10.288. Find the area of a trapezoid whose diagonals are 7 and 8 cm, and the bases are 3 and 6 cm.
Solution.

Let in trapezoid $A B C D$ (Fig. 10.81) $B C \| A D, B C=3 \text{ cm}, A D=6$ cm, $A C=7 \text{ cm}, B D=8$ cm. Draw $C E \| B D$. Then $A E=A D+D E=$ $=A D+B C=9$ cm, $C E=B D$. Therefore, trapezoid $A B C D$ and triangle $A C E$ have equal areas (they have equal heights and the base of the triangle is equal to the sum of the bases of the trapezoid). By Heron's formula, the area of the triangle $S=\sqrt{12 \cdot 5 \cdot 4 \cdot 3}=12 \sqrt{5} \text{ cm}^{2}$.

Answer: $12 \sqrt{5} \text{ cm}^{2}$.