RD Sharma solution class 7 chapter 20 Mensuration I Exercise 20.4

Exercise 20.4

Page-20.26

Question 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

Answer 1:

We know that the area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$
(i) Here, base = 18 cm and height = 3.5 cm
     ∴ Area of the triangle = $\left(\frac{1}{2}\times 18\times 3.5\right)=31.5 {\mathrm{cm}}^2$
(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 = $\left(\frac{1}{2}\times 80\times 15\right)=600 {\mathrm{cm}}^2$

Question 2:

Find the altitude of a triangle whose area is 42 cm² and base is 12 cm.

Answer 2:

We have,
Altitude of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Base}}$
Here, base = 12 cm and area = 42 cm²

∴ Altitude = $\frac{2\times 42}{12}= 7 \mathrm{cm}$

Question 3:

The area of a triangle is 50 cm². If the altitude is 8 cm, what is its base?

Answer 3:

We have,
Base of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Altitude}}$2×AreaBase
Here, altitude = 8 cm and area = 50 cm²

∴ Altitude =$\frac{2\times 50}{8}=12.5 \mathrm{cm}$
2×4212= 7 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.

Answer 4:

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 14.7 m.
Then,
Area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

                              = $\frac{1}{2}\times 20.8\times 14.7 = 152.88 {\mathrm{m}}^2$

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.

Answer 5:

For the first triangle, we have,
Base = 15 cm and altitude = 7 cm
Thus, area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Altitude}$

                                       = $\frac{1}{2}\times 15\times 7=52.5 {\mathrm{cm}}^2$
It is given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 cm²
One side of the second triangle = 10.5 cm
Therefore,
The other side of the second triangle = $\frac{2\times \mathrm{Area}}{\mathrm{One} \mathrm{side} \mathrm{of} \mathrm{a} \mathrm{triangle}}$

                                                     = $\frac{2\times 52.5}{10.5}=10 \mathrm{cm}$
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?

Answer 6:

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 m²

Area of one right triangular flower bed = $\frac{1}{2}\times 12 \mathrm{m}\times 5 \mathrm{m} = 30 {\mathrm{m}}^2$
Therefore,

Required number of right triangular flower beds = $\frac{960 {\mathrm{m}}^2}{30 {\mathrm{m}}^2}= 32$

Question 7:

In Fig. 29, 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.

Answer 7:

We have,
AC = 84 cm, DL = 16.5 cm and BM = 12 cm
Area of Δ ADC = $$$\frac{1}{2}$x AC x DL
                          = $$$\frac{1}{2}$x 84 cm x 16.5 cm = 693 cm²
Area of Δ ABC = $$$\frac{1}{2}$x AC x BM
                          = $$$\frac{1}{2}$x 84 cm x 12 cm = 504 cm²
Hence,
Area of quadrilateral ABCD = Area of Δ ADC  + Area of Δ ABC
                                                          = (693 + 504) cm²
                                                                             = 1197 cm²

Page-20.27

Question 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.

Answer 8:

We have,
Diagonal AC = 48 cm and diagonal BD = 32 m
∴ Area of a quadrilateral = $\frac{1}{2}$x Product of diagonals
                                      = $\frac{1}{2}$x AC x BD
                                      = ($\frac{1}{2}$ x 48 x 32) m² = (24 x 32) m²  = 768 m²

Question 9:

In Fig 31, 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.

Answer 9:

We have,
Area of the rectangle = AB x BC
                             = 32 m x 18 m
                             = 576 m²
Area of the triangle  = $\frac{1}{2}$x AD x FE
                           = $\frac{1}{2}$x BC x FE       [Since AD = BC]
                           = $\frac{1}{2}$x 18 m x 14 m
                           = 9 m x 14 m = 126 m²
∴ Area of the shaded region = Area of the rectangle − Area of the triangle
                                      =(576 − 126) m²
                                                  = 450 m²

Question 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.

Answer 10:

We have,
Join points PR and SQ.
These two lines bisect each other at point O.

Here, AB = DC = SQ = 40 cm and AD = BC =RP = 25 cm
Also OP = OR = $\frac{RP}{2}=\frac{25}{2}=12.5 \mathrm{cm}$
From the figure we observed that,
Area of Δ SPQ = Area of Δ SRQ
Hence, area of the shaded region = 2 x (Area of Δ SPQ)
                                                       = 2 x ($\frac{1}{2}$ x SQ x OP)
                                                       = 2 x ($\frac{1}{2}$ x 40 cm x 12.5 cm)
                                                       = 500 cm²

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
Area of ΔABC = $\frac{1}{2}$ x AC x OB
                          = $\frac{1}{2}$ x AC x (BD − OD)
                          = $\frac{1}{2}$ x 28 cm x (42 cm − 12 cm) = $\frac{1}{2}$ x 28 cm x 30 cm = 14 cm x 30 cm = 420 cm²
Area of Δ ADC = $\frac{1}{2}$ x AC x OD
                          = $\frac{1}{2}$ x 28 cm x 12 cm = 14 cm x 12 cm = 168 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

                                         = (Side of square)² + $\frac{1}{2}\times \mathrm{Base}\times \sqrt{{\left(\mathrm{Equal} \mathrm{side}\right)}^2 -\frac{1}{4}\times {\left(\mathrm{Base}\right)}^2}$
                                        = (8)² + $\frac{1}{2}\times 8\times \sqrt{{\left(5\right)}^2 -\frac{1}{4}{\left(8\right)}^2}$
                                        = $64+4 \times \sqrt{25-16}$
                                        = $64 +4 \times \sqrt{9}$
                                        = $64+4\times 3$
                                        = 64 + 12 = 76 cm²

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 APB = $\frac{1}{2}$x AB x AP
                               = $\frac{1}{2}$x 20 cm x 25 cm = 250 cm²
      Area of Δ PDC = Area of Δ APB = 250 cm²

      Area of Δ RPQ = $\frac{1}{2}$x Base x Height
                                = $\frac{1}{2}$x 25 cm x 10 cm = 125 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
Area of Δ CED = $\frac{1}{2}$x EC x ED
                          = $\frac{1}{2}$x 30 m x 25 m
                          = 375 m²
Hence,
Area of the quadrilateral ABCD  = (750  + 375) m²
                                                     = 1125 m²

Page-20.28

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
Area of ΔABE = $\frac{1}{2}$x BE x  height of the triangle
                          = $\frac{1}{2}$x 10 cm x  (20 − 12) cm
                          = $\frac{1}{2}$x 10 cm x 8 cm = 40 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 = $\frac{1}{2}\times h\times 3h=\frac{3h^2}{2}{\mathrm{m}}^2$..........(i)

The rate of cultivating the field is Rs 24.60 per hectare.
Therefore,
Area of the triangular field = $\frac{ 332.10}{ 24.60}=13.5 \mathrm{hectare}$
                                                            = 135000 m²         [Since 1 hectare = 10000 m² ]..........(ii)
From equation (i) and (ii) we have,
$\frac{3h^2}{2}=135000 {\mathrm{m}}^2$
3h² = 135000 x 2 = 270000 m²

 h² = $\frac{270000 }{3}\mathrm{m}$²=90000 m2 = (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 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
                              = (0.8 m x 0.5 m) + ($\frac{1}{2}$ x 0.8 m x 0.2 m)      [Since 1 m = 100 cm]
                              = 0.4 m² + 0.08 m² = 0.48 m²
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 m²
Cost of painting the wall per m² = 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.)