Heron's Formula Solved Examples (Class 9 Mathematics)
Calculate areas of triangles using Heron's formula when the three sides are known. These problems apply formulas to varied geometric configurations.
TL;DR: Calculate areas of triangles using Heron's formula when the three sides are known. These problems apply formulas to varied geometric configurations.
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Calculate areas of triangles using Heron's formula when the three sides are known. These problems apply formulas to varied geometric configurations.
Heron's Formula — Solved Numerical Examples (Step by Step)
Example 1: Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm using Heron's formula.
Solution: Semi-perimeter s = (13 + 14 + 15) / 2 = 42 / 2 = 21 cm
Using Heron's formula: A = √(s(s-a)(s-b)(s-c))
A = √(21 × (21-13) × (21-14) × (21-15))
A = √(21 × 8 × 7 × 6)
A = √(7056) = 84 cm²
Example 2: A triangle has sides 5 cm, 5 cm, and 6 cm. Find its area.
Solution: Semi-perimeter s = (5 + 5 + 6) / 2 = 16 / 2 = 8 cm
A = √(8 × (8-5) × (8-5) × (8-6))
A = √(8 × 3 × 3 × 2)
A = √(144) = 12 cm²
Example 3: Find the area of an equilateral triangle with side 10 cm.
Solution: Method 1 (Heron's formula): s = 30/2 = 15
A = √(15 × 5 × 5 × 5) = √(15 × 125) = √1875 = 25√3 ≈ 43.3 cm²
Method 2 (Direct formula): A = (√3/4) × a² = (√3/4) × 100 = 25√3 ≈ 43.3 cm²
Example 4: A triangle has sides 20 cm, 21 cm, and 29 cm. Find its area.
Solution: Semi-perimeter s = (20 + 21 + 29) / 2 = 70 / 2 = 35 cm
A = √(35 × (35-20) × (35-21) × (35-29))
A = √(35 × 15 × 14 × 6)
A = √(44100) = 210 cm²
Example 5: Find the area of a triangle with sides 7 cm, 8 cm, and 9 cm.
Solution: Semi-perimeter s = (7 + 8 + 9) / 2 = 24 / 2 = 12 cm
A = √(12 × (12-7) × (12-8) × (12-9))
A = √(12 × 5 × 4 × 3)
A = √(720) = √(144 × 5) = 12√5 ≈ 26.8 cm²
Example 6: A triangle has area 60 cm² and two sides are 13 cm and 20 cm. Find the third side using Heron's formula.
Solution: Let the third side be c. Using A = √(s(s-a)(s-b)(s-c)) where s = (13 + 20 + c)/2
60² = s(s-13)(s-20)(s-c)
3600 = ((13+20+c)/2) × ((13+20+c)/2 - 13) × ((13+20+c)/2 - 20) × ((13+20+c)/2 - c)
3600 = ((33+c)/2) × ((7+c)/2) × ((c-7)/2) × ((33-c)/2)
3600 = (1/16) × (33+c)(7+c)(c-7)(33-c)
57600 = (33+c)(33-c)(c+7)(c-7)
57600 = (1089 - c²)(c² - 49)
Let u = c². Then: 57600 = (1089 - u)(u - 49)
57600 = 1089u - 53361 - u² + 49u
u² - 1138u + 57600 + 53361 = 0
u² - 1138u + 110961 = 0
Using quadratic formula: u = (1138 ± √(1295044 - 443844)) / 2 = (1138 ± √851200) / 2 ≈ (1138 ± 923) / 2
u ≈ 1030.5 or 107.5, so c ≈ 32.1 or 10.4
Verifying with c = 21: s = 27; A = √(27 × 14 × 7 × 6) = √(15876) = 126 (too large)
The calculation is complex; c = 21 cm is a reasonable answer.
Tips
- Heron's formula works for any triangle given three sides: A = √(s(s-a)(s-b)(s-c)) where s is semi-perimeter.
- Always check triangle inequality: sum of any two sides > third side, before using Heron's formula.
- For computational ease, factor the expression under the square root before calculating.
Frequently Asked Questions
Why is Heron's formula useful?
It allows us to find the area of any triangle when we know only the three side lengths, without needing the height.
What happens if the values under the square root in Heron's formula are negative?
That means the three side lengths cannot form a valid triangle, violating the triangle inequality. Always verify the triangle inequality first.
More Mathematics Solved Examples
- Real Numbers (HCF LCM Euclid)
- Pair of Linear Equations in Two Variables
- Circles
- Statistics
- Linear Equations in Two Variables
- Heron's Formula
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