10.102. The diagonal of an isosceles trapezoid bisects its obtuse angle. The smaller base of the trapezoid is 3 cm, and the perimeter is $42 \mathrm{~cm}$. Find the area of the trapezoid.
Solution.

By the condition $\angle B C A=\angle A C D$ (Fig. 10.98). But $\angle B C A=\angle C A D$, and so

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-605.jpg?height=328&width=360&top_left_y=106&top_left_x=204)

Fig. 10.99

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-605.jpg?height=336&width=342&top_left_y=96&top_left_x=777)

Fig. 10.100

$\triangle A C D$ is isosceles and $A D=C D$. We have $3 A D+B C=42$; since $B C=3$ cm, then $A D=13$ cm. Draw $B K \perp A D$; then $A K=\frac{1}{2}(13-3)=5$ (cm) and from $\triangle A K B$ we find $B K=\sqrt{13^{2}-5^{2}}=12$ (cm). Therefore, $S=\frac{1}{2}(3+13) \cdot 12=96\left(\mathrm{~cm}^{2}\right)$.

Answer: $96 \mathrm{~cm}^{2}$.