12 HERON’S FORMULA CHAPTER

CHAPTER
12
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HERON’S FORMULA
Points to Remember :
1
 Base  height
2
3
(side) 2
2. Area of an Equilateral triangle 
4
a
3. Area of a isosceles triangle 
4b 2  a 2 , where, a is base and b represents equal sides.
4
4. Heron’s Formula : If a, b, c denote the lengths of the sides of a triangle, then its Area,
1. Area of right triangle 
A
s ( s  a )( s  b )( s  c )
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abc
2
5. Area of a quadrilateral can be calculated by dividing the quadrilateral into triangles and using heron’s
formula for calculating area of each triangle.
where,
s
B
ILLUSTRATIVE EXAMPLES
Example 1. Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm respectively.
Solution.
Let a = 13 cm, b = 14 cm, c = 15 cm
a  b  c 13  14  15

 21 cm
 s
2
2
Area  s( s  a )( s  b )( s  c)
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 21( 21  13)( 21  14)(21  15) sq cm.
 21 8  7  6 sq. cm  3  7  2  2  2  7  2  3 sq. cm
= 2× 2 × 3 × 7 sq. cm. = 84 sq. cm.
 Area of triangle = 84 sq. cm. Ans.
Example 2. 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 it perimeter is 180 cm.
—NCERT
Solution.
Given perimeter of equilateral triangle = 180 cm.
Side of an equilateal triangle is a cm, then its perimeter is 3a cm.
 3a = 180  a = 60 cm
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Now, s  a  a  a  60  60  60  180  90 cm.
2
2
2
Using heron’s formula,
Area
 s( s  a )( s  a )( s  a )
 90(90  60)(90  60)(90  60) cm 2
 90  30  30  30 cm 2  3  3  3  3  3 10000 cm 2
 3  3 100 3 cm 2  900 3 cm 2 Ans.
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135
Example 3. A triangular park ABC has sides 120 m, 80 m and 50 m (see figure). A gardener Dhania has to put
a fence all around it and also plant grass inside. How much area does she need to plant? Find the
cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3 m wide for a gate
on one side.
Solution.
(i) For Area of the park :
Let a = 120 m, b = 80 m, c = 50 m
a  b  c 120  80  50

 125 m
2
2

s

Area  s( s  a )( s  b )( s  c)
B
 125 (125  120)(125  80)(125  50) m 2
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 125  5  45  75 m 2  375 15 m 2
Also, Perimeter of the park = AB + BC + AC = 250 m
 length of wire needed for fencing = 250 m – 3 m (to left for gate)
= 247 m
and cost of fencing = 247 × 20 Rs = Rs. 4940 Ans.
Example 4. 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). 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?
—NCERT
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MATHEMATICS–IX
Solution.
We will calculate area of triangular wall
here, a = 122 m, b = 22 m, c = 120 m
s

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a  b  c 122  22  120
264

m
m  132 m
2
2
2
Area  s( s  a )( s  b)( s  c)
 132 (132  122) (132  22) (132  120) m 2
 132 10 110 12 m 2
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 11 12 10 10  11 12 m 2  10  11 12 m 2  1320 m 2
Now, Rent charges = Rs. 5000 per m2 per year
 Rent charged from company for 3 months
3
12
= Rs. 16,50,000 Ans.
Example 5. There is a slide in a park. One of the its side walls has been painted, in colour with a message
‘‘KEEP THE PARK GREEN AND CLEAN’’, (see figure). If the sides of the wall are 15 m, 11 m and
6 m, find the area painted in colour.
—NCERT
 Rs. 5000  1320 
Solution.
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here,
a = 15 m, b = 11m, c = 6m

s
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Area
B
a  b  c 15  11  6
32

