CHAPTER – 12
HERON‟S FORMULA
Exercise - 12.1
Q1.
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:
Side of an equilateral triangle = a.
Using Heron’s Formula:
Area of triangle =
ss  as  bs  c 
Where s is the semi-perimeter of triangle =
abc
2
And a, b, c are the sides of triangle.
Now, semi perimeter of an equilateral triangle is s 
Area of an equilateral triangle =
ss  as  as  a

3a  3a
 3a
 3a

 a 
 a 
 a

2  2
 2
 2


3a  3a  2a  3a  2a  3a  2a 




2  2  2  2 

3a a a a
3 2
  

a ……………(1)
2 2 2 2
4
Now perimeter of an equilateral triangle = 3a
Given, perimeter of an equilateral triangle = 180 cm
 3a  180 cm
a 
180
cm
3
 a  60 cm
a  a  a 3a

2
2
So, area of an equilateral triangle 
3 2
a
4
 3
2

 60 cm2
 4

 3

 
 60  60 cm2
 4

 900 3 cm2
Q2.
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 figure 12.9). The
advertisements yield an earning of Rs. 5000 per m2 per year.
A company hired one of its walls for 3 months. How much rent did it pay?
1 22
m
22 m
120 m
Fig. 12.9
Sol:
The sides of triangle are 122 m, 22 m and 120 m.
Perimeter of triangle = Sum of sides
 122 m  22 m  120 m
 264 m
Semi-perimeter of  , s 

Perimeter
2
264
m  132 m
2
By Heron’s Formula:
Area of triangle =
ss  as  bs  c 
 132132  122132  22132  120 m2
 132  10  110  12 m2
 11  12  10  11  10  12 m2
 11  12  10  m 2
 1320 m2
Rent of 1m2 per year = Rs. 5000.
 Rent of 1m2 per month = Rs.
5000
12
 Rent of 1320 m2 per month = Rs.
( 1 year = 12 months)
5000 1320
12
 Rent of 1320 m2 for three months = Rs.
5000  1320  3
12
 Rs.550000 3
 Rs.1650000
So, company paid Rs. 16,50,000 as rent.
Q3.
There is a slide in a park. One of its side walls has been painted in some
color with a message “KEEP THE PARK GREEN AND CLEAN” (See fig.
12.10) If the sides of the walls are 15 m, 11 m and 6 m, find the area painted
in color.
6m
11 m
KEEP THE PARK
GREEN & CLEAN
15 m
Fig. 12.10
Sol:
Here the area painted is in the shape of a triangle with sides 11m, 6 m and 15 m.
Perimeter of triangle = 11 m  6 m  15 m  32 m
Semi-perimeter of triangle =
Perimeter 32

 16 m
2
2
By Heron’s Formula
The area of triangle  ss  as  bs  c 
 1616  1116  616  15 m2
 16  5  10  1 m 2
 4  4  5  5  2 m2
 4  5 2 m2
 20 2 m 2
So, 20 2 m 2 is the required area.
Q4.
Find the area of a triangle, two sides of which are 18 cm and 10 cm and
perimeter is 42 cm.
Sol:
Now two sides of  are 18 cm and 10 cm.
Let the third side of given triangle be x.
Perimeter of triangle = 42 cm
 18  10  x  cm  42 cm
 28 cm  x  42 cm
 x  42  28  cm
 x  14 cm
 Semi-perimeter of triangle s =
Perimeter
2

42
 21cm
2
By Heron’s Formula,
Area of the triangle  ss  as  bs  c 
 2121 1821 1021 14 cm2
 21  3  11  7 cm2
 3  7  3  11  7 cm2
 21 11 cm2
So, area of the triangle is 21 11 cm2 .
Q5.
Sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540 cm.
Find its area.
Sol:
The ratio of sides is 12:17:25
Let the common ratio of given sides be x.
Then the sides are 12x,17x and 25x respectively.
Perimeter of triangle = 540 cm
 12x  25x  17x  540 cm
 54x  540 cm
x
540
cm
54
 x  10 cm
So, a  12x  12  10  120 cm
b  17x  17  10  170 cm
c  25x  25  10  250 cm
s
Perimeter 540

