Class 9 Mathematics
Chapter 10: Heron’s Formula
Exercise 10.1 – Stepwise Solutions
Important Formula Used
Heron’s Formula:
If the sides of a triangle are a, b, c, then
s = (a + b + c) / 2
Area = √[s(s − a)(s − b)(s − c)]
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?
Step 1: Since the triangle is equilateral, all sides are equal.
So, side = a
Step 2: Semi-perimeter
s = (a + a + a) / 2 = 3a/2
Step 3: Using Heron’s formula
Area = √[s(s − a)(s − a)(s − a)]
= √[(3a/2)(a/2)(a/2)(a/2)]
= √[3a4/16]
= (a2/4)√3
Thus, area of an equilateral triangle = (√3/4)a2
Step 4: Perimeter = 180 cm
3a = 180
a = 60 cm
Step 5: Area = (√3/4) × 60 × 60
= (√3/4) × 3600
= 900√3 cm²
Answer: 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. The advertisements yield an earning of ₹ 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?
Given: Sides = 122 m, 22 m, 120 m
Step 1: Semi-perimeter
s = (122 + 22 + 120) / 2
= 264 / 2 = 132 m
Step 2: Area using Heron’s formula
Area = √[132(132 − 122)(132 − 22)(132 − 120)]
= √[132 × 10 × 110 × 12]
= √[1742400]
= 1320 m²
Step 3: Rent for 1 year
= 1320 × 5000
= ₹ 66,00,000
Step 4: Rent for 3 months = 1/4 of yearly rent
= 66,00,000 / 4
= ₹ 16,50,000
Answer: ₹ 16,50,000
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”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Given: Sides = 15 m, 11 m, 6 m
Step 1: Semi-perimeter
s = (15 + 11 + 6) / 2
= 32 / 2 = 16 m
Step 2: Area using Heron’s formula
Area = √[16(16 − 15)(16 − 11)(16 − 6)]
= √[16 × 1 × 5 × 10]
= √800
= 20√2 m²
Answer: 20√2 m²
Q4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Step 1: Find the third side
Third side = 42 − (18 + 10)
= 42 − 28 = 14 cm
Step 2: Semi-perimeter
s = 42 / 2 = 21 cm
Step 3: Area using Heron’s formula
Area = √[21(21 − 18)(21 − 10)(21 − 14)]
= √[21 × 3 × 11 × 7]
= √4851
Now, 4851 = 9 × 49 × 11
So,
Area = 21√11 cm²
Answer: 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.
Step 1: Let the sides be 12x, 17x and 25x
Perimeter = 12x + 17x + 25x = 54x
54x = 540
x = 10
Step 2: So sides are
120 cm, 170 cm, 250 cm
Step 3: Semi-perimeter
s = 540 / 2 = 270 cm
Step 4: Area using Heron’s formula
Area = √[270(270 − 120)(270 − 170)(270 − 250)]
= √[270 × 150 × 100 × 20]
= √81000000
= 9000 cm²
Answer: 9000 cm²
Q6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Step 1: Let the unequal side be x
12 + 12 + x = 30
x = 30 − 24 = 6 cm
Step 2: Semi-perimeter
s = 30 / 2 = 15 cm
Step 3: Area using Heron’s formula
Area = √[15(15 − 12)(15 − 12)(15 − 6)]
= √[15 × 3 × 3 × 9]
= √1215
= 9√15 cm²
Answer: 9√15 cm²