Find the area of an isosceles triangle whose equal sides are 13 cm and whose base is 24 cm.

Step-by-step solution

1. 1

   Drop the altitude to the base. In an isosceles triangle the altitude from the apex to the base also bisects the base, splitting the triangle into two congruent right triangles. Each has hypotenuse $13$ and one leg

   $\frac{24}{2}=12\ \text{cm}$

2. 2

   Find the height with the Pythagorean theorem.

   $h=\sqrt{13^{2}-12^{2}}=\sqrt{169-144}=\sqrt{25}=5\ \text{cm}$

   (the familiar $5$–$12$–$13$ right triangle).

3. 3

   Apply the area formula.

   $A=\frac12\times\text{base}\times\text{height}=\frac12\times 24\times 5$

4. 4

   Evaluate.

   $A=60\ \text{cm}^{2}$

   $\boxed{60\ \text{cm}^{2}}$

5. 5

   Check with Heron's formula. The semiperimeter is $s=\tfrac{13+13+24}{2}=25$, so

   $A=\sqrt{25(25-13)(25-13)(25-24)}=\sqrt{25\cdot 12\cdot 12\cdot 1}=\sqrt{3600}=60\ \text{cm}^{2}$

   matching the first method ✓.

6. 6

   Confirm the triangle exists. The triangle inequality requires $13+13=26>24$ ✓ — only just, which is why the triangle is so flat: a height of only $5$ cm over a base of $24$ cm.