# NCERT Solutions For Class 9th Maths Chapter 12 : HERONS Formula

### NCERT Solutions for Class IX Maths : Chapter 12 – HERONS Formula

CBSE NCERT Solutions For Class 9th Maths Chapter 12 : HERONS Formula. NCERT Solutins For Class 9 Mathematics HERONS Formula. Exercise 12.1, Exercise 12.2

## NCERT Solutions for Class 9th Maths : Chapter 12 – HERONS Formula

Page No: 202

**Exercise 12.1**

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?

**Answer**

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

Using heron’s formula,

Area of the signal board = √s (s-a) (s-b) (s-c)

= √(3a/2) (3a/2 – a) (3a/2 – a) (3a/2 – a)

= √3a/2 × a/2 × a/2 × a/2

= √3a^{4}/16

= √3a^{2}/4

= √3/4 × 60 × 60 = 900√3 cm^{2}

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. 12.9). The advertisements yield an earning of ₹5000 per m^{2} per year. A company hired one of its walls for 3 months. How much rent did it pay?

**Answer**

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) m^{2}

= √132 × 10 × 110 × 12 m^{2}

= 1320 m^{2}

Rent of advertising per year = ₹ 5000 per m^{2}

Rent of one wall for 3 months = ₹ (1320 × 5000 × 3)/12 = ₹ 1650000

Page No: 203

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. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

**Answer**

Sides of the triangular wall are 15 m, 11 m and 6 m.

Semi perimeter of triangular wall (s) = (15 + 11 + 6)/2 m = 16 m

Using heron’s formula,

Area of the message = √s (s-a) (s-b) (s-c)

= √16(16 – 15) (16 – 11) (16 – 6) m^{2}

= √16 × 1 × 5 × 10 m^{2 }= √800 m^{2}

= 20√2 m^{2}

4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

**Answer**

Two sides of the triangle = 18cm and 10cm

Perimeter of the triangle = 42cm

Third side of triangle = 42 – (18+10) cm = 14cm

Semi perimeter of triangle = 42/2 = 21cm

Using heron’s formula,

Area of the triangle = √s (s-a) (s-b) (s-c)

= √21(21 – 18) (21 – 10) (21 – 14) cm^{2}

= √21 × 3 × 11 × 7 m^{2}

= 21√11 cm^{2}

5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

**Answer**

Ratio of the sides of the triangle = 12 : 17 : 25

Let the common ratio be x then sides are 12x, 17x and 25x

Perimeter of the triangle = 540cm

12x + 17x + 25x = 540 cm

⇒ 54x = 540cm

⇒ x = 10

Sides of triangle are,

12x = 12 × 10 = 120cm

17x = 17 × 10 = 170cm

25x = 25 × 10 = 250cm

Semi perimeter of triangle(s) = 540/2 = 270cm

Using heron’s formula,

Area of the triangle = √s (s-a) (s-b) (s-c)

= √270(270 – 120) (270 – 170) (270 – 250)cm^{2}

= √270 × 150 × 100 × 20 cm^{2}

= 9000 cm^{2}

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

**Answer**

Length of the equal sides = 12cm

Perimeter of the triangle = 30cm

Length of the third side = 30 – (12+12) cm = 6cm

Semi perimeter of the triangle(s) = 30/2 cm = 15cm

Using heron’s formula,

Area of the triangle = √s (s-a) (s-b) (s-c)

= √15(15 – 12) (15 – 12) (15 – 6)cm^{2}

= √15 × 3 × 3 × 9 cm^{2}

= 9√15 cm^{2}

Page No: 206

**Exercise 12.2**

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

**Answer**

∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

BD is joined.

In ΔBCD,

By applying Pythagoras theorem,

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

⇒ BD^{2} = 12^{2 }+ 5^{2 }

⇒ BD^{2} = 169

⇒ BD = 13 m

Area of ΔBCD = 1/2 × 12 × 5 = 30 m^{2}

Now,

Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m

Using heron’s formula,

Area of ΔABD = √s (s-a) (s-b) (s-c)

= √15(15 – 13) (15 – 9) (15 – 8) m^{2}

= √15 × 2 × 6 × 7 m^{2}

= 6√35 m^{2 }= 35.5 m^{2} (approx)^{
}

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m^{2 }+^{ }35.5m^{2} = 65.5m^{2 }

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**

AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm

In ΔABC,

By applying Pythagoras theorem,

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

⇒ 5^{2} = 3^{2 }+ 4^{2 }

⇒ 25 = 25

Thus, ΔABC is a right angled at B.

Area of ΔBCD = 1/2 × 3 × 4 = 6 cm^{2}

Now,

Semi perimeter of ΔACD(s) = (5 + 5 + 4)/2 cm = 14/2 cm = 7 m

Using heron’s formula,

Area of ΔABD = √s (s-a) (s-b) (s-c)

= √7(7 – 5) (7 – 5) (7 – 4) cm^{2}

= √7 × 2 × 2 × 3 cm^{2}

= 2√21 cm^{2 }= 9.17 cm^{2} (approx)^{
}

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD = 6 cm^{2 }+^{ }9.17 cm^{2} = 15.17 cm^{2 }

3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

**Answer**

Length of the sides of the triangle section I = 5cm, 1cm and 5cm

Perimeter of the triangle = 5 + 5 + 1 = 11cm

Semi perimeter = 11/2 cm = 5.5cm

Using heron’s formula,

Area of section I = √s (s-a) (s-b) (s-c)

= √5.5(5.5 – 5) (5.5 – 5) (5.5 – 1) cm^{2}

= √5.5 × 0.5 × 0.5 × 4.5 cm^{2}

= 0.75√11 cm^{2 }= 0.75 × 3.317cm^{2} = 2.488cm^{2} (approx)

Length of the sides of the rectangle of section I = 6.5cm and 1cm

Area of section II = 6.5 × 1 cm^{2} =^{ }6.5 cm^{2}

Section III is an isosceles trapezium which is divided into 3 equilateral of side 1cm each.

