# Problem of the Week Problem C and Solution Altitude Change

## Problem

In acute $$\triangle ABC$$, two altitudes have been drawn in. Point $$M$$ lies on $$AB$$ so that $$CM$$ is an altitude of $$\triangle ABC$$, and point $$N$$ lies on $$AC$$ so that $$BN$$ is an altitude of $$\triangle ABC$$.

Suppose $$CM=32$$ cm, $$AB=36$$ cm, and $$AC= 40$$ cm. Determine the length of altitude $$BN$$.

Note: An altitude of a triangle is the line segment drawn from a vertex of the triangle perpendicular to the opposite side.

## Solution

The area of a triangle is determined using the formula $\text{area} = \frac{\text{base} \times \text{height}}{2}$ The height of the triangle is the length of an altitude and the base of the triangle is the length of the side to which a particular altitude is drawn.

Thus, \begin{aligned} \text{Area } \triangle ABC&=\frac{AB\times CM}{2}\\ &=\frac{36\times 32}{2}\\ &=576\text{ cm}^2\end{aligned}

Also, \begin{aligned} \text{ Area } \triangle ABC&=\frac{AC\times BN}{2}\\ 576&=\frac{40\times BN}{2}\\ 1152&=40\times BN\\ BN&= 28.8 \text{ cm}\end{aligned}

Therefore, the length of altitude $$BN$$ is $$28.8$$ cm.