NCERT Solutions Class 9 Mathematics Chapter 12 Herons Formula

Exercise 12.1

Q.1) A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the
area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area
of the signal board?
Sol.1) Length of the side of equilateral triangle = a
Perimeter of the signal board = 3𝑎 = 180 𝑐𝑚
∴ 3𝑎 = 180 𝑐𝑚 ⇒ 𝑎 = 60 𝑐𝑚
Semi perimeter of the signal board (s) = 3𝑎/2

Q.2) The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 𝑚, 22 𝑚 and 120 𝑚 (see Fig.). The advertisements yield on earning of 𝑅𝑠 5000 𝑝𝑒𝑟 𝑚² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Sol.2) The sides of the triangle are 122 𝑚, 22 𝑚 and 120 𝑚.
Perimeter of the triangle is 122 + 22 + 120 = 264𝑚
Semi perimeter of triangle (𝑠) = 264/2
= 132 𝑚
Using heron’s formula,
Area of the advertisement = √𝑠 (𝑠 − 𝑎) (𝑠 − 𝑏) (𝑠 − 𝑐)
= √132(132 – 122)(132 – 22)(132 – 120) 𝑚²
= √132 × 10 × 110 × 12 𝑚²
= 1320 𝑚²
Rent of advertising per year = 𝑅𝑠. 5000 𝑝𝑒𝑟 𝑚²
Rent of one wall for 3 months = 𝑅𝑠. 1320 × 5000 × 3/12 = 𝑅𝑠. 1650000

Q.3) There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig.). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Q.4) Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Sol.4) Here 𝑎 = 18 𝑐𝑚, 𝑏 = 10 𝑐𝑚, 𝑐 = ?
Perimeter of the triangle = 42 𝑐𝑚
⇒ 𝑎 + 𝑏 + 𝑐 = 42
⇒ 18 + 10 + 𝑐 = 42
⇒ 𝑐 = 42 – 28 = 14
Using heron’s formula,

Q.5) Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Sol.5) Let the sides of the triangle be 12𝑥 𝑐𝑚 ,17𝑥 𝑐𝑚 and 25𝑥 𝑐𝑚.
Perimeter of the triangle = 540 cm
∴ 12𝑥 + 17𝑥 + 25𝑥 = 540
⇒ 54 𝑥 = 540
⇒ 𝑥 = 540/54 = 10
∴ Sides of the triangle are (12 × 10) 𝑐𝑚, (17 × 10) 𝑐𝑚 and (25 × 10) 𝑐𝑚 i.e.,
120 𝑐𝑚, 170 𝑐𝑚 and 250 𝑐𝑚.
Now, suppose 𝑎 = 120 𝑐𝑚, 𝑏 = 170 𝑐𝑚, 𝑐 = 250 𝑐𝑚,
∴ 𝑠 = 𝑎+𝑏+𝑐/2 = 540/2 𝑐𝑚 = 270 𝑐𝑚
Using heron’s formula,

Q.6) An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Sol.6) Here, 𝑎 = 𝑏 = 12 𝑐𝑚,
Also, 𝑎 + 𝑏 + 𝑐 = 30
⇒ 12 + 12 + 𝑐 = 30
⇒ 𝑐 = 30 – 24 = 6

Exercise 12.2

Q.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?
Sol.1) ∠𝐶 = 90°, 𝐴𝐵 = 9 𝑚, 𝐵𝐶 = 12 𝑚, 𝐶𝐷 = 5 𝑚 and 𝐴𝐷 = 8 𝑚
BD is joined.
In 𝛥𝐵𝐶𝐷,
By applying Pythagoras theorem,
𝐵𝐷² = 𝐵𝐶² + 𝐶𝐷²
⇒ 𝐵𝐷² = 122 + 52
⇒ 𝐵𝐷² = 169
⇒ 𝐵𝐷 = 13 𝑚
Area of 𝛥𝐵𝐶𝐷 = 1/2 × 12 × 5 = 30 𝑚²
Now, In Δ𝐴𝐵𝐷, 𝑎 = 9 𝑚, 𝑏 = 8 𝑚, 𝑐 = 13 𝑚

= 6√35 𝑚² = 35.5 𝑚² (approx).
Area of quadrilateral 𝐴𝐵𝐶𝐷 = Area of 𝛥𝐵𝐶𝐷 + Area of 𝛥𝐴𝐵𝐷 = 30 𝑚² + 35.5𝑚²
= 65.5𝑚²

Q.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.
Sol.2) 𝐴𝐵 = 3 𝑐𝑚, 𝐵𝐶 = 4 𝑐𝑚, 𝐶𝐷 = 4 𝑐𝑚, 𝐷𝐴 = 5 𝑐𝑚 and 𝐴𝐶 = 5 𝑐𝑚
In 𝛥𝐴𝐵𝐶,
By applying Pythagoras theorem,
𝐴𝐶² = 𝐴𝐵2 + 𝐵𝐶²
⇒ 52 = 32 + 42
⇒ 25 = 25
Thus, 𝛥𝐴𝐵𝐶 is a right angled at B.
Area of 𝛥𝐵𝐶𝐷 = 1/2 × 3 × 4 = 6 𝑐𝑚²
Now,

= 2√21 𝑐𝑚² = 9.17 𝑐𝑚2 (𝑎𝑝𝑝𝑟𝑜𝑥)
Area of quadrilateral 𝐴𝐵𝐶𝐷 = Area of 𝛥𝐴𝐵𝐶 + Area of 𝛥𝐴𝐵𝐷 = 6 𝑐𝑚² + 9.17 𝑐𝑚²
= 15.17 𝑐𝑚²

Q.3) Radha made a picture of an aeroplane with coloured paper as shown in the figure. Find the total area of the paper used.