Area of the trapezium = 3 × √3/4 × 1^{2 }cm^{2 }= 1.3 cm^{2 }(approx)

Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm

Area of region IV and V = 2 × 1/2 × 6 × 1.5cm^{2 }= 9cm^{2}

Total area of the paper used = (2.488 + 6.5 + 1.3 + 9)cm^{2 }=^{ }19.3 cm^{2}

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.

**Answer**

Given,

Area of the parallelogram and triangle are equal.

Length of the sides of the triangle are 26 cm, 28 cm and 30 cm.

Perimeter of the triangle = 26 + 28 + 30 = 84 cm

Semi perimeter of the triangle = 84/2 cm = 42 cm

Using heron’s formula,

Area of the triangle = √s (s-a) (s-b) (s-c)

= √42(42 – 26) (46 – 28) (46 – 30) cm^{2}

= √46 × 16 × 14 × 16 cm^{2}

= 336 cm^{2}Let height of parallelogram be h.

Area of parallelogram = Area of triangle

28cm × h = 336 cm^{2}

⇒^{ }h = 336/28 cm

⇒^{ }h = 12 cm

The height of the parallelogram is 12 cm.

Page No: 207

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**

Diagonal AC divides the rhombus ABCD into two congruent triangles of equal area.

Semi perimeter of ΔABC = (30 + 30 + 48)/2 m = 54 m

Using heron’s formula,

Area of the ΔABC = √s (s-a) (s-b) (s-c)

= √54(54 – 30) (54 – 30) (54 – 48) m^{2}

= √54 × 14 × 14 × 6 cm^{2}

= 432 m^{2}

Area of field = 2 × area of the ΔABC = (2 × 432)m^{2 }= 864 m^{2}

Thus,

Area of grass field which each cow will be getting = 864/18^{ }m^{2 }=^{ }48 m^{2}

6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

**Answer**

Semi perimeter of each triangular piece = (50 + 50 + 20)/2 cm = 120/2 cm = 60cm

Using heron’s formula,

Area of the triangular piece = √s (s-a) (s-b) (s-c)

= √60(60 – 50) (60 – 50) (60 – 20) cm^{2}

= √60 × 10 × 10 × 40 cm^{2}

= 200√6 cm^{2}

Area of triangular piece = 5 × 200√6 cm^{2 }= 1000√6 cm^{2}

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 Fig. 12.17. How much paper of each shade has been used in it?

**Answer**

We know that,

As the diagonals of a square bisect each other at right angle.

Area of given kite = 1/2 (diagonal)^{2}

= 1/2 × 32 × 32 = 512 cm^{2}

Area of shade I = Area of shade II

⇒ 512/2 cm^{2 }= 256^{ }cm^{2}

So, area of paper required in each shade = 256 cm^{2}

For the III section,

Length of the sides of triangle = 6cm, 6cm and 8cm

Semi perimeter of triangle = (6 + 6 + 8)/2 cm = 10cm

Using heron’s formula,

Area of the III triangular piece = √s (s-a) (s-b) (s-c)

= √10(10 – 6) (10 – 6) (10 – 8) cm^{2}

= √10 × 4 × 4 × 2 cm^{2}

= 8√6 cm^{2}

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 Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm^{2} .

**Answer**

Semi perimeter of the each triangular shape = (28 + 9 + 35)/2 cm = 36 cm

Using heron’s formula,

Area of the each triangular shape = √s (s-a) (s-b) (s-c)

= √36(36 – 28) (36 – 9) (36 – 35) cm^{2}

= √36 × 8 × 27 × 1 cm^{2}

= 36√6 cm^{2 }= 88.2 cm^{2}

Total area of 16 tiles = 16 × 88.2 cm^{2 }= 1411.2 cm^{2}Cost of polishing tiles = 50p per cm^{2}

Total cost of polishing the tiles = Rs. (1411.2 × 0.5) = Rs. 705.6

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**

Let ABCD be the given trapezium with parallel sides AB = 25m and CD = 10mand the non-parallel sides AD = 13m and BC = 14m.

CM ⊥ AB and CE || AD.

In ΔBCE,

BC = 14m, CE = AD = 13 m and

BE = AB – AE = 25 – 10 = 15m

Semi perimeter of the ΔBCE = (15 + 13 + 14)/2 m = 21 m

Using heron’s formula,

Area of the ΔBCE = √s (s-a) (s-b) (s-c)

= √21(21 – 14) (21 – 13) (21 – 15) m^{2}

= √21 × 7 × 8 × 6 m^{2}

= 84 m^{2 }

also, area of the ΔBCE = 1/2 × BE × CM = 84 m^{2 }

⇒ 1/2 × 15 × CM = 84 m^{2 }

⇒ CM = 168/15 m^{2 }

⇒ CM = 56/5 m^{2 }

Area of the parallelogram AECD = Base × Altitude = AE × CM = 10 × 84/5 = 112 m^{2 }

Area of the trapezium ABCD = Area of AECD + Area of ΔBCE

= (112+ 84) m^{2 }= 196 m^{2 }