Indicate the given distances using the identity of tangents drawn from a certain point.

Draw the two height lines of the trapezoid from points C and D.

These two height lines (blue dashed lines) divide the trapezoid into a rectangle (center) and a right-angled triangle (right-left ) share. Write your Pythagorean theorem for both right triangles.

The Pythagorean Theorem for the left triangle:

$${\left[15+\left(8-x\right)\right]}^2=h^2+{\left[15-\left(8-x\right)\right]}^2$$

$${\left[15+\left(8-x\right)\right]}^2-{\left[15-\left(8-x\right)\right]}^2=h^2$$

$${\left(23-x\right)}^2-{\left(7+x\right)}^2=h^2$$

$$529-46x+x^2-\left(49+14x+x^2\right)=h^2$$

$$529-46x+x^2-49-14x-x^2=h^2$$

$$-60x+480=h^2$$

The Pythagorean Theorem for the right triangle:

$${\left(10+x\right)}^2={\mathrm{h}}^2+{\left(10-x\right)}^2$$

$${10}^2+2·10·x+x^2=h^2+{10}^2-2·10·x+x^2$$

$$\cancel{{10}^2}+20x+\bcancel{x^2}=h^2+\cancel{{10}^2}-20x+\bcancel{x^2}$$

$$40x=h^2$$

Izjedančite obe vrednosit h ² s.

$$-60x+480=40x$$

$$480=40x+60x$$

$$480=100x$$

$$x=\frac{480}{100}$$

$$x=\frac{24}{5}$$

$$h=\sqrt{192}$$

$$h=8·\sqrt{3}$$

$$A=m·h$$

$$=\frac{33}{\cancel{2}}·\cancel{2}·4\sqrt{3}$$

$$A=132\sqrt{3}$$