**Exercise 12.2**

## 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

BD

^{2}= BC^{2}+ CD^{2 }
= (12)

BD

BD = 13 m

^{2}+ (5)^{2 }= 144 + 25BD

^{2}= 169BD = 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 = 2s = 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 = 2s = (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 2s = 84 cm

s = 42 cm

By Heron’s formula,

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

Area of parallelogram = Area of triangle

h × 28 cm = 336 cm

^{2 }
h = 12 cm

Therefore, the height of the
parallelogram is 12 cm.

##
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,

For ΔBCD,

Semi-perimeter,

= 54 m

By Heron’s formula,

Area of triangle

Therefore, area of ΔBCD

Area for grazing for 1 cow = 864/18=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= 1/2 (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,

For ΔBEC,

Semi-perimeter,

By Heron’s formula,

Area of triangle

Area of ΔBEC

m

^{2}= 84 m^{2}
Area of ΔBEC

Area of ABED = BF × DE = 11.2 ×
10 = 112 m

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