Through the vertex of the smaller base, draw a line parallel to the lateral side of the trapezoid, and apply the Law of Cosines.

## Solution

Let $K$ be the midpoint of the larger base $AB$ of trapezoid $ABCD$. Assume that $\angle DAB = 60^\circ$. Since

$$
\angle ADK = \angle KDC = \angle AKD,
$$

triangle $ADK$ is equilateral, so $DK = AK = KB$. Therefore, $\angle ADB = 90^\circ$, and $\angle DBA = 30^\circ$. But

$$
\angle DBA = 60^\circ,
$$

which is impossible. Therefore, $\angle ABC = 60^\circ$.

Let $BC = MC = MD = x$, and $AD = AK = KB = y$. Then $x + y = 10$. Draw a line through vertex $C$ parallel to $AD$, intersecting the base $AB$ at point $P$. In triangle $BCP$, we know that
$BC = x$, $CP = AD = y$, $BP = AB - AP = AB - DC = 2(y - x)$, and $\angle CBP = 60^\circ$.

By the Law of Cosines,

$$
y^2 = x^2 + 4(y - x)^2 - 2x(y - x)
$$

From the system of equations

$$
\left\{
\begin{array}{l}
x + y = 10 \\
y^2 = x^2 + 4(y - x)^2 - 2x(y - x)
\end{array}
\right.
$$

we find that $x = 3$, $y = 7$. Therefore, the height of the trapezoid is $x \sin 60^\circ = \frac{3\sqrt{3}}{2}$. Thus,

$$
S_{ABCD} = (x + y) \cdot \frac{3\sqrt{3}}{2} = 15\sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-26.jpg?height=298&width=578&top_left_y=780&top_left_x=317)

Fig. 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-26.jpg?height=315&width=786&top_left_y=777&top_left_x=953)

Fig. 2.

## Answer

$15\sqrt{3}$.

Submit a comment
\begin{aligned} & {\left[\begin{array}{l}\text { Law of Cosines }\end{array}\right]} \\ & {[\text { Transfer of sides, diagonals, etc. }]}\end{aligned}

In trapezoid $A B C D$, point $K$ is the midpoint of base $A B$, and point $M$ is the midpoint of base $C D$. Find the area of the trapezoid, given that $D K$ is the bisector of angle $D$, $B M$ is the bisector of angle $B$, the largest angle at the lower base is $60^{\circ}$, and the perimeter is 30.