 270 cm
2
2
By Heron’s Formula:
Area of the triangle  ss  as  bs  c 
 270270  120270  170270  250 cm2
 270  150  100  20 cm2
 3  3  3  10  3  5  10  10  10  2  2  5 cm2
 3  3  10  10  2  5
 9  1000  9000 cm2
So, area of the given triangle is 9000 cm2 .
Q6.
An isosceles triangle has perimeter 30 cm and each of equal sides is 12
cm. Find the area of the triangle.
Sol:
Equal sides of isosceles triangle is 12 cm each.
Let the third side of given triangle be x.
Perimeter of triangle = 30 cm
 12  12  x  cm  30 cm
 24 cm  x  30 cm
 x  30  24  cm
 x  6 cm
 Third side of the  = 6 cm
Semi- perimeter of given triangle is =
30
cm  15 cm
2
By Heron’s formula:
Area of triangle  ss  as  bs  c 
 1515  1215  1215  6 cm2
 15  3  3  9 cm2
 15  3  3  3  3 cm2
 3  3 15 cm2  9 15 cm2
So, area of the given triangle is 9 15 cm2 .
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?
Sol : Park is in the shape of a quadrilateral.
A
9m
8m
13 m
B
Join BD.
Consider  BCD, BCD  90o
By Pythagoras Theorem
BD 2  BC 2  CD 2
 BD   12 m  5 m
2
2
 BD   144 m 2  25 m 2
2
 BD   144  25  m 2
2
 BD  169 m2
2
 BD  13 m
2
2
Taking square root on both sides, we get:
BD2

5m
C
12 m
2
D
13 m
 BD  13 m
By Heron’s formula:
Area of ABD  ss  as  bs  c 
Where s = semi-perimeter of 
a, b, c are sides of triangle.
So, s 
AB  BD  AD  9  13  8 

m
2
2


 30 
s    m  15 m
 2 
a  AB  9 m
b  BD  13 m
c  AD  8 m
ar  ABD  ss  as  bs  c 
 1515  915  1315  8 m2
 15  6  2  7 m2
 3  5  3  2  2  7 m2
 6 35 m 2
Consider BCD
a  BC  12 m
b  CD  5 m
c  BD  13 m
 BC  CD  BD  12 m  5 m  13 m 30 m
s


2
2
2


= 15 m
ar  BCD  ss  as  bs  c 
(By Heron’s formula)
 1515  1215  515  13 m2
 15  3  10  2 m 2
 3  5  3  2  5  2 m2
 3  5  2 m2
 30 m2
Now, from figure it is clear:
ar ABCD   ar  ABD   ar  BCD


 6 35  30 m 2
 6  5.91  30  m2
( 35  5.91 )
 65.46 m2 (approx)
So, area of the park is 65.46 m2 (approx).
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.
Sol : Given: Quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm
and AC = 5 cm
A
5 cm
B
4c
m
3 cm
5
D
cm
4c
m
C
Consider ABC ,
a  AB  3 cm
b  BC  4 cm
c  AC  5 cm
Semi-perimeter, s 
AB  BC  AC  3  4  5 

cm
2
2



areaABC  ss  as  bs  c 
 66  36  46  5 cm2
 6  3  2  1 cm2
 3  2  3  2 cm2
12
 6 cm
2
 3  2 cm2  6 cm2
Consider ACD
a  AC  5 cm
b  CD  4 cm
c  AD  5 cm
Semi-perimeter, s 
AC  CD  AD 5 cm  4 cm  5 cm

2
2

14
cm  7 cm
2
areaACD  ss  as  bs  c 
 77  57  47  5 cm2
 7  2  3  2 cm2
 2 21 cm2
Now, from figure it is clear that :
area ABCD   area ACD   area ABC


 2 21  6 cm2
 2  4.58   6 cm2
[
21  4.58 ]
 9.16  6  cm2
 15.16 cm2 (approx)
So, area of quadrilateral ABCD is 15.16 cm2 (approx).
Q3.
Radha made a picture of an aero plane with colored paper as shown in Fig.
12.15. Find the total area of the paper used.
5 cm
I
IV
6 cm
6.5 cm
1.5 cm
V
II
1c
m
III
m
1c
1 cm
2 cm
Fig. 12.15
Sol : From figure we observe that there are five parts of aero plane named as I, II, III,
IV and V, where IV and V are identical regions.
Firstly, we will calculate the area of paper used in part I.
In part I, we have a triangle with sides 5 cm, 5 cm and 1 cm.
area of triangle I =
ss  as  bs  c 
Here a  5 cm , b  5 cm , c = 1 cm
s
a  b  c  5  5  1
11

