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Rectangle in an isosceles triangle

There is a rectangle inside an isosceles triangle. One side of the rectangle is on the base of the isosceles triangle. The 2 corners on the other side of the rectangle are touching the equal lengths of the 2 isosceles triangle.

Image:

http://www.mathhelpforum.com/math-he...e-triangle.png

Attachment 19526

*"Write down, in terms of x the height, h, of the rectangle."*

Here's what I came up with:

One of the longer lengths of the triangle is $\displaystyle a + b$ which equals $\displaystyle \sqrt{10^2 + 2.5^2} = \sqrt{106.25}$

Length a + length b = sqrt(106.25)

The equation becomes:

$\displaystyle \underbrace{\sqrt{ \bigg(\frac{1}{2}x \bigg)^2 + (10-h)^2}}_{\text{length a}} + \underbrace{\sqrt{h^2 + \bigg(\frac{5-x}{2} \bigg)^2}}_{\text{length b}} = \sqrt{106.25}$

That doesn't seem correct and seems way too complex. I cannot simplify it down to just h=...

Can anyone help me?