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Heron's formula

When you know all three side lengths of a triangle but no angles or height, Heron's formula gets you the area directly:

\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}, \qquad s = \dfrac{a+b+c}{2}

This is one of the most common area tools on the AMC — memorize it cold.

Brahmagupta's formula

The analogous formula for a cyclic quadrilateral (one inscribed in a circle) with sides $a,b,c,d$ :

\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}, \qquad s = \dfrac{a+b+c+d}{2}

Less common than Heron's, but it appears whenever a problem specifically tells you a quadrilateral is cyclic and gives all four sides.

Stewart's theorem

Given a cevian (a segment from a vertex to the opposite side) of length $d$ that splits side $a$ into pieces $m$ and $n$ (with $m+n=a$ ):

b^2 m + c^2 n = a(d^2 + mn)

A useful mnemonic is “man + dad = bmb + cnc” matching up which side pairs with which segment. Stewart's theorem is the general tool behind median-length and angle-bisector-length formulas — if you ever forget one of those, you can rederive it from Stewart's.

Worked example

Problem: A triangle has sides 13, 14, 15. Find its area.

Solution: Semi-perimeter $s = \tfrac{13+14+15}{2}=21$ . By Heron's formula:

\text{Area} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21\cdot8\cdot7\cdot6}

=\sqrt{7056} = 84

The 13-14-15 triangle is a classic AMC configuration precisely because it has this clean integer area.

Practice problems

1.Find the area of a triangle with sides 9, 10, 17.

2.Find the area of a cyclic quadrilateral with sides 3, 4, 5, 6.

3.A triangle has sides 5, 5, 6. Find its area using Heron's formula.

4.Explain why Heron's formula alone can't be used to find the area of a general (non-cyclic) quadrilateral given all four sides.

Area Techniques

Heron's formula

Brahmagupta's formula

Stewart's theorem

Practice problems