Measuring Space: Perimeter and Area · Lesson 6 of 10
Heron's Formula and Brahmagupta's Formula
“When the height goes missing, Heron and Brahmagupta arrive with backup.”
• Understand semi-perimeter. • Apply Heron's formula step by step. • Verify Heron's formula in familiar triangle types. • Understand Brahmagupta's cyclic-quadrilateral formula. • Understand special cases and generalisation.
Heron's Formula
The familiar formula A = 1/2 × b × h works very well when we know a triangle’s base and its corresponding perpendicular height. However, the height is not always given directly, and finding it may require extra construction or calculation.
Sometimes, we are given only the lengths of the three sides of a triangle. In such cases, Heron’s formula gives us a direct way to find the area without first calculating the height. By using the three side lengths and their semiperimeter, we can determine the area of the triangle in a systematic and convenient way.
Half the perimeter of a polygon.
How to Use Heron's Formula
| Step | Action |
|---|---|
| 1 | Check the three side lengths. |
| 2 | Calculate s=(a+b+c)/2. |
| 3 | Calculate s−a, s−b and s−c. |
| 4 | Multiply s(s−a)(s−b)(s−c). |
| 5 | Take the square root. |
Problem
Find the area of a triangle with sides 3, 4 and 5.
- 1.s=(3+4+5)/2=6.
- 2.A=√[6(6−3)(6−4)(6−5)].
- 3.A=√(6×3×2×1).
- 4.A=√36=6 square units.
Equilateral Triangle as a Special Case
For an equilateral triangle of side a, s=3a/2. Substituting into Heron's formula gives √3 a²/4, the familiar equilateral-triangle area formula.
Beyond Triangles: Cyclic Quadrilaterals
Four side lengths alone do not determine the area of an arbitrary quadrilateral. The shape can flex while keeping all sides unchanged. But if the quadrilateral is cyclic—its four vertices lie on one circle—the extra geometric condition is strong enough to determine its area.
A Rectangle as a Special Case
Every rectangle is cyclic. If its sides are a,b,a,b, then s=a+b. Brahmagupta's formula becomes √(b·a·b·a)=ab, exactly the rectangle area.
Special Cases and Generalisation
A result obtained from a more general statement by adding an extra condition.
A broader statement that includes a previously known result as one of its special cases.
Brahmagupta's formula generalises Heron's formula. A triangle can be viewed as a limiting cyclic quadrilateral whose fourth side has length 0. Substituting d=0 reduces Brahmagupta's expression to Heron's formula.
Practice Problems
- Use Heron's formula to find the area of a triangle with sides 7 cm, 24 cm and 25 cm.
- An isosceles triangle has equal sides 15 cm and base 10 cm. Find its area using Heron's formula.
- A triangle has sides in the ratio 2:3:4 and perimeter 45 cm. Find its area.
- Verify Brahmagupta's formula for a rectangle of sides 8 cm and 5 cm.
- Explain why four side lengths do not determine the area of every quadrilateral.
- Explain how Heron's formula appears as a special case of Brahmagupta's formula.
Key Takeaways
• Heron's formula finds triangle area from three sides. • Semi-perimeter is half the perimeter. • Familiar triangle formulas appear as special cases. • Four side lengths alone do not determine a general quadrilateral area. • Brahmagupta's formula works for cyclic quadrilaterals. • Brahmagupta's formula generalises Heron's formula.
Next, we explore an ancient geometric problem: converting a rectangle into an equal-area square.