The document RD Sharma Solutions (Part - 2) - Ex-20.4, Mensuration - I, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**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 âŠ¥ BD.**

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

= 14 cm x 30 cm = 420 cm^{2}

Hence,

Area of the quadrilateral *ABCD* = Area of Î”* ABC* + Area of Î” *ADC*

= (420 + 168) cm^{2} = 588 cm^{2}

**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.**

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

â‡’ 2*x* = (18 âˆ’ 8) cm = 10 cm

â‡’ *x *= 5 cm

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

**Find the area of Fig. 34 in the following ways: (i) Sum of the areas of three triangles (ii) Area of a rectangle âˆ’ sum of the areas of five triangles**

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 Î”* APB* = Area of Î” *PDC*

Area of Î” *PDC* = Area of Î” *APB* = 250 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 âˆ’ Sum of the areas of three triangles (There is a mistake in the question; it should be area of three triangles)

= (1000 âˆ’ 625 ) cm^{2} = 375 cm^{2}

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

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}

Hence,

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

= 1125 m^{2}

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

Join *BE*.

Area of the rectangle *BCDE* =* CD* x *DE*

= 10 cm x 12 cm = 120 cm^{2}

Hence,

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

**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.**

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 m^{2} [Since 1 hectare = 10000 m^{2} ]..........(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

**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}.**

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

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

Area of the remaining wall (leaving windows ) = (13.5 âˆ’ 0.96 )m^{2} = 12.54 m^{2}ost 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