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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 which equals

Length a + length b = sqrt(106.25)

The equation becomes:

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

Can anyone help me?