Lesson: Heron's Formula

Exercise 12.1

Question: 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?

Solution

Let the length of the side of the equilateral triangle be

∴ The perimeter of the equilateral triangle

  3a 180 cm

cm60a  

Semi perimeter of the signal board



 

Using Heron's formula,

the area of the signal board =

   

( )

Sides of an equilateral triangle are equal

3 3 3 3

2 2 2 2

900 3

a a a a

a a a

a a

s s a s

a a

b s c

   

   

   

   

   

  

 

  







  





Question: 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 Rs 5000 per m

per year. A

company hired one of its walls for 3 months. How

much rent did it pay?

Fig.

12.9

Solution

The sides of the triangle are 122 m, 22 m and 120 m.

Hence,

, , and122 m 22 m 120 m.a b c  

The perimeter of the triangle is

m.122 22 120 264  

The semi perimeter of the triangle (s)

m264 / 2 132 

 s 132

By Heron's formula,

Area of the triangular wall =

 

  

   

   



s s a s b s c

132(132 122)(132 22)(132 120)

132 10 110 12 m

1320 m

Earning due to advertising per year =

5000Rs

per m

Hence, the rent of one wall for 3 months

Rs Rs1320 5000 3( / 12 165000) 0   

Question: 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.

Fig. 12.10

Solution

The sides of the triangular wall are 15 m, 11 m and 6

Hence, the semi perimeter of the triangular

wall

m m( )15 11 6 / 2 16   

16 s

By Heron's formula,

the area painted=

 

16(16 15)(16 11)(16 6)

16 1 5 10 m

800 m

20 m

s s a s b s c  

   

   



Question: 4

Find the area of a triangle two sides of which

are 18 cm and 10 cm and the perimeter is 42 cm.

Solution

The perimeter of the triangle

42

The two sides of the triangle

18

cm and 10 cm.

Hence, the third side of the triangle

cm cm.( )42 18 10 14   

The semi perimeter of the triangle

 

cm21 s 

By Heron's formula,

Area of the triangle =

 

3 7

21 11

  

   

   



s s a s b s c

21(21 18)(21 10)(21 14)

21 11 cm

Question: 5

Sides of a triangle are in the ratio of

12:17: 25

and

its perimeter is 540 cm. Find its area.

Solution

Ratio of the sides of the triangle

12:17: 25

Let the common ratio be x, then the sides of the

triangle will be

12 , 17x x

and

25 ,x

The perimeter of the triangle

540

cm.

cm12 17 25 540x x x   

cm54 540x 

10x 

Hence, the sides of the triangle will be,

cm12 12 10 120x   

cm17 17 10 170x   

cm25 25 10 250x   

The semi perimeter of triangle

540/ 2 270 

cm.

270 cms 

By Heron's formula,

Area of the triangle =

 

s s a s b s c  

 

270 270 270 270a b c   

120   270(270 )(270 170)(270 250)

270 150 100 20 cm   

cm9000

Question :6

An isosceles triangle has perimeter 30 cm and each of

the equal sides is 12 cm. Find the area of the triangle.

Solution

The length of the equal sides of the isosceles triangle

cm12

The perimeter of the triangle

cm.30

Hence, the length of the third side

cm cm30 12 12) 6(   

The semi perimeter of the triangle

cm cm30/ 2 15 

cm15 s 

By Heron's formula,

Area of the triangle =

 

s s a s b s c  

   

15 15 12 15 12 15 6   

cm15 3 3 9   

cm9 15

Exercise 12.2

Question: 1

A park, in the shape of a quadrilateral ABCD, has

m90º, 9 ,C AB  

m m12 , 5BC CD 

and

m8AD 

. How much area does it occupy?

Solution

Given,

m90º, 9 ,C AB  

m m12 , 5BC CD 

and

m8 .AD 

Let us join BD.

Now, in ΔBCD,

2 2 2

BD BC CD 

(Pythagoras theorem)

2 2 2

12 5BD  

169BD 

m13BD 

Area of

1/ 2 12 5 30BCD    

Now,

Semi perimeter of ΔABD (s)

m m m(8 9 13 / 2 3) 0/ 2 15    

m15s 

Using Heron's formula,

Area of ΔABD =

 

s s a s b s c  

   

15 15 13 15 9 15 8   

2 6 7 m15   

m6 35

m (approx.)35.5

Hence, the area of the quadrilateral ABCD = Area of

ΔBCD + Area of

m m m

2 2 2

30 35.5 65.5 ABD   

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.

Solution

The sides of the quadrilateral ABCD are,

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

The diagonal AC = 5 cm.

Now, in ΔABC,

2 2 2

AC AB BC 

(Pythagoras theorem).

2 2 2

5 3 4  

25 25 

Hence, ΔABC is a right angled at B.

∴ Area of ΔABC = 1/2 × 3 × 4 = 6 cm

Now,

The semi perimeter of ΔACD ,s = (5 + 5 + 4)/2 cm =

14/2 cm = 7 cm

cm7 s 

By Heron's formula,

Area of ΔACD =

 

5 4

7 2 2 3

cm (approx.)

7 7 5 7 7

2 21

9.17

s s a s b s c  

   

   



Hence, the area of the quadrilateral ABCD = Area of

ΔABC + Area of ΔACD = 6 cm

+ 9.17 cm

= 15.17

Question: 3

Radha made a picture of an airplane with coloured

paper as shown in Fig 12.15. Find the total area of the

paper used.

Fig 12.15

Solution

As shown in the figure, there are five sections in the

picture of the aeroplane. Hence, the total area will be

sum of the areas of the five sections.

