Squeeze a side to zero
Heron's formula as a special case
A 4-gon with a tiny side
Take a triangle with sides a, b, c. Think of it as a 4-gon whose fourth side d has length 0: the corners A and D coincide. A circle can be drawn through any three points that are not in a straight line, so every triangle is cyclic. That means Brahmagupta's formula must apply!
Put d = 0 in Brahmagupta's formula √((s − a)(s − b)(s − c)(s − d)). What do we get?
What this lesson covers
The idea
Any triangle is cyclic and can be treated as a cyclic 4-gon whose fourth side d is zero; putting d = 0 in Brahmagupta's formula gives Heron's formula, so Brahmagupta's formula generalises it.
A 4-gon with a tiny side
Take a triangle with sides a, b, c. Think of it as a 4-gon whose fourth side d has length 0: the corners A and D coincide. A circle can be drawn through any three points that are not in a straight line, so every triangle is cyclic. That means Brahmagupta's formula must apply!
Put d = 0 in Brahmagupta's formula √((s − a)(s − b)(s − c)(s − d)). What do we get?
- Heron's formula
- Zero
- A brand new formula
Shrink d to 0
Pick a triangle. Then shrink the fourth side d with the − button, all the way to 0. Do it for all three triangles.
Heron is a special case
Any triangle is cyclic and can be treated as a cyclic 4-gon whose fourth side d is zero. Putting d = 0 in Brahmagupta's formula gives Heron's formula, so Brahmagupta's formula generalises Heron's formula.
With d = 0, s = (a + b + c) ÷ 2, which is the semi-perimeter of the triangle. And s − d = s, so √((s − a)(s − b)(s − c)(s − d)) = √(s(s − a)(s − b)(s − c)). That is exactly Heron's formula!
Notes
Any triangle is a cyclic 4-gon with fourth side d = 0. Putting d = 0 in Brahmagupta's formula gives Heron's formula, so Brahmagupta's formula generalises it.
Check yourself
A triangle has sides 3, 4, 5. Treat it as a 4-gon with d = 0. What is the semi-perimeter s?
Answer: 6
s = (3 + 4 + 5 + 0) ÷ 2 = 12 ÷ 2 = 6.
Put d = 0 in Brahmagupta's formula, and suppose s = 9. What is s − d?
Answer: 9
s − d = 9 − 0 = 9. So the factor s − d is just s.
A triangle has sides 5, 5, 6. Treat it as a 4-gon with d = 0, so s = 8. Use √((s − a)(s − b)(s − c)(s − d)). What is the area?
Answer: 12
Area = √(3 × 3 × 2 × 8) = √144 = 12. This is Heron's formula.
Why can a triangle be treated as a cyclic 4-gon?
- Because every triangle has four sides. A triangle has three sides. We give it a fourth side of length 0.
- Any three points not on a line lie on a circle, so every triangle is cyclic. Its fourth side is 0 — correct. Yes. So Brahmagupta's formula applies with d = 0.
- Because a triangle has the same area as a rectangle. That is not true, and it is not the reason. The reason is that every triangle has a circumcircle.