cm
cm 
2
2
2


Area of triangle I =

11  11
 11
 11 
2
  5   5   1 cm
22
 2
 2

11 1 1 9
   cm 2
2 2 2 2

11  1 1 3  3
cm 2
2222

3
11 cm2
2 2

3
 3.32 cm2
4

9.96
cm2  2.49 cm2
4
[ 11  3.32 ]
Area of quadrilateral II:
The quadrilateral II is a rectangle with length 6.5 cm and breadth of 1 cm.
So, area of rectangle = length breadth
 Area of II = 6.5  1 cm2 = 6.5 cm2
Area of quadrilateral III:
Quadrilateral III is a trapezium with equal non parallel sides.
We divide it into a parallelogram and an equilateral triangle.
1c
m
h
m
1c
1 cm
1 cm
1 cm
1 cm
2 cm
Perpendicular height of parallelogram =
12  0.52
 1 0.25
 0.75 

Area of parallelogram = height base
5
3 cm
10
75

100
553
10  10
 5


3  1cm2
 10

 0.5 3 cm2
Area of equilateral triangle 
3 2
a
4
where a is the side of triangle.
3
2
 1 cm2
4
Area of given triangle 

3
 1 cm 2
4

3
cm 2
4

1
3 cm2
4
 0.25 3 cm2
Area of trapezium = area of ||gm + area of equilateral 


 0.5 3  0.25 3 cm2
 0.5  0.25  3 cm2
 0.75 3 cm2
 0.75  1.73 cm2
[
 1.2975 cm 2
Area of triangle IV: [It is a right angled triangle.]
Here, height = 6 cm
Base of triangle = 1.5 cm
Area of triangle =
1
 base  height
2
3  1.73 ]
1

   1.5  6 cm2
2

 4.5 cm2
Area of triangle V:
area  IV   area  V 
 area of triangle V = 4.5 cm2
So,
area of paper used = area of triangle I + area of quadrilateral II
+ area of quadrilateral III + area of triangle IV + area of
triangle V
 2.49  1.2975  6.5  4.5  4.5 cm 2
 19.2875 cm2
 19.29 cm 2 (approx)
So, 19.29 cm 2 paper is used for aero plane.
Q4.
A triangle and a parallelogram have the same base and 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 parallelogram.
Sol :
26 c
m
A E
B
Area of triangle ABC,
a  AB  26 cm
b  BC  28 cm
c  AC  30 cm
30
28 cm D
cm
28 cm
C
F
Semi-perimeter s 
AB  BC  CA
2

26  28  30
cm
2

84
cm  42 cm
2
areaABC  4242  2642  2842  30 cm2
(By using Heron’s formula)
 42  16  14  12 cm2
 2  3  7  2  2  2  2  2  7  2  2  3 cm2
 2  3  7  2  2  2 cm2
 336 cm2
Area of parallelogram = height base
area BCDE   DF  BC
 DF 28 cm
According to question,
area BCDE   area ABC
DF  28 cm  336 cm2
 336 
DF  
cm
 28 
[Divide by 28 on both sides]
 DF  12 cm
So, the height of parallelogram is 12 cm.
Q5.
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 : Let ABCD be a rhombus shaped field.
30 m
D
30 m
48
C
m
30 m
A
B
30 m
Area of ABC :
a  AB  30 m
b  BC  30 m
c  CA  48 m
Semi-perimeter, s 
AB  BC  CA 30  30  48

m
2
2

108
m  54 m
2
areaABC  ss  as  bs  c 
[By Heron’s formula]
 5454  3054  3054  48 m2
 54  24  24  6 m2
 3  3  2  3  2  2  2  3  2  2  2  3  2  3 m2
 3  3  3  2  2  2  2 m 2
 432 m2
As we know that a diagonal divides the rhombus into two triangles of equal area.
 Area of rhombus  2  area of a triangle
 area ABCD   2  area ABC
 2  432 m2
 864 m2
Therefore, area for grazing for 18 cows = 864 m2
 Area for grazing for 1 cow =
864 2
m = 48 m 2
18
So, each cow will graze in 48 m2 of field.
Q6.
An umbrella is made by stitching 10 triangular pieces of clothes of two
different colors (see figure 12.16), each piece measuring 20 cm, 50 cm and
50 cm. How much cloth of each color is required for the umbrella?
50
cm
20
cm
Fig. 12.16
Sol : For each triangular piece of umbrella,
a = 50 cm
b = 50 cm
c = 20 cm
Semi-perimeter, s 
a  b  c 50  50  20