𝐴𝐷 = 𝐹𝐶 = 1 𝑐𝑚, 𝐷𝐶 = 𝐴𝐹 = 1 𝑐𝑚
∴ 𝐵𝐹 = 𝐵𝐶 – 𝐹𝐶 = (2 – 1) 𝑐𝑚 = 1 𝑐𝑚
Hence, Δ𝐴𝐵𝐹 𝑖s equilateral.
Also, E is the mid-point of BF.
∴ 𝐵𝐸 = 1/2 𝑐𝑚 = 0.5𝑐𝑚
Also, 𝐴𝐵² = 𝐴𝐸² + 𝐵𝐸² [Pythagoras theorem]
⇒ 𝐴𝐸² = 12 – (0.5)² = 0.75
⇒ 𝐴𝐸 = 0.9 𝑐𝑚 (𝑎𝑝𝑝𝑟𝑜𝑥. )
Area of the trapezium = 1/2
(sum of the parallel sides) × distance between them.
= 1/2 × (𝐵𝐶 + 𝐴𝐷) × 𝐴𝐸 = 1 2 × (2 + 1) × 0.9 𝑐𝑚² = 1.4𝑐𝑚²
For the triangle marked IV :
It is a right-triangle
∴ Area of the triangle = 1/2 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
= 1/2 × 6 × 1.5 𝑐𝑚² = 4.5 𝑐𝑚²
For the triangle marked V :
This triangle is congruent to the triangle marked IV.
Hence, area of the triangle = 4.5 𝑐𝑚²
Total area of the paper used = (2.5 + 6.5 + 1.4 + 4.5 + 4.5)𝑐𝑚2 = 19.4 𝑐𝑚²

Q.4) A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Sol.4) In the figure, ABCD is a parallelogram and ABE is the triangle which stands on the base AB
For the triangle 𝐴𝐵𝐸, 𝑎 = 30 𝑐𝑚, 𝑏 = 28 𝑐𝑚, 𝑐 = 26 𝑐𝑚.

⇒ ℎ = 12 𝑐𝑚
The height of the parallelogram is 12𝑐𝑚

Q.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?
Sol.5) Clearly, the diagonal AC of the rhombus divides it into two congruent triangles.
For triangle ABC, a = b = 30 m, c = 48 m.

= 432 𝑚²
Area of field = 2 × area of the 𝛥𝐴𝐵𝐶 = (2 × 432)𝑚² = 864𝑚²
Thus, Number of cows = 18
Area of grass field which each cow will be getting = (864/18)𝑚² = 48𝑚²

Q.6) An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.), each piece measuring 20 cm, 50 cm, and 50 cm. How much cloth of each colour is required for the umbrella?
Sol.6) First we find the area of one triangular piece.
Here, 𝑎 = 𝑏 = 50 𝑐𝑚, 𝑐 = 20 𝑐𝑚

Q.7) A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure.
How much paper of each shade has been used in it?
Sol.7 ABCD is a square.
So, 𝐴𝑂 = 𝑂𝐶 = 𝑂𝐵 = 𝑂𝐷
and ∠𝐴𝑂𝐵 = 90° [Diagonals of a square bisect each other at right angles]
𝐵𝐷 = 32 𝑐𝑚 (Given)
⇒ 𝑂𝐴 = 32/2 𝑐𝑚 = 16 𝑐𝑚
ΔABD is a right triangle.
So, area of Δ𝐴𝐵𝐷 = 1/2 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
= 1/2 × 32 × 16 𝑐𝑚² = 256 𝑐𝑚²
Thus, area of Δ𝐵𝐶𝐷 = 256 𝑐𝑚²
For triangle 𝐶𝐸𝐹, 𝑎 = 𝑏 = 6 𝑐𝑚, 𝑐 = 8 𝑐𝑚.

Hence, paper needed for shade I= 256 𝑐𝑚²
for shade II = 256 𝑐𝑚²
and for shade III = 17.92 𝑐𝑚²

Q.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 figure). Find the cost of polishing the tiles at the rate of 50 𝑝 𝑝𝑒𝑟 𝑐𝑚².

Sol.8) We have lengths of the sides of 1 triangular tile are 𝑎 = 35 𝑐𝑚, 𝑏 = 28 𝑐𝑚, 𝑐 = 9 𝑐𝑚.

Area of 16 such tiles = 16 × 88.2 = 1411.2
Cost of polishing tiles 1𝑐𝑚2 = 50𝑝 𝑝𝑒𝑟 𝑐𝑚2 = 𝑅𝑒. 0.50
Total cost of polishing the floral design = 𝑅𝑠. (1411.2 × 0.5) = 𝑅𝑠. 705.6

Q.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.
Sol.9) In the figure ABCD is the field.
Draw 𝐶𝐹 || 𝐷𝐴 and 𝐶𝐺 ⊥ 𝐴𝐵.
𝐷𝐶 = 𝐴𝐹 = 10 𝑚, 𝐴𝐷 = 𝐹𝐶 = 13 𝑚
For Δ𝐵𝐶𝐹, 𝑎 = 15 𝑚, 𝑏 = 14 𝑚, 𝑐 = 13 𝑚