## EXERCISE 12.2

#### Page No 206:

#### Question 1:

A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

#### Answer:

Let us join BD.

In ΔBCD, applying Pythagoras theorem,

BD^{2} = BC^{2} + CD^{2}

= (12)^{2} + (5)^{2}

= 144 + 25

BD^{2} = 169

BD = 13 m

Area of ΔBCD

For ΔABD,

By Heron’s formula,

Area of triangle

Area of ΔABD

Area of the park = Area of ΔABD + Area of ΔBCD

= 35.496 + 30 m^{2 }= 65.496 m^{2} = 65.5 m^{2} (approximately)

#### Question 2:

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

#### Answer:

For ΔABC,

AC^{2} = AB^{2} + BC^{2}

(5)^{2} = (3)^{2} + (4)^{2}

Therefore, ΔABC is a right-angled triangle, right-angled at point B.

Area of ΔABC

For ΔADC,

Perimeter = 2*s* = AC + CD + DA = (5 + 4 + 5) cm = 14 cm

*s* = 7 cm

By Heron’s formula,

Area of triangle

Area of ABCD = Area of ΔABC + Area of ΔACD

= (6 + 9.166) cm^{2} = 15.166 cm^{2} = 15.2 cm^{2} (approximately)

#### Question 3:

Radha made a picture of an aeroplane with coloured papers as shown in the given figure. Find the total area of the paper used.

#### Answer:

**For triangle I**

This triangle is an isosceles triangle.

Perimeter = 2*s* = (5 + 5 + 1) cm = 11cm

Area of the triangle

**For quadrilateral II**

This quadrilateral is a rectangle.

Area = *l × b* = (6.5 × 1) cm^{2 }= 6.5 cm^{2}

**For quadrilateral III**

This quadrilateral is a trapezium.

Perpendicular height of parallelogram

Area = Area of parallelogram + Area of equilateral triangle

= 0.866 + 0.433 = 1.299 cm^{2}

Area of triangle (IV) = Area of triangle in (V)

Total area of the paper used = 2.488 + 6.5 + 1.299 + 4.5 × 2

= 19.287 cm^{2}

#### Question 4:

A triangle and a parallelogram have the same base and the same area. If the sides of triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

#### Answer:

**For triangle**

Perimeter of triangle = (26 + 28 + 30) cm = 84 cm

2*s* = 84 cm

*s* = 42 cm

By Heron’s formula,

Area of triangle

Area of triangle

= 336 cm^{2}

Let the height of the parallelogram be *h*.

Area of parallelogram = Area of triangle

*h* × 28 cm = 336 cm^{2}

*h* = 12 cm

Therefore, the height of the parallelogram is 12 cm.

#### Page No 207:

#### Question 5:

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

#### Answer:

Let ABCD be a rhombus-shaped field.

For ΔBCD,

Semi-perimeter, = 54 m

By Heron’s formula,

Area of triangle

Therefore, area of ΔBCD

Area of field = 2 × Area of ΔBCD

= (2 × 432) m^{2} = 864 m^{2}

Area for grazing for 1 cow = 48 m^{2}

Each cow will get 48 m^{2} area of grass field.

#### Question 6:

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

#### Answer:

For each triangular piece,

Semi-perimeter,

By Heron’s formula,

Area of triangle

Since there are 5 triangular pieces made of two different coloured cloths,

Area of each cloth required

#### Question 7:

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangles of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it?

#### Answer:

We know that

Area of square (diagonal)^{2}

Area of the given kite

Area of 1^{st} shade = Area of 2^{nd} shade

Therefore, the area of paper required in each shape is 256 cm^{2}.

**For III**^{rd}** triangle**

Semi-perimeter,

By Heron’s formula,

Area of triangle

Area of III^{rd} triangle

Area of paper required for III^{rd} shade = 17.92 cm^{2}

#### Question 8:

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see the given figure). Find the cost of polishing the tiles at the rate of 50p per cm^{2}.

#### Answer:

It can be observed that

Semi-perimeter of each triangular-shaped tile,

By Heron’s formula,

Area of triangle

Area of each tile

= (36 × 2.45) cm^{2}

= 88.2 cm^{2}

Area of 16 tiles = (16 × 88.2) cm^{2}= 1411.2 cm^{2}

Cost of polishing per cm^{2} area = 50 p

Cost of polishing 1411.2 cm^{2} area = Rs (1411.2 × 0.50) = Rs 705.60

Therefore, it will cost Rs 705.60 while polishing all the tiles.

#### Question 9:

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

#### Answer:

Draw a line BE parallel to AD and draw a perpendicular BF on CD.

It can be observed that ABED is a parallelogram.

BE = AD = 13 m

ED = AB = 10 m

EC = 25 − ED = 15 m

For ΔBEC,

Semi-perimeter,

By Heron’s formula,

Area of triangle

Area of ΔBEC

m^{2}= 84 m^{2}

Area of ΔBEC

⇒ 84 = 12 × 15 × BF⇒ BF = 16815 = 11.2 m

Area of ABED = BF × DE = 11.2 × 10 = 112 m^{2}

Area of the field = 84 + 112 = 196 m^{2}