m
m  16 m
2
2
2
 s ( s  a )( s  b)( s  c)
 16(16  15)(16  11)(16  6) m 2  16  1 5  10 m 2
 4  4  5  2  5 m 2  4  5 2 m 2  20 2 m 2
 Area painted in a colour = Area of side wall
= 20 2 m 2 Ans.
Example 6. An isosceles triangle has perimeter 30 m and each of the equal sides is 12 cm. Find the area of the
triangle.
—NCERT
Solution.
here, a = b = 12 cm.
also, a + b + c = 30 m  12 + 12 + c = 30 cm  c = 6 cm
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137
abc
30
cm 
cm  15 cm
2
2
 Area of the triangle
s
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 s ( s  a )( s  b)( s  c)
 15(15  12)(15  12)(15  6) cm 2
 15  3  3  9 cm 2  5  3  3  3  3  3 cm 2
 3  3  5  3 cm 2  9 15 cm 2 Ans.
Example 7. The perimeter of a triangle is 450 m and its sides are in the ratio of 13 : 12 : 5. Find the area of the
triangle.
Solution.
Let the sides of the triangle by 13 x, 12 x and 5x.
Given, perimeter of triangle = 450 m
 13 x + 12 x + 5 x = 450
 30x = 450

x = 15
 Sides are 13 × 15 m, 12 × 15 m and 5 × 15 m
i.e. 195 m, 180 m and 75 m.
Let a = 195 m, b = 180 m, c = 75 m
s
B
a  b  c 195  180  175 450


m  225 m
2
2
2
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 Area  s( s  a)(s  b)(s  c)
 225(225  195)(225  180)(225  75) sq. m
 225  30  45 150 sq. m.
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 15 15  2 15  3 15  5  5  2  3 sq. m.
= 15 × 15 × 2 × 3 × 5 sq. m.
= 6750 sq. m Ans.
Example 8. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5m and
AD = 8 m. How much area does it occupy?
—NCERT
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Solution.
1
1
 BC  CD   12  5 m 2  30 m 2
2
2
Now, In BCD, BD2 = BC2 + CD2
Area of ΔBCD 
 BD  12 2  52 m  144  25 m  169 m  13 m
For ABD, let a = 13 m, b = 8 m, c = 9 m.
Now, s 
a  b  c 13  8  9
30

m
m  15 m
2
2
2
 Area of ΔABD 
s ( s  a )( s  b )( s  c )
 15(15  13)(15  8)(15  9) m 2
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HERON’S FORMULA
MATHEMATICS–IX
 15  2  7  6 m 2  2  2  3  3  5  7 m 2
 2  3 35 m 2  6 35 m 2  6  5.9 m 2 (approx.)
= 35.4 m2 (approx.)
 Required Area = ar (ABD) + ar (BCD)
= 35.4 m2 + 30 m2 = 65.4 m2 (approx.) Ans.
Example 9. A parallelogram, the length of whose sides are 60 m and 25 m has one diagonal 65 m long. Find the
area of the parallelogram.
Solution.
Let ABCD be the given parallelogram.
Area of parallelogram ABCD
= area of ABC + area of ACD
= 2 area of ABC
Let, a = 60 m, b = 65 m, c = 25 m
s
60  65  25
m  75 m
2
 area of ABC  s( s  a )( s  b )( s  c)
 75(75  60)(75  65)(75  25) m 2
2
B
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 75 15 10  50 m
= 5 × 3 × 5 × 2 × 5 m2 = 750 m2
 Area of parallelogram ABCD
= 2 × area of ABC = 2 × 750 m2
= 1500 m2 Ans.
Example 10. 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.
—NCERT
Solution.
For the triangle ;
Let a = 26 cm, b = 28 cm, c = 30 cm
then, s 
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a  b  c 26 cm  28 cm  30 cm 84


cm  42 cm
2
2
2
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Now, Area of triangle  s( s  a )( s  b )( s  c)
 42(42  26)(42  28)(42  30) cm 2
 42  16 14 12 cm 2
 2  3  7  4  4  2  7  2  2  3 cm 2
 2  2  2  2  3  3  4  4  7  7 cm 2
= 2 × 2 × 3 × 4 × 7 cm2 = 336 cm2
For the parallelogram :
Area = base × height

