NCERT Solutions for Class 9 Maths Chapter 10 Heron’s Formula

NCERT
SOLUTIONS
CLASS - 9th

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Class : 9th
Subject : Maths
Chapter : 12
Chapter Name : Heron’s Formula

Exercise 12.1

Q1 A traf c 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. Side of traf c signal board = a
Perimeter of traf c signal board
3
2s = 3a ⇒ s = a
2

Area of given triangle = √s(s − a)(s − b)(s − c)
3 3 3 3
Area of given triangle = √ a( a − a) ( a − a) ( a − a)
2 2 2 2

= √(
3
a) (
a
)(
a
)(
a
)
....(1)
2 2 2 2

√3
2
= a
4

Perimeter of traf c signal board = 180 cm
Side of traf c signal board (a) = (
180
) cm = 60cm
3

√3
2
Using equation (1), area of traffic signal board = (60cm)
4

3600 2 2
= ( √3) cm = 900√3cm
4

Page : 202 , Block Name : Exercise 12.1

Q2 The triangular side walls of a yover have been used for advertisements. The sides of the
walls are 122 m, 22 m and 120 m (see Fig). 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?

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Answer. The sides of the triangle (i.e., a, b, c) are of 122 m, 22 m, and 120 m respectively.
Perimeter of triangle = (122 + 22 + 120) m
2s = 264 m
s = 132 m
Area of triangle = √s(s − a)(s − b)(s − c)
2
Area of given triangle = [√132(132 − 122)(132 − 22)(132 − 120)]m

2 2
= [√132(10)(110)(12)]m = 1320m

Rent of 1 m area per year = Rs 5000
2

Rent of 1 m area per month = Rs
2 5000

12

Rent of 1320 m area per 3 month = Rs(
2 5000

12
× 3 × 1320)

= Rs (5000 x 330) = Rs 1650000
Therefore, the company had to pay Rs 1650000.

Page : 202 , Block Name : Exercise 12.1

Q3 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, nd the area painted in colour.

Answer. Sides of the triangular wall are 15 m, 11m, an 6m.
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)

2
= √16(16 − 15)(16 − 11)(16 − 6)m

2 2
= √16 × 1 × 5 × 10m = √800m
2
= 20√2m

Page : 203 , Block Name : Exercise 12.1

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Q4 Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is
42cm.

Answer. Let the third side of the triangle be x.
Perimeter of the given triangle 42 cm
18 cm + 10 cm + x = 42
x = 14 cm
Perimeter 42cm
s = = = 21cm
2 2

Area of a triangle == √s(s − a)(s − b)(s − c)
2
Area of the given triangle = (√21(21 − 18)(21 − 10)(21 − 14))cm

2
= (√21(3)(11)(7))cm

2
= 21√11cm

Page : 203 , Block Name : Exercise 12.1

Q5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

Answer. Let the common ratio between the sides of the given triangle be x.
Therefore, the side Of the triangle will be 12x, 17x, and 25x.
Perimeter of this triangle = 540 cm
12x + 17x + 25x = 540 cm
54x = 540 cm
x = 10 cm
Sides of the triangle will be 120 cm, 170 cm, and 250 cm.
Perimeter of triangle 540cm
s = = = 270cm
2 2

Area of triangle = √s(s − a)(s − b)(s − c)

2
= [√270(270 − 120)(270 − 170)(270 − 250)]cm

2
= [√270 × 150 × 100 × 20]cm

Therefore, the area Of this triangle is 9000 cm . 2

Page : 203 , Block Name : Exercise 12.1

Q6 An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the
area of the triangle.

Answer. Let the third side Of this triangle be x.
Perimeter of triangle = 30 cm
12 cm + 12 cm + x = 30 cm

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x = 6 cm
Perimeter of triangle 30cm
s = = = 15cm
2 2

Area of triangle = √s(s − a)(s − b)(s − c)

2
= [√15(15 − 12)(15 − 12)(15 − 6)]cm

2
= [√15(3)(3)(9)]cm

2
= 9√15cm

Page : 203 , Block Name : Exercise 12.1

Exercise 12.2

Q1 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. Let us join BD.
In A3CD, applying Pythagoras theorem,
2 2 2
BD = BC + CD

2 2
= (12) + (5)

= 144 + 25

2
BD = 169

BD = 13m

Area of ∆ BCD
1 1 2 2
= × BC × CD = ( × 12 × 5) m = 30m
2 2

For ΔABDr

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Perimeter (9+8+13)m
s = = = 15m
2 2

Area of triangle = √s(s − a)(s − b)(s − c)

2
Area of ΔABD = [√15(15 − 9)(15 − 8)(15 − 13)]m

2
= (√15 × 6 × 7 × 2)m

2
= 6√35m

2
= (6 × 5.916)m

2
= 35.496m

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

2 2 2
= 35.496 + 30m = 65.496m = 65.5m ( approximately )

Page : 206 , Block Name : Exercise 12.2

Q2 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 △ABC ,

2 2 2
AC = AB + BC

2 2 2
(5) = (3) + (4)

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

Area of ∆ ABC, = 1

2
× AB × BC =
1

2
× 3 × 4 = 6cm
2

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

For ΔADC,

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

s = 7cm

Area of triangle = √s(s − a)(s − b)(s − c)

2
Area of ΔADC = [√7(7 − 5)(7 − 5)(7 − 4)]cm

2
= (√7 × 2 × 2 × 3)cm

2
= 2√21cm

2
= (2 × 4.583)cm

2
= 9.166cm

Area of ABCD = Area of ΔABC + Area of ΔACD

2 2 2
= (6 + 9.166)cm = 15.166cm = 15.2cm ( approximately )

Page : 206 , Block Name : Exercise 12.2

Q3 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.

