Four corners on a circle

The idea

A cyclic quadrilateral with sides a, b, c, d and semi-perimeter s = (1/2)(a + b + c + d) has area √((s – a)(s – b)(s – c)(s – d)).

Cyclic 4-gons

The sides alone do not fix the area of a 4-gon. But a key property can: being cyclic, which means all four corners lie on one circle. In 628 CE Brahmagupta found a formula for the area of a cyclic 4-gon.

A cyclic 4-gon has four equal sides 3, 3, 3, 3. What do you expect its area to be?

• 9
• 12
• 6

Try the formula

Choose the four sides. The toy draws the 4-gon with its corners on a circle, computes the area with the formula, and measures it on the drawing. Try three different sets of sides.

It looks like Heron's formula

A cyclic quadrilateral with sides a, b, c, d and semi-perimeter s = (a + b + c + d) ÷ 2 has area √((s − a)(s − b)(s − c)(s − d)).

This is Brahmagupta's formula. It looks a lot like Heron's formula. Test it with sides 3, 3, 3, 3: s = 6, and the area is √(3 × 3 × 3 × 3) = √81 = 9, the area of a square of side 3.

Notes

A cyclic 4-gon with sides a, b, c, d and s = (a + b + c + d) ÷ 2 has area √((s − a)(s − b)(s − c)(s − d)) (Brahmagupta's formula).

Check yourself

A cyclic 4-gon has sides 2, 5, 5, 8, so s = 10. What is its area?

Answer: 20

Area = √(8 × 5 × 5 × 2) = √400 = 20.

A cyclic 4-gon has sides 3, 4, 8 and 11. What is its area?

Answer: 30

s = 13. Area = √(10 × 9 × 5 × 2) = √900 = 30.

A cyclic 4-gon has all four sides equal to 7. Use Brahmagupta's formula. What is its area?

Answer: 49

s = 14. Area = √(7 × 7 × 7 × 7) = √2401 = 49, the area of a square of side 7.

Brahmagupta's formula gives the area of a 4-gon that is…

• any 4-gon with the given sides. The sides alone do not fix the area of a 4-gon. The formula needs the cyclic property.
• a rectangle only. It works for rectangles, but also for every cyclic 4-gon.
• cyclic: all four corners lie on one circle — correct. Yes. The cyclic property is the extra information that fixes the area.