Area
( Area of parallelogram = Area of triangle)
base
336

cm  12 cm. Ans.
28
height 
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Example 11. Find the area of a quadrilateral ABCD, in which B = 90°, AB = 15 cm, BC = 36 cm, CA = 24 cm and
DA = 21 cm.
Solution.
Join A to C.
In ABC, B = 90°
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 36  15 cm 2
2
= 270 cm2
 Area of ABC 
also, In ABC,
AC2 = AB2 + BC2 = 152 + 362 = 225 + 1296 = 1521
 AC  1521 cm  39 cm
Now, For area of ACD, we will use hero’s formula.
let, a = 39 cm, b = 24 cm, c = 21 cm.
s
a  b  c 39  24  21

 42 cm
2
2
 Area of ACD  s( s  a )( s  b )( s  c)
 42 (42  39)(42  24)(42  21) cm 2
2
B
2
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2
 42  3  18  21 cm  126 3 cm  218.232 cm
 Area of ABCD = 270 cm2 + 218.232 cm2 = 488.232 cm2 Ans.
Example 12. Radha made a picture of an aeroplane with coloured paper as shown in the figure. Find the total
area of the paper used.
—NCERT
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Solution.
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Area I = area of isosceles triangle with a = 1 cm and b = 5 cm
a

4b 2  a 2
4
1

4  25  (1) 2 cm 2
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MATHEMATICS–IX
99
cm 2  2.5 cm 2 (approx)
4
Area II = area of rectangle = length × breadth
= 6.5 × 1 cm2 = 6.5 cm2
Area III = Area of Trapezium
= 3 × area of equilateral triangle with a = 1 cm

3
3  1.732
5.196
 (1) 2 cm 2 
cm 2 
cm 2
4
4
4
= 1.299 cm2  1.3 cm2 (approx)
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= 3
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1
2
2
Area of IV and V  2   6 1.5 cm  9 cm
2
 Total area of paper used
= area I + area II + area III + area IV + area V
= 2.5 cm2 + 6.5 cm2 + 1.3 cm2 + 9 cm2
= 19.3 cm2 (approx) Ans.
Example 13. An Umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure),
each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the
umbrella?
—NCERT
Solution.
In one triangular piece,
a = 20 cm, b = 50 cm, c = 50 cm.
Now, s 
B
20  50  50 cm 120

cm  60 cm
2
2
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 Area  s ( s  a )( s  b)( s  c)
 60 (60  20)(60  50)(60  50) cm 2
2
 60  40  10 10 cm  200 6 cm 2
 Area of 5, Ist coloured triangles = 5 × 200 6 cm2 = 1000 6 cm2
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and area of 5, IInd coloured triangles = 5 × 200 6 cm2 = 1000 6 cm2
Example 14. 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.
—NCERT
Solution.
Each diagonal of square = 32 cm
1
 CO   32 cm  16 cm
2
 Area of each shaded portion I and II
1
  base  height
2
1
  32 16 cm 2  256 cm 2
2
And, ara of IIIrd portion with sides 6 cm, 6 cm and 8 cm.
a
8

4b 2  a 2 
4(6) 2  (8) 2 cm 2
4
4
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 2  144  64 cm 2  2  80 cm 2
 8 5 cm 2  17.92 cm 2 Ans.
Example 15. 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.
—NCERT
Solution.
For each triangular tile,
Let a = 9 cm, b = 28 cm, c = 35 cm
s
a  b  c 9  28  35
72

cm 
cm  36 cm
2
2
2
 Area of each triangular tile 
s ( s  a )( s  b )( s  c )
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 36 (36  9)(36  28)(36  35) cm 2
 36  4  2  3  9 cm 2  6  2  3 6 cm 2  36 6 cm 2
 Total area of floral design  16  36 6 cm 2
 Cost of polishing the 16 triangular tiles
B
1
 16  36 6  Rs. 288 6  Rs. 705.45 Ans.
2
Example 16. Find the area of a trapezium ABCD in which AB || DC, AB = 77 cm, BC = 25 cm, CD = 60 cm and
DA = 26 cm.
Solution.
Draw DE || BC and DF  AB
the, DE = BC= 25 cm.
AE = AB – EB = AB – DC = 77 cm – 60 cm = 17 cm.
In DAE,
Let a = 17 cm, b = 25 cm, c = 26 cm
 Rs.

