Applications to Special Triangles

Heron's formula works for any triangle, but when the triangle has special symmetry , equilateral, isosceles, or right , the formula simplifies nicely. This lesson works through these special cases, both to give you fluency and to provide ways to check your answer using a shortcut.

Equilateral triangle

Side $a$. By Heron's formula with $a = b = c$, semi-perimeter $s = \tfrac{3a}{2}$, and each $s - a = \tfrac{a}{2}$. $\text{Area} = \sqrt{\tfrac{3a}{2} \cdot \tfrac{a}{2} \cdot \tfrac{a}{2} \cdot \tfrac{a}{2}} = \sqrt{\tfrac{3 a^4}{16}} = \tfrac{\sqrt{3}}{4} a^2.$

This is the closed-form equilateral-triangle area. Memorise: Area $= \tfrac{\sqrt{3}}{4} a^2$.

Quick examples.

• Side $4$: Area $= \tfrac{\sqrt{3}}{4} \cdot 16 = 4\sqrt{3} \approx 6.93$ square units.
• Side $10$: Area $= 25\sqrt{3} \approx 43.30$ square units.
• Side $1$: Area $= \tfrac{\sqrt{3}}{4} \approx 0.433$ square units.

Isosceles triangle

Two equal sides $b$, base $a$. With $s = \tfrac{2b + a}{2} = b + \tfrac{a}{2}$:

$s - a = b - \tfrac{a}{2}, \quad s - b = \tfrac{a}{2}$ (twice).

$\text{Area} = \sqrt{(b + \tfrac{a}{2})(b - \tfrac{a}{2}) \cdot (\tfrac{a}{2})^2}$ $= \tfrac{a}{2} \sqrt{b^2 - \tfrac{a^2}{4}}$ $= \tfrac{a}{4} \sqrt{4b^2 - a^2}$.

This matches the formula in lesson 1, derived from "base $\times$ height $\div 2$". Equivalent to Heron's formula, just simplified using the isosceles symmetry.

Quick example. Base $a = 10$, equal sides $b = 13$. Area $= \tfrac{10}{4}\sqrt{4 \cdot 169 - 100} = \tfrac{5}{2}\sqrt{576} = \tfrac{5}{2} \cdot 24 = 60$ square units.

Right triangle

Legs $a$ and $b$, hypotenuse $c$ with $a^2 + b^2 = c^2$. Direct area: $\tfrac{1}{2} ab$.

By Heron's formula with $s = \tfrac{a + b + c}{2}$, you can verify: $s - a = \tfrac{-a + b + c}{2}, \quad s - b = \tfrac{a - b + c}{2}, \quad s - c = \tfrac{a + b - c}{2}$.

Multiplying these out and using $c^2 = a^2 + b^2$ eventually gives $\tfrac{1}{2} ab$. But the direct formula is much easier , use it.

Quick example. Legs $5$ and $12$, hypotenuse $13$. Direct area: $\tfrac{1}{2} \times 5 \times 12 = 30$. Verify with Heron's: $s = 15$, $s - a = 10, s - b = 3, s - c = 2$. Product: $15 \times 10 \times 3 \times 2 = 900$. Area: $\sqrt{900} = 30$. ✓

Recognising the type

Given three side lengths, here is how to choose:

1. Are all three equal? Use equilateral formula.
2. Are two equal? Use isosceles formula (or Heron's).
3. Do the lengths satisfy $a^2 + b^2 = c^2$ for the longest $c$? Right triangle , use $\tfrac{1}{2}$ leg $\times$ leg.
4. Otherwise: Use Heron's formula.

Worked examples

Example 1. Find the area of an equilateral triangle of side $8$.

$\tfrac{\sqrt{3}}{4} \cdot 64 = 16\sqrt{3} \approx 27.71$ sq units.

Example 2. Find the area of an isosceles triangle with base $12$ and equal sides $10$ each.

Use isosceles formula: $\tfrac{12}{4}\sqrt{400 - 144} = 3\sqrt{256} = 3 \cdot 16 = 48$ sq units.

Example 3. Find the area of a right triangle with legs $9$ and $12$.

$\tfrac{1}{2} \cdot 9 \cdot 12 = 54$ sq units.

Example 4. Find the area of a triangle with sides $3, 3, 3$ using both equilateral formula and Heron's. Confirm they match.

Equilateral: $\tfrac{\sqrt{3}}{4} \cdot 9 = \tfrac{9\sqrt{3}}{4}$. Heron: $s = 4.5$, $s - a = 1.5$ three times. Product: $4.5 \cdot 1.5^3 = 4.5 \cdot 3.375 = 15.1875$. Area: $\sqrt{15.1875} = \tfrac{9\sqrt{3}}{4}$ (verified by exact computation: $\tfrac{81 \cdot 3}{16} = \tfrac{243}{16}$, $\sqrt{\cdot} = \tfrac{9\sqrt{3}}{4}$). ✓

Example 5. Find the area of an isosceles right triangle with legs $5$.

Legs are equal (isosceles), and one angle is right. Use $\tfrac{1}{2} \cdot 5 \cdot 5 = \tfrac{25}{2} = 12.5$ sq units. Or via Heron's, with sides $5, 5, 5\sqrt{2}$.

Try it yourself

1. Find the area of an equilateral triangle of side $6$.
2. Find the area of an isosceles triangle with base $14$ and equal sides $25$.
3. Find the area of a right triangle with legs $6$ and $8$.
4. Find the area of an isosceles right triangle with legs $10$.
5. Find the area of an equilateral triangle of side $\sqrt{12}$.
6. Find the area of an isosceles triangle with sides $13, 13, 10$.
7. Verify the isosceles formula by computing it via Heron's for the triangle with base $a = 6$, equal sides $b = 5$.
8. Find the area of a triangle with sides $5, 12, 13$.
9. Find the area of an equilateral triangle whose perimeter is $36$.
10. An isosceles triangle has perimeter $32$ and base $12$. Find its area.

Pitfalls / Insight

• Identify the type first. Don't always default to Heron's formula , special-case formulas are faster.
• Memorise $\tfrac{\sqrt{3}}{4} a^2$ for equilateral triangles. It comes up often.
• For right triangles, use legs. No need to compute the hypotenuse if both legs are known.

Insight. Heron's formula is universal, but it is often more work than necessary. For the most common special cases , equilateral and right , we have one-line formulas. Always look for the shortcut first.