10.039. The lines containing the non-parallel sides of an isosceles trapezoid intersect at a right angle (Fig. 10.39). Find the lengths of the sides of the trapezoid if its area is \(12 \mathrm{~cm}^{2}\) and the height is \(2 \mathrm{~cm}\).

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-574.jpg?height=346&width=594&top_left_y=111&top_left_x=95)

Fig. 10.39

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-574.jpg?height=382&width=398&top_left_y=77&top_left_x=801)

Fig. 10.40
## Solution.

Given $S_{\triangle A B C D}=12 \mathrm{~cm}^{2}, B E=2 \mathrm{~cm} ; \angle O A D=\angle O D A=\frac{90^{\circ}}{2}=45^{\circ}$, $\frac{B E}{\sin 45^{\circ}}=A B, A B=\frac{2}{\sqrt{2}} \cdot 2=2 \sqrt{2}=C D, A E=B E=2, \frac{B C+A D}{2} \cdot B E=12$, $\frac{B C+A D}{2} \cdot 2=12, B C+A D=12$, then $2 B C+2 A E=12, B C+A E=6$, $B C+2=6, B C=4 \mathrm{~cm}, A D=4+4=8 \mathrm{~cm}$.

Answer: 4 and 8 cm; $2 \sqrt{2}$ and $2 \sqrt{2}$ cm.