10.061. A perpendicular line drawn from the vertex of a parallelogram to its diagonal divides this diagonal into segments of length 6 and $15 \mathrm{~cm}$. The difference in the lengths of the sides of the parallelogram is 7 cm (Fig. 10.60). Find the lengths of the sides of the parallelogram and its diagonals.
## Solution.

By the condition $A E=6 \mathrm{~cm}, E C=15 \mathrm{~cm}, B C-A B=7$ cm. Let $B C=x$, $A B=x-7 \cdot \mathrm{B} \triangle B E A \angle B E A=90^{\circ} \Rightarrow B E^{2}=x^{2}-14 x+49-36$. In $\triangle B E C$ $\angle B E C=90^{\circ} \Rightarrow B E^{2}=x^{2}-225$. Therefore, $x^{2}-14 x+49-36=x^{2}-225$; $x=17$, so $B C=17 \mathrm{~cm}, A B=17-7=10 \mathrm{~cm}, A C=6+15=21$ (cm); $A O=\frac{21}{2}, E O=\frac{21}{2}-A E=\frac{21}{2}-6=\frac{9}{2}, B E=\sqrt{289-225}=8, B D=2 \cdot \sqrt{64+\frac{81}{4}}$ $\left(\triangle B E O\right.$, where $\left.\angle B E O=90^{\circ}\right), B D=\sqrt{337}$.

Answer: $10 ; 17 ; 21 ; \sqrt{337} \mathrm{cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-585.jpg?height=376&width=386&top_left_y=80&top_left_x=207)

Fig. 10.61

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-585.jpg?height=368&width=516&top_left_y=94&top_left_x=638)

Fig. 10.62