Given diagonals, lower base, and height, find the legs and upper base of isosceles trapezoid Given an the height, base, and diagonals of an isosceles trapezoid, how am I to find the upper base and the legs?
I know I can find the area of the triangles made by the diagonals, but how is that helpful? Do I have to use trigonometry (law of sines or cosines)?
 A: Assuming you need geometric construction.


*

*Starting from the lower base $AB$, draw the perpendicular upwards (with length of the height) at point $B$, say $BC$.

*Then draw another perpendicular to $BC$ at point $C$, call this line $\ell$.

*Now draw a circle $C_A$ from point $A$ with radius of the diagonal, and from $B$ - another circle $C_B$ with the same radius (both diagonals should have the same length since this is an isosceles trapezoid).

*Each of $C_A$ and $C_B$ should intersect $\ell$ in two points. Call the right intersection of $C_A$ and $\ell$ as $K$, and the left intersection of $C_B$ and $\ell$ - as $D$.

*You are looking for the trapezoid $ABKD$.

A: Begin by drawing a sketch of the situation. Below is a crude image, but it will do for our purposes.

We know that $DX = \sqrt{d^{2} - x^{2}}$ by Pythagoras. Therefore, we have that $XC = b_{2} - DX = b_{2} - \sqrt{d^{2} - x^{2}}.$
Since $ABCD$ is isosceles, we have that $AB = b_{2} - 2XC = \boxed{2\sqrt{d^{2} - x^{2}} - b_{2}}.$
For the legs, use Pythagoras once again. Each leg has length
$$BC = \sqrt{x^{2} + \left(b_{2} - \sqrt{d^{2} - x^{2}}\right)^{2}}$$
$$= \sqrt{x^{2} + b_{2}^{2} - 2b_{2}\sqrt{d^{2} - x^{2}} + d^{2} - x^{2}}$$
$$\boxed{\sqrt{b_{2}^{2} - 2b_{2}\sqrt{d^{2} - x^{2}} + d^{2}}}.$$
