Complete notes with formula, applications, and solved problems
Heron's formula is a powerful method to find the area of a triangle when all three sides are known. It was discovered by Hero (or Heron) of Alexandria, a Greek mathematician and engineer.
Traditional formula: Area = ½ × base × height
Problem: Height is not always known or easy to find
Solution: Heron's formula needs only the three sides!
This makes it extremely useful for real-world problems where measuring height is difficult.
Area of triangle = √[s(s-a)(s-b)(s-c)]
Where:
• a, b, c are the three sides of the triangle
• s is the semi-perimeter = (a + b + c)/2
• s = half of the perimeter
Q: Find the area of a triangle with sides 3 cm, 4 cm, and 5 cm.
Solution:
Given: a = 3 cm, b = 4 cm, c = 5 cm
Step 1: Calculate s
s = (a + b + c)/2 = (3 + 4 + 5)/2 = 12/2 = 6 cm
Step 2: Calculate (s-a), (s-b), (s-c)
s - a = 6 - 3 = 3
s - b = 6 - 4 = 2
s - c = 6 - 5 = 1
Step 3: Apply Heron's formula
Area = √[s(s-a)(s-b)(s-c)]
= √[6 × 3 × 2 × 1]
= √36
= 6 cm²
Verification: This is a 3-4-5 right triangle, so Area = ½ × 3 × 4 = 6 cm² ✓
Q: Find the area of a triangle with sides 7 m, 8 m, and 9 m.
Solution:
a = 7 m, b = 8 m, c = 9 m
s = (7 + 8 + 9)/2 = 24/2 = 12 m
s - a = 12 - 7 = 5
s - b = 12 - 8 = 4
s - c = 12 - 9 = 3
Area = √[12 × 5 × 4 × 3]
= √720
= √(144 × 5)
= 12√5
≈ 26.83 m²
Heron's formula can be extended to find areas of quadrilaterals by dividing them into triangles.
Q: A quadrilateral ABCD has sides AB = 5 cm, BC = 6 cm, CD = 7 cm, DA = 8 cm, and diagonal AC = 9 cm. Find its area.
Solution:
Divide into triangles ABC and ACD
For triangle ABC:
Sides: 5, 6, 9
s₁ = (5 + 6 + 9)/2 = 10
Area₁ = √[10(10-5)(10-6)(10-9)] = √[10×5×4×1] = √200 = 10√2 cm²
For triangle ACD:
Sides: 9, 7, 8
s₂ = (9 + 7 + 8)/2 = 12
Area₂ = √[12(12-9)(12-7)(12-8)] = √[12×3×5×4] = √720 = 12√5 cm²
Total Area = 10√2 + 12√5 ≈ 14.14 + 26.83 = 40.97 cm²
Q: Find area of equilateral triangle with side 6 cm.
Method 1: Using direct formula
Area = (√3/4) × 6² = (√3/4) × 36 = 9√3 ≈ 15.59 cm²
Method 2: Using Heron's formula
s = (6 + 6 + 6)/2 = 9
Area = √[9(9-6)(9-6)(9-6)] = √[9×3×3×3] = √243 = 9√3 cm² ✓
| Triangle Type | When to Use Heron's | Alternative Method |
|---|---|---|
| General Triangle | Only sides known | ½ × base × height (if height known) |
| Right Triangle | Verification | ½ × base × height (easier) |
| Equilateral | Can use | (√3/4)a² (faster) |
| Scalene | Best method | Need to find height first |
| Isosceles | Good option | Can find height geometrically |
Area = √[s(s-a)(s-b)(s-c)]
s = (a+b+c)/2
Heart of the chapter!
s = (a + b + c)/2
s = Perimeter/2
Calculate this first
Area = ½ × base × height
Use when height is known
Area = (√3/4) × a²
Derived from Heron's
All sides = a
Area = ½ × base × height
Or use Heron's formula
Both give same result
Divide into 2 triangles
Area = Area₁ + Area₂
Need diagonal length
1. Find s = (a+b+c)/2
2. Calculate (s-a)(s-b)(s-c)
3. Multiply by s
4. Take square root
• Always find s first
• Check triangle inequality
• Simplify under √ if possible
• Verify with other methods
• Memorize the formula: Area = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2
• Always calculate semi-perimeter (s) as your FIRST step.
• Check if triangle is valid: sum of any two sides > third side.
• Look for perfect squares under the square root for easier calculation.
• For equilateral triangles, the direct formula (√3/4)a² is faster.
• Practice simplifying square roots - very important for this chapter!
• For quadrilaterals, draw the diagonal to visualize two triangles clearly.
• Verify your answer by checking if it makes sense (positive value, reasonable size).
Q1: Triangle with sides 5, 12, 13. Find area.
Hint: This is a right triangle! Verify using both methods.
Q2: Triangle with sides 10, 10, 12. Find area.
Hint: Isosceles triangle.
Q3: Equilateral triangle with perimeter 30 cm. Find area.
Hint: Each side = 10 cm. Use (√3/4)a²
Answers:
Q1: 30 units² (s=15, Area = √[15×10×3×2] = √900 = 30)
Q2: 48 units² (s=16, Area = √[16×6×6×4] = √2304 = 48)
Q3: 25√3 ≈ 43.3 cm²