# Problem of the Week

Problem E and Solution

The Other Side

## Problem

In \(\triangle STU\), a median is drawn from vertex \(S\), meeting side \(TU\) at point \(M\).
The length of side \(ST\) is \(7\) cm, the length of side \(SU\) is \(9\) cm, and the length of the median \(SM\) is \(7\) cm.

Determine the length of \(TU\).

## Solution

**Solution 1**

Since \(ST=SM=7\), \(\triangle STM\) is isosceles. In \(\triangle STM\), draw an altitude from vertex \(S\) to \(TM\), intersecting \(TM\) at \(N\). Let \(TN=x\). In an isosceles triangle, the altitude drawn to the base bisects the base. Therefore, \(NM = TN=x\). Since \(SM\) is a median in \(\triangle STU\), it follows that \(MU=TM=2x\). Let \(SN=h\).

Since \(\triangle SNM\) is a right-angled triangle, we can use the Pythagorean Theorem as follows.
$$
\begin{align}
SN^2&=SM^2-NM^2\\
h^2&=7^2-x^2\\
h^2&=49-x^2 \tag{1}
\end{align}
$$
Since \(\triangle SNU\) is a right-angled triangle, we can use the Pythagorean Theorem as follows.
$$
\begin{align}
SN^2&=SU^2-NU^2\\
h^2&=9^2-(x+2x)^2\\
h^2&=81-(3x)^2\\
h^2&=81-9x^2 \tag{2}
\end{align}
$$
In both equations \((1)\) and \((2)\), the left side is \(h^2\). Therefore, the right side of equation \((1)\) must equal the right side of equation \((2)\).
$$
\begin{align}
49-x^2&=81-9x^2\\
-x^2+9x^2&=81-49\\
8x^2&=32\\
x^2&=4
\end{align}
$$
Since \(x>0\), it follows that \(x=2\).

Therefore, \(TU=TN+NM+MU=x+x+2x=4x=4(2)=8\) cm.

**Solution 2**

This solution is presented for students who have done some trigonometry and know the Cosine Law. Since \(SM\) is a median, let \(TM=MU=y\). Then \(TU=2y\).
Using the Cosine Law in \(\triangle STM\),
$$
\begin{align}
SM^2&=ST^2+TM^2-2(ST)(TM)\cos{T}\\
7^2&=7^2+y^2-2(7)(y)\cos{T}\\
49&=49+y^2-14y\cos{T}\\
14y\cos{T}&=y^2\tag{1}
\end{align}
$$
Using the Cosine Law in \(\triangle STU\),
$$
\begin{align}
SU^2&=ST^2+TU^2-2(ST)(TU)\cos{T}\\
9^2&=7^2+(2y)^2-2(7)(2y)\cos{T}\\
81&=49+4y^2-28y\cos{T}\\
28y\cos{T}&=4y^2-32\\
14y\cos{T}&=2y^2-16\tag{2}
\end{align}
$$
Subtracting equation \((2)\) from equation \((1)\) allows us to solve for \(y\).
$$
\begin{align}
14y\cos{T}&=y^2\tag{1}\\
14y\cos{T}&=2y^2-16\tag{2}\\
0&=-y^2+16\\
y^2&=16
\end{align}
$$
Since \(y>0\), it follows that \(y=4\).

Therefore, the length of \(TU\) is \(2(4)=8\) cm.