How to find the length of one of the sides of a triangle given the area 
The triangles are drawn to scale. The first triangle has side lengths of 17, 17, 16 while the second triangle has side lengths of 17,17,$k$. The triangles have the same area. 
Find the value of $k$ algebraically. 
So for the first triangle, I know that the height of the triangle splits the base into two parts of 8 each. sO then using pythagorean theorem, I get the height to be 15 and then the area is $\frac{1}{2} 16 *15 = 120$ 
For the second triangle, I'm not sure what to do and I don't know what solving for $k$ algebraically means. 
 A: HINT
Exactly the same idea. Split in two parts using the height, and in the half-triangle you have the hypotenuse of $17$ and one of the legs is $k/2$.


*

*By the Pythagorean theorem, height $h$ satisfies $h^2 + (k/2)^2 = 17^2$, can you find $h(k)$?

*Now the area of the big triangle is $k \cdot h(k) /2$, but you already know this is $120$, can you solve for $k$?


Remark It's obvious one of the answers will be $k=16$ because then the triangles are identical. Are there other values?
Update
You have $$k = \sqrt{4\left(17^2 - h^2\right)} = 2\sqrt{17^2 - h^2},$$ hence the final equation is 
$$
120 = k(h) \cdot h /2
    = \frac{h}{2} \cdot 2\sqrt{17^2 - h^2}
    =  h \sqrt{17^2 - h^2}
$$
To solve this, square both sides to get
$$
120^2 = h^2 \left(17^2 - h^2\right)
$$
and let $z = h^2$ to get a quadratic in $z$.
A: Don’t calculate, cut the triangle along the height instead to get two triangles with sides $8-17-15$. Now glue them together at their common sides of length $8$ to get the wanted triangle; its sides are $17-17-30$.
Alternatively call the angle on top $\phi$.  Then the triangle’s area equals $0.5\cdot 17^2\cdot\sin(\phi)$. Firstly, this shows that there’s exactly one other solution, provided that $\phi\neq\pi/2$.  This other solution has an angle of $\pi-\phi$ on top. For this solution holds that 
$0.5k/17=\sin(\pi/2-\phi/2)=\cos(\phi/2)$. As $\sin(\phi/2)=8/17$ we have
$(8/17)^2+(0.5k/17)^2=1$, from which $k=30$ follows immediately.
