[Transfer of side, diagonal, etc.]

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

#
Through the vertex of the trapezoid, draw a line parallel to the diagonal.

## Solution

Through vertex $C$ of the smaller base $BC$ of trapezoid $ABCD$ ($BC=3, AD=6, BD=8, AC=7$), draw a line parallel to diagonal $BD$, intersecting line $AD$ at point $K$. The sides of triangle $ACK$ are: $AC=7$, $CK=BD=8$, and $AK=AD+DK=AD+BC=6+3=9$. By Heron's formula, $S_{ACK}=\sqrt{12 \cdot 5 \cdot 4 \cdot 3}=12 \sqrt{5}$, and $S_{ABCD}=S_{ACK}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-44.jpg?height=340&width=715&top_left_y=2288&top_left_x=677)

## Answer

$12 \sqrt{5}$

In trapezoid $ABCD$, angles $A$ and $D$ at the base $AD$ are $60^\circ$ and $90^\circ$ respectively. Point $N$ lies on the base $BC$, such that $BN:BC=2:3$. Point $M$ lies on the base $AD$, and line $MN$ is parallel to the side $AB$ and divides the area of the trapezoid in half. Find $AB:BC$.

## Hint

Let $NC=a$ and express the height of the trapezoid in terms of $a$.

## Solution

Let $NC=a$, $AB=x$. Then $BN=2a$, $AM=BN=2a$. If $NK$ is the height of the trapezoid, then

$$
DK=NC=a, MK=\frac{1}{2}MN=\frac{1}{2}AB=\frac{x}{2}.
$$

From the equality of the areas of parallelogram $ABNM$ and trapezoid $MNCD$, we have

$$
2AM=NC+MD, \text{ or } 4a=a+a+\frac{x}{2}.
$$

From this, we find that $x=4a$. Therefore,

$$
\frac{AB}{BC}=\frac{x}{3a}=\frac{4a}{3a}=\frac{4}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-45.jpg?height=309&width=323&top_left_y=1206&top_left_x=868)

## Answer

$4:3$.