# 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.