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.
"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?