s
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a  b  c 17  25  26

cm  34 cm
2
2
 Area of DAE 
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s ( s  a ) ( s  b) ( s  c)
 34 (34  17)(34  25)(34  26) cm 2
 17  3  4 cm 2  204 cm 2
1
 AE  DF
2
1
  17  DF cm 2
2
1
from (1) and (2), we get  17  DF  204
2
2  204
 DF 
cm  24 cm
17
1
 area of trapezium ABCD  (AB  DC)  DF
2
...(1)
Also, area of ΔDAE 
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...(2)
MATHEMATICS–IX
1
(77  60)  24 cm 2
2
= 1644 cm2 Ans.
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PRACTICE EXERCISE
1. Find the area of the triangle whose sides are :
(i) 9 cm, 12 cm and 15 cm
(ii) 150 cm, 120 cm and 2 m
(iii) 6 m, 11 m and 15 m
(iv) 26 cm, 28 cm, 30 cm
2. Find area of a equilateral triangle of side 16 cm.
3. An isosceles triangle has a perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the
triangle.
4. Calculate the area of the shaded portion, in the given figure.
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5. The sides of a triangle are in the ratio 13:14:15 and its perimeter is 210 cm. Find its area.
6. If the area of an equilateral triangle is 49 3 cm2, find its perimeter..
7. Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find
(i) The area of the field.
(ii) the length of perpendicular from the opposite vertex on the side measuring 154 m.
8. ABC is a right triangle right angled at A with AB = 6 cm, AC = 8cm. A circle with centre O and radius r is
inscribed in the triangle. Find r.
9. The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, find the length of the
perpendicular on the side of length 50 m from the opposite vertex. Calculate also the cost of watering it
at Rs. 3.50 per 100 sq. m.
10. A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest altitude.
11. 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.
12. The area of a rhombus is 72 cm2. If one of the diagonals is 18 cm long, find the length of the other
diagonal.
13. The diagonal of a four sided field is 40 m. The perpendiculars from the opposite vertices on this diagonal
are 20 m and 14 m. Find the area of the field.
14. A field is in the shape of a trapezium whose parallel sides are 24 m and 52 m. The non-parallel sides are
26 m and 30 m. Find the area of the field.
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15. Find area of the following polygons :
(i)
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(ii)
PRACTICE TEST
MM : 20
Time : 1 hour
General Instructions :
Each Question carry 4 marks.
B
1. The perimeter of a right triangle is 450 cm and its sides are in the ratio 13 : 12 : 5. Find the area of the
triangle.
2. Find the area of a rhombus, one of whose sides is 25 cm and one of whose diagonal is 48 cm.
3. In a ABC, AB = 4 cm, BC = 9 cm and AC = 7 cm. Find the length of the perpendicular from A to BC.
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.
5. A field is in the shape of a trapezium whose parallel sides are 55 m, 40 m and non-parallel sides are 20 m,
25 m. Find the area of the field.
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ANSWERS OF PRACTICE EXERCISE
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1. (i) 54 cm2
2.
64 3 cm 2
6. 42 cm
9. 67.2 cm, Rs. 58.80
13. 680 m2
(ii) 8969 cm2
(iii) 20 2 m 2
(iv) 336 cm2
3. 9 15 cm 2
4. 384 cm2
5. 2100 cm2
7. (i) 2772 m2
10. 939.14 cm2, 15.39 cm
14. 912 m2
(ii) 36 m
8. 2 cm
2
11. 15.2 cm
12. 8 cm
15. (i) 249.4 cm2 (ii) 444 m2
ANSWERS OF PRACTICE TEST
1. 9000 m2
144
2. 336 cm2
3.
4
5 cm
3
4. 12 cm
HERON’S FORMULA
9. 950 m2
MATHEMATICS–IX