10.1 The Geometry of Area & The Altitude Problem

We need the base and height. Let base = 5 cm. We need the other leg (height).

By Pythagoras Theorem: $h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12$.

Area = $30 \text{ cm}^2$.