Now,

The length of the sides of the triangle in section I are

5 cm, 1 cm and 5 cm.

The perimeter of the triangle = 5 + 5 + 1 = 11 cm.

The semi perimeter = 11/2 cm = 5.5 cm

cm5.5 s 

By Heron's formula,

Area of section I =

 

5 1

5.5 0.5 0.5 4.5

0.75 3.317

2.488 ( )

cm approx.

5.5 5.5 5 5.5 5.5

0.75 11

s s a s b s c  

   

   



 



The lengths of the sides of the rectangle of section II

are 6.5 cm and 1 cm.

Hence, the area of section II = 6.5 × 1 = 6.5 cm

Section III is an isosceles trapezium which is divided

into three equilateral triangles of side 1 cm each.

Hence, the area of the trapezium = 3 × √3/4 ×

= 1.3 cm

(approx)

Section IV and V are 2 congruent right angled

triangles with base 6 cm and height 1.5 cm.

Area of the region IV and V = 2 × 1/2 × 6 × 1.5

= 9 cm

Hence, the total area of the paper = (2.488 + 6.5 + 1.3

+ 9) cm

= 19.3 cm

Question: 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.

Solution

The lengths of the sides of the triangles are 26 cm, 28

cm and 30 cm.

The perimeter of the triangle = 26 + 28 + 30 = 84 cm.

The semi perimeter of the triangle = 84/2 cm = 42 cm

42 cm s 

By Heron's formula,

Area of the triangle =

 

   

26 28 30

16 14 12

  

   

   



s s a s b s c

42 42 42 42

336

Let the height of parallelogram be h.

Area of the parallelogram and area of the triangle are

equal.

⟹ Area of parallelogram = Area of triangle

28 cm × h = 336 cm

⟹ h = 336/28 cm

⟹ h = 12 cm

Hence, 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?

Solution

Let the rhombus be ABCD, with sides AB, BC, CD and

DA.

The diagonal AC divides the rhombus ABCD into two

congruent triangles of equal area.

The semi perimeter of ΔABC = (30 + 30 + 48)/2 m =

54 m

54 m s 

By Heron's formula,

Area of the ΔABC =

   

30 30 48

14 14 6

252

  

   

   



s s a s b s c

54 54 54 54

Thus, the area of field = 2 × area of the ΔABC = (2 ×

252) m

= 504 m

Hence, the area of the grass field each cow will get =

504/18 m

= 28 m

Question: 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?

Fig.12.16

Solution

The lengths of the sides of the triangle are 20 cm, 50

cm and 50 cm.

Thus, the semi perimeter of each triangular piece, s =

(50 + 50 + 20)/2 cm = 120/2 cm = 60 cm.

By Heron's formula,

the area of the triangular piece =

 

   

50 50 20

10 10 40

200

60 60 60 60

s s a s b s c  

   

   



∴ The area of the triangular piece of one colour= 5 ×

200√6 cm

= 1000√6 cm

Hence, 1000√6 cm

cloth of each colour is required for

the umbrella.

Question: 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?

Fig.

12.17

Solution

Diagonals of a square bisect each other at right angle.

Hence, the area of the square = 1/2 (diagonal)

= 1/2 × 32 × 32 = 512 cm

∵ Area of shade I = Area of shade II

⟹ 512/2 cm

= 256 cm

So, the area of paper required in each shade = 256

For III section,

The length of the sides of the triangle are 6 cm, 6 cm

and 8 cm.

The semi perimeter of the triangle =

(6 )6

8 

cm =

10 cm

10 cm s 

By Heron's formula,

Area of the III triangular pieces =

 

   

6 6 8

4 4 2

8 5

  

   

   



s s a s b s c

10 10 10 10

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 Fig. 12.18). Find the cost of

polishing the tiles at the rate of 50p per cm

Fig. 12.18

Solution

The lengths of the sides of the triangles are 9 cm, 28

cm and 35 cm.

The semi perimeter of each triangular shape = (28 + 9

+ 35)/2 cm = 36 cm

36 s cm

Using Heron's formula,

the area of each triangular shape =

 

   

28 9 35

8 27 1

88.2

36 36 36 36

s s a s b s c  

   

   



Thus, the total area of 16 tiles

16 88.2 1411.2  

Cost of polishing the tiles = 50p per cm

Hence, the total cost of polishing the tiles = Rs

(1411.2 × 0.5) = Rs 705.6

Question: 9

A field is in the shape of a trapezium whose parallel

sides are 25 m and 10 m long. The non-parallel sides

are 14 m and 13 m. Find the area of the field.

Solution

Let ABCD be the given trapezium with the parallel

sides as 25 m and 10 m and the non-parallel sides as

14 m and 13 m as shown in the figure above.

Let the parallel sides are AB = 25 m and CD = 10 m

and the non-parallel sides are AD = 13 m and BC =

14 m.

We draw a perpendicular line from vertex C to AB

such that CM ⊥ AB. Also, draw a line parallel to AD

from point C such that it intersects AB at point E.

Now, in ΔBCE,

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

m25 10 15 .BE AB AE    

The semi perimeter of the ΔBCE =

15 13( )

14 

21 m

21 ms 

By Heron's formula,

Area of the ΔBCE =

 

   

14 13 15

7 8 6

21 21 21 21

s s a s b s c  

   

   



Also, the area of the

BCE BE CM    

15 84

CM   

168

CM 

Area of the parallelogram AECD = Base × Altitude

10 112

AE CM    

Area of the trapezium ABCD = Area of AECD + Area

of ΔBCE



(112 + 84) m

= 196 m