Heron’s formula questions with answers are provided here. Class 9 students can practise the questions based on Heron’s formula to prepare for the exams. These extra questions are prepared by our subject experts, as per the NCERT curriculum and latest CBSE syllabus (2022-2023). Learn Heron’s formula in detail at BYJU’S.
Definition: Heron’s formula is a formula used to find the area of a triangle. According to this formula;
Area of triangle = √(s(s-a)(s-b)(s-c))
Where a, b and c are the sides of a triangle and s is the semiperimeter of triangle.
s = (a+b+c)/2
Heron’s Formula Questions and Solutions
Q.1: Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm.
Solution: Given the sides of a triangle are:
a = 12 cm
b = 6 cm
c = 15 cm
According to Heron’s formula;
Area of triangle = √(s(s-a)(s-b)(s-c))
s = (12 + 6 + 15)/2 = 33/2 = 16.5
Area = √(16.5(16.5-12)(16.5-6)(16.5-15))
= √(16.5 x 4.5 x 0.5 x 1.5)
= 34.2 cm2
Q.2: Find the Area of a Triangle whose two sides are 18 cm and 10 cm, respectively and the perimeter is 42 cm.
Solution: Given two sides of a triangle are 18 cm and 10 cm, respectively.
a = 18cm, b = 10 cm
Perimeter of triangle = 42 cm
a+b+c = 42
18+10+c = 42
c = 42 – 28 = 14 cm
Using Heron’s formula, we have;
Area of triangle = √(s(s-a)(s-b)(s-c))
Semiperimeter, s = 42/2 = 21
Area = √(21(21-18)(21-10)(21-14))
= √(21 x 3 x 11 x 7)
= 69.7 cm2
Q.3: A triangular park has sides 120 m, 80 m and 50 m. A gardener has to put a fence all around it and also plant grass inside. How much area does he need to plant?
Solution: Given,
Sides of triangular park are 120m, 80m and 50m.
Semiperimeter, s = (120 + 80 + 50)/2 = 125 m
Using Heron’s formula, we have;
Area of triangle = √(s(s-a)(s-b)(s-c))
Area = √(125 (125-120) (125 – 80) (125 – 50)
= √(125 x 75 x 45 x 5)
= 375√15 m2
Q.4: The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.
Solution: Given,
Ratio of the sides of the triangle is 12: 17: 25
Let the sides of triangle be 12x, 17x and 25x
Given, perimeter of the triangle = 540 cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540cm
So, x = 10
Thus, the sides of the triangle are:
12 x 10 = 120 cm
17 x 10 = 170 cm
25 x 10 = 250 cm
Semiperimeter, s = 540/2 = 270 cm
Using Heron’s formula,
Area of the triangle = √[s (s-a) (s-b) (s-c)]
= √[270 (270 – 120) (270 – 170) (270 – 250)]
= √(270 x 150 x 100 x 20)
= 9000 cm2
Q.5: Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 20.5 cm.
Solution: Given,
Side a = 4.5 cm
Side b = 10 cm
Perimeter of triangle = 20.5 cm
a+b+c = 20.5
4.5+10+c = 20.5
14.5+c = 20.5
c = 20.5 – 14.5
c = 6 cm
Semiperimeter, s = (4.5+10+6)/2 = 20.5/2 = 10.25 cm
Using Heron’s formula,
Area of the triangle = √[s (s-a) (s-b) (s-c)]
Area = √[10.25(10.25-4.5) (10.25-10) (10.25-6)]
Area = 7.91 cm2
Q.6: What is the area of a triangle whose sides are 9 cm, 12 cm and 15 cm?
Solution: Given, the sides of a triangle are:
a = 9 cm
b = 12 cm
c = 15 cm
Semiperimeter, s = (9 + 12 + 15)/2 = 36/2 = 18 cm
According to Heron’s formula;
Area of the triangle = √[s (s-a) (s-b) (s-c)]
Area = √[18 (18 – 9) (18 – 12) (18 – 15)]
Area = √[18 (9) (6) (3)]
Area = 54 cm2
Q.7: The perimeter of a right triangle is 300m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.
Solution: Given,
Perimeter of right triangle = 300 m
Ratio of sides of triangle is 3 : 5 : 7
Let the sides of triangle be:
a = 3x
b = 5x
c = 7x
Thus,
3x + 5x + 7x = 300
15x = 300
x = 300/15 = 20
Thus, the sides of triangle are:
a = 3x = 3 (20) = 60 m
b = 5x = 5 (20) = 100 m
c = 7x = 7 (20) = 140 m
Semiperimeter, s = 300/2 = 150 m
Using Heron’s formula, we have;
Area of the triangle = √[s (s-a) (s-b) (s-c)]
Area = √(150 (150 – 60) (150 – 100) (150 – 140))
Area = 1500√3 m2
Q.8: The sides of a triangle are 7 cm, 9 cm, and 14 cm. What is the area of the triangle?
Solution: Given,
a = 7cm, b = 9 cm and c = 14 cm
Semiperimeter, s = (7 + 9 + 14)/2 = 15 cm
Area of triangle = = √[s (s-a) (s-b) (s-c)] By Heron’s formula.
Area = √[15(15-7) (15-9) (15-14)]
= √[(15) (8) (6) (1)]
= 12√5 cm2
Q.9: The sides of a triangle are 11 m, 60 m and 61 m. What is the altitude to the smallest side?
Solution: Given, sides of the triangle are 11 m, 60 m and 61 m.
The smallest side is 11 m.
Thus,
Area of triangle = ½ (base) (height)
A = ½ (11) h ….(i)
We can find the area of the triangle using Heron’s formula here.
Area of triangle = = √[s (s-a) (s-b) (s-c)]
s = (11+60+61)/2 = 66 m
Area = √[66(66-11) (66-60) (66-61)]
Area = √(66 x 55 x 6 x 5) = 330 m2
Now putting the value of area in equation (i), we get;
330 = ½ (11) h
h = (2 x 330)/11 = 60m
Therefore, the altitude to the smallest side is 60m.
Q.10: If every side of a triangle is doubled, by what percentage is the area of the triangle increased?
Solution: Let a, b and c be the sides of a triangle.
Semiperimeter, s = (a+b+c)/2
Now, if each of the side is doubled, then the new sides of a triangle are:
A = 2a, B = 2b, C = 2c
Semiperimeter, S = (A+B+C)/2 = (2a + 2b + 2c)/2 = 2s
By Heron’s formula,
Area of triangle, A’ = √[S (S-A) (S-B) (S-C)]
A’ = √[2s(2s-2a)(2s-2b)(2s-2c)]
A’ = √[2s2(s-a)2(s-b)2(s-c)]
= 4√s(s-a)(s-b)(s-c)
A’ = 4A
Increase in Area = (4A – A)/A x 100% = 300%
Hence, the area is increased by 300%, if the sides of the triangle are doubled.
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