## Task A-3.3.

A trapez with legs of length 4 and 5 can have a circle inscribed in it, and the sum of the measures of the angles at the longer base is $120^{\circ}$. Calculate the area of this trapezoid.
## Solution.

Let $a$ and $c$ be the lengths of the bases $(a>c)$, and $v$ be the length of the height of the trapezoid.

Since a circle can be inscribed in the trapezoid, it follows that $a+c=4+5=9$.

Draw a line parallel to one of the legs through the vertex where the other leg and the shorter base intersect. We obtain a triangle with sides of lengths $a-c$, 4, and 5, and the angle between the sides of lengths 4 and 5 is $180^{\circ}-120^{\circ}=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a374c05c2f1967f14be6g-14.jpg?height=397&width=554&top_left_y=2183&top_left_x=680)

By the cosine rule, we have

$$
(a-c)^{2}=4^{2}+5^{2}-2 \cdot 4 \cdot 5 \cdot \cos 60^{\circ}=16+25-20=21
$$

i.e., $a-c=\sqrt{21}$.

Expressing the area in two ways, we get

$$
\frac{(a-c) v}{2}=\frac{4 \cdot 5 \cdot \sin 60^{\circ}}{2}
$$

Substituting $a-c=\sqrt{21}$, we get $v=\frac{10 \sqrt{7}}{7}$.

Finally, the area of the trapezoid is

$$
P=\frac{a+c}{2} \cdot v=\frac{45 \sqrt{7}}{7}
$$