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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 $frac12 16 *15 = 120$

For the second triangle, I'm not sure what to do and I don't know what solving for $k$ algebraically means.

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 $k/2$.

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


2. 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 $$k = sqrt4left(17^2 - h^2right) = 2sqrt17^2 - h^2,$$ hence the final equation is
$$
120 = k(h) cdot h /2
= frach2 cdot 2sqrt17^2 - h^2
= h sqrt17^2 - h^2
$$

To solve this, square both sides to get
$$
120^2 = h^2 left(17^2 - h^2right)
$$
and let $z = h^2$ to get a quadratic in $z$.

My first inclination would be to use Heron's formula for the area of $Delta ABC = (ABC)$ whose side lengths $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ are $a=k$, $b=17$, and $c=17$ and whose semiperimeter $s$ therefore is $s = frac34+k2$.

That is, $$(ABC) = s(s-a)(s-b)(s-c) = 120$$ by Heron's formula, using the values given above for the parameters, and the area calculated using the other triangle.

This is an expression in $k$ which is cubic, but it will have at most 1 real solution (that is 0 real solutions and 3 imaginary, or 1 real and 2 imaginary). Solve it!