ncert solution for class 9 maths Chapter 12 – Heron's Formula Exercise 12.2

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,
BD² = BC² + CD²

= (12)² + (5)²
= 144 + 25
BD² = 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 m² = 65.496 m² = 65.5 m² (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,

AC² = AB² + BC²

(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) cm² = 15.166 cm² = 15.2 cm² (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) cm² = 6.5 cm²
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 cm²

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 cm²

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 cm²

Let the height of the parallelogram be h.
Area of parallelogram = Area of triangle
h × 28 cm = 336 cm²

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,

Semi-perimeter,      

= 54 m

By Heron’s formula,

Area of triangle

Therefore, area of ΔBCD

Area of field = 2 × Area of ΔBCD
= (2 × 432) m² = 864 m²

Area for grazing for 1 cow   = 864/18=48 m²

Each cow will get 48 m² 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 cm². 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 cm²

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 cm².

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) cm²
= 88.2 cm²

Area of 16 tiles = (16 × 88.2) cm²= 1411.2 cm² Cost of polishing per cm² area = 50 p

Cost of polishing 1411.2 cm² 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

m²= 84 m²

Area of ΔBEC

Area of ABED = BF × DE = 11.2 × 10 = 112 m² Area of the field = 84 + 112 = 196 m²