# 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.