cm
2
2

120
cm
2
 60 cm
Area of each piece  ss  as  bs  c 
[By Heron’s formula]
 6060  5060  5060  20 cm2
 60  10  10  40 cm2
 2  3  10  10  10  2  2  10 cm2
 2  10  10 2  3 cm2
 200 6 cm2
Here are total 10 pieces of triangular shapes of 2 different colors.
 There are 5 pieces of each color.
 Area of cloth of one color = 5  area of one triangle
 5  200 6 cm2
 1000 6 cm2
Q7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles
triangle of base 8 cm and equal sides 6 cm each is to be made of three
different shades as shown in figure 12.17. How much paper of each shade
has been used in it?
I
II
III
Fig. 12.17
Sol:
In square, we are given diagonal = 32 cm
Area of square =
1
diagonal2
2

1
2
 32 cm2
2

1
 1024 cm2
2
 512 cm2
We know that a diagonal of square divides it into two triangles of equal area.
 Area of shaded region I = area of shaded region II.

1
area of square
2

1
 512 cm2
2
 256 cm2
So, area of shaded region I = 256 cm2
Area of shaded region II = 256 cm2
Area of triangle III,
a = 6 cm
b = 6 cm
c = 8 cm
Semi-perimeter, s 
abc 668

cm
2
2


 20 
  cm  10 cm
 2 
By Heron’s formula:
Area of triangle =
ss  as  bs  c 
 1010  610  610  8 cm2
 10  4  4  2 cm2
 2  5  4  4  2 cm2
 2  4 5 cm2
 8 5 cm 2
Area of shaded region III = 8 5 cm2
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 12.18). Find
the cost of polishing the tiles at the rate of 50 paise per cm2.
35
28
cm
cm c m
28
9 cm
Q8.
9 cm
Fig. 12.18
Sol:
For each triangular tile,
a = 28 cm
b = 9 cm
c = 35 cm
Semi perimeter, s 
abc
2

28  9  35
cm
2

72
cm
2
 36 cm
By Heron’s formula:
Area of a triangular tile  ss  as  bs  c  cm2
 3636  2836  936  35 cm2
 36  8  27  1 cm2
 6  6  2  2  2  3  3  3 cm2
 6  2  3 2  3 cm2
 36 6 cm2
 36  2.45  cm2
[ 6  2.45 ]
 88.2 cm2
Area of 1 triangular tile = 88.2 cm 2
Area of 16 triangular tiles  16 88.2 cm2
 1411.2 cm2
Cost of polishing per cm2 area = 50 p
Cost of polishing 1411.2 cm2 area = 1411.2  50 p
 70560.0 p
 Rs.
70560
 Rs.705.60
100
Therefore, it will cost Rs.705.60 for polishing the tiles.
Q9.
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 : Let ABCD be a trapezium shaped field.
A
14 m
10 m
B
14 m
E
D
13 m
F
C
15 m
10 m
25 m
We divide it into a parallelogram ABED and a triangle BEC.
We draw BF  DC .
So, area of BEC 
1
 base  height
2

1
 EC  BF
2

1
 15  BF m2 ………………….(1)
2
By Heron’s formula:
areaBCE  ss  as  bs  c 
Where, a  BE  14 m
b  EC  15 m
c  BC  13 m
s
a  b  c 14  15  13
42

m
m  21m
2
2
2
So,
areaBCE  2121 1421 1521 13 m2
 21  7  6  8 m 2
 3  7  7  2  3  2  2  2 m2
 3  7  2  2 m2
 84 m2 …………………….(2)
From (1) and (2),
1
 15  BF  84 m2 ……………….(2)
2
2 

BF =  84   m
15 

[multiply by
2
on both sides]
15
56
 168 

 11 .2 m
m
5
 15 
Now BF is the perpendicular height between two parallel lines.
So, area of parallelogram ABED  base height
 DE BF
 10  11 .2 m 2
 112 m2
Now, area ABCD   area ABED  area BEC
 112  84  m 2
 196 m2
So, area of field is 196 m2 .