = 10 cm x 12 cm = 120 cm²

Hence,
Area of the pentagon ABCDE = (120 + 40) cm² = 160 cm²

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.

Answer 16:

Let altitude of the triangular field be h m
Then base of the triangular field is 3h m.

Area of the triangular field = 

The rate of cultivating the field is Rs 24.60 per hectare.
Therefore,
Area of the triangular field =  

= 135000 m²         [Since 1 hectare = 10000 m² ]..........(ii)
From equation (i) and (ii) we have,

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 Fig. 37. Find the cost of painting the wall (leaving windows) at the rate of Rs 15 per m₂.

Answer 17:

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 m²
From the figure we observed that,
Area of the window = Area of the rectangle + Area of the triangle

Area of  two windows = 2 x 0.48 = 0.96 m²
Area of the remaining wall (leaving windows ) = (13.5 − 0.96 )m² = 12.54 m²ost of painting the wall per m² = Rs. 15
Hence, the cost of painting on the wall = Rs. (15 x 12.54) = Rs. 188.1