 # Problem of the Week Problem D and Solution There are Two Sides

## Problem

A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side.

In $$\triangle ABC$$, a median is drawn from vertex $$A$$, meeting side $$BC$$ at point $$D$$. The length of $$BD$$ is $$6$$ cm and the length of the median $$AD$$ is $$13$$ cm. The area of $$\triangle ABC$$ is $$72 \text{cm}^2$$. Determine the lengths of sides $$AB$$ and $$AC$$.

## Solution

First we will draw the altitude from vertex $$A$$, meeting side $$BC$$ at point $$E$$. Let $$h$$ be the length of the altitude $$AE$$. Let $$x$$ be the length of $$DE$$. Since $$AD$$ is a median, $$DC=BD = 6$$. Since $$E$$ is on $$DC$$ and the length of $$DE$$ is $$x$$, the length of $$EC$$ is $$6-x$$. We know the area of $$\triangle ABC$$ is $$72 \text{cm}^2$$. Also, since $$BD=DC=6$$ cm, it follows that $$BC=12$$ cm. Thus, \begin{aligned} \frac{BC\times AE}{2}&=72\\ \frac{12h}{2}&=72\\ h&=12 \end{aligned} Since $$\triangle AED$$ is right-angled, we can use the Pythagorean Theorem as follows. \begin{aligned} DE^2+AE^2&=AD^2\\ x^2+12^2&=13^2\\ x^2&=13^2-12^2\\ x^2&=169-144=25 \end{aligned} Since $$x>0$$, it follows that $$x=5$$ cm. Thus, $$BE=6+x=6+5=11$$ cm, and $$EC=6-x=6-5=1$$ cm.

Since $$\triangle AEB$$ is right-angled, we can use the Pythagorean Theorem as follows. \begin{aligned} AE^2+BE^2&=AB^2\\ 12^2+11^2&=AB^2\\ AB^2&=144+121=265 \end{aligned} Since $$AB>0$$, it follows that $$AB=\sqrt{265}$$ cm.

Since $$\triangle AEC$$ is right-angled, we can use the Pythagorean Theorem as follows. \begin{aligned} AE^2+EC^2&=AC^2\\ 12^2+1^2&=AC^2\\ AC^2&=144+1=145 \end{aligned} Since $$AC>0$$, it follows that $$AC=\sqrt{145}$$ cm.

Therefore, the lengths of sides $$AB$$ and $$AC$$ are $$\sqrt{265}$$ cm and $$\sqrt{145}$$ cm, respectively. These are approximately equal to $$16.3$$ cm and $$12.0$$ cm.