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

This triangle is an isosceles triangle.

Perimeter = 2s = (5 + 5 + 1)cm = 11cm
11cm
s = = 5.5cm
2

−√s(s−a)(s−b)(s−c)
Area of the triangle

2
= [√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√11cm

2
= (0.75 × 3.317)cm

2
= 2.488cm (approximately)

For quadrilateral II
This quadrilateral is a rectangle.
Area = I × b = (6.5 × 1)cm = 6.5cm 2 2

For quadrilateral Ill
This quadrilateral is a trapezium.
Perpendicular height of parallelogram = (√1 2 2
− (0.5) ) cm

= √0.75cm = 0.866cm

Area = Area of parallelogram + Area of equilateral triangle
√3
2 2
= (0.866)1 + (1) = 0.866 + 0.433 = 1.299cm
4

Area of triangle (IV) = Area of triangle in (V)
1 2 2
= ( × 1.5 × 6) cm = 4.5cm
2

Total area of the paper used = 2.488 + 6.5 + 1.299 + 4.5 x 2
2
= 19.287cm

Page : 206 , Block Name : Exercise 12.2

Q4 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, nd the
height of the parallelogram.

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Answer. For triangle
Perimeter of triangle = (26 + 28 + 30) cm = 84 cm
2s 84 cm
s = 42 cm
Area of triangle = = √s(s − a)(s − b)(s − c)
Area of triangle = [√42(42 − 26)(42 − 28)(42 − 30)]cm 2

= [√42(16)(14)(12)]cm
2
= 336cm
2
= 336 cm 2

Let the height of the parallelogram be h.
Area of parallelogram = Area of triangle
h x 28 cm = 336 cm 2

h = 12 cm
Therefore, the height of the parallelogram is 12 cm.

Page : 206 , Block Name : Exercise 12.2

Q5 A rhombus shaped eld 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 eld will each cow be getting?

Answer.

(48+30+30)cm
s = = 54m
2

Area of triangle = √s(s − a)(s − b)(s − c)

2
Therefore, area of ΔBCD = [√54(54 − 48)(54 − 30)(54 − 30)]m

2
= √54(6)(24)(24) = 3 × 6 × 24 = 432m

Area Of eld = 2 x Area Of ABCD
2 2
= (2 × 432)m = 864m

864 2
Area for grazing for 1 cow = = 48m
18

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

Page : 207 , Block Name : Exercise 12.2

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Book : Mathematics Ncert Solutions | Chapter-12 Maths

Q6 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?

Answer. For each triangular piece,
(20 + 50 + 50)cm
s = = 60cm
2

Semi-perimeter,

Area of triangle = √s(s − a)(s − b)(s − c)

2
Area of each triangular piece = [√60(60 − 50)(40 − 50)(60 − 20)]cm

2 2
= [√60(10)(10)(40)]cm = 200√6cm

Since there are 5 triangular pieces made of two different coloured cloths,
2
Area of each cloth required = (5 × 200√6)cm

2
= 1000√6cm

Page : 207 , Block Name : Exercise 12.2

Q7 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. How much paper
of each shade has been used in it?

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Answer. We know that
1 2
Area of square = ( diagonal )
2

1 2 2
Area of the given kite = (32cm) = 512cm
2

nd
Area of 1st shade = Area of 2 shade
2
512cm 2
= = 256cm
2

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

For III triangle

(6+6+8)cm
Semi-perimeter, s = = 10cm
2

Area of triangle = √s(s − a)(s − b)(s − c)

Area of IIIrd triangle = √10(10 − 6)(10 − 6)(10 − 8)

2
= (√10 × 4 × 4 × 2)cm

2
= (4 × 2√5)cm

2
= (4 × 2√5cm

2
= (8 × 2.24)cm

2
= 17.92cm

Area of paper required for I I I rd
shade = 17.92 cm 2

Page : 207 , Block Name : Exercise 12.2

Q8 A oral design on a oor is made up of 16 tiles which are triangular, the sides of the
triangle being 9 cm, 28 cm and 35 cm (see Fig.). Find the cost of polishing the tiles at the rate
of 50p per cm . 2

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Answer. It can be observed that
(35+28+9)cm
Semi-perimeter of each triangle-shaped tile,s = 2
= 36cm

By Heron’s formula,
Area of triangle= = √s(s − a)(s − b)(s − c)
2
Area of each tile = [√36(36 − 35)(36 − 28)(36 − 9)]cm

2
= [√36 × 1 × 8 × 27]cm

2
= 36√6cm

2
= (36 × 2.45)cm

2
= 88.2cm

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

Cost of polishing per cm area = 50 p
2

Cost of polishing 1411.2 cm area = Rs (1411.2 x 0.50) = Rs 705.60
2

Therefore. It will cost Rs 705.60 while polishing all the tiles.

Page : 207 , Block Name : Exercise 12.2

Q9 A eld 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 eld.

Answer.

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Draw a line BE parallel to AD and draw a perpendicular 3F on CO.
It can be observed that ABED is a parallelogram.
BE = AD = 13 m
ED = AB = 10 m
EC = 25 - ED = 15m
For ∆ BEC,
(13+14+15)m

Semi-perimeter, s=
2 = 21m

By Heron’s formula,
Area of triangle = √s(s − a)(s − b)(s − c)

2
Area of ΔBEC = [√21(21 − 13)(21 − 14)(21 − 15)]m

= [√21(8)(7)(6)]m2 =84m2

1
Area of ΔBEC = × CE × BF
2

2 1
84cm = × 15cm × BF
2

168
BF = ( ) cm = 11.2cm
15

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

2
Area of the field = 84 + 112 = 196m

Page : 207 , Block Name : Exercise 12.2

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