[ Transfer of side, diagonal, etc. ] [Pythagorean theorem (direct and inverse).]

In trapezoid $ABCD$, the lateral sides $AB=27, CD=28$, the base $BC=5$, and $\cos \angle BCD = -\frac{2}{7}$. Find the diagonal $AC$.
Since $\cos \angle B C D=-\frac{2}{7}<0$, angle $B C D$ is obtuse, so angle $A D C$ is acute and $\cos \angle A D C=\frac{2}{7}$. Drop a perpendicular $C H$ to the base $A D$. Then

$$
\begin{gathered}
D H=C D \cos \angle A D C=28 \cdot \frac{2}{7}=8 \\
C H=C D \sin \angle A D C=28 \sqrt{1-\frac{1}{19}}=12 \sqrt{5}
\end{gathered}
$$

Draw a line through vertex $C$ parallel to the side $A B$, intersecting line $A D$ at point $P$. Then

$$
P H=\sqrt{C P^{2}-C H^{2}}=\sqrt{A B^{2}-C H^{2}}=\sqrt{2 \bar{\gamma}^{2}-(12 \sqrt{5})^{2}}=3
$$

If point $P$ lies between $A$ and $H$ (Fig.1), then

$$
A H=A P+P H=5+3=8 .
$$

In this case

$$
A C=\sqrt{A H^{2}+C H^{2}}=\sqrt{64+144 \cdot 5}=28 .
$$

If point $P$ lies between $D$ and $H$ (Fig.2), then

$$
A H=A P-P H=5-3=2
$$

In this case

$$
A C=\sqrt{A H^{2}+C H^{2}}=\sqrt{4+144 \cdot 5}=2 \sqrt{181}
$$

## Answer

$28,2 \sqrt{181}$.

In trapezoid $A B C D$ ($B C \| A D$), it is known that $A B=c$ and the distance from the midpoint of segment $C D$ to line $A B$ is $d$. Find the area of the trapezoid.

## First Solution

Let $M$ be the midpoint of the side $C D$ of trapezoid $A B C D$ (Fig.1), $M N$ be the perpendicular dropped from point $M$ to line $A B, M N=d, h$ be the height of the trapezoid. Then

$$
S_{\triangle A M D}+S_{\triangle B M C}=\frac{1}{2} A D \cdot \frac{1}{2} h+\frac{1}{2} B C \cdot \frac{1}{2} h=\frac{1}{2} \cdot \frac{1}{2}(A D+B C) h=\frac{1}{2} S_{A B C D}
$$

Therefore,

$$
S_{\triangle A B M}=S_{A B C D}-S_{\triangle A M D}-S_{\triangle B M C}=\frac{1}{2} S_{A B C D}
$$

Thus,

$$
S_{A B C D}=2 S_{\triangle A B M}=2 \cdot \frac{1}{2} A B \cdot M N=A B \cdot M N=c d
$$

## Second Solution

Let $M$ be the midpoint of the side $C D$ of trapezoid $A B C D$ (Fig.2), $M N$ be the perpendicular dropped from point $M$ to line $A B, M N=d$. Draw a line through point $M$ parallel to the side $A B$. Let this line intersect lines $B C$ and $A D$ at points $K$ and $L$ respectively. Then triangles $C K M$ and $D L M$ are equal by side and two adjacent angles, so the area of trapezoid $A B C D$ is equal to the area of parallelogram $A B K L$, i.e., $A B \cdot M N=c d$.

## Answer

$c d$.