7. A circle is inscribed in trapezoid $A B C D$, touching the lateral side $A B$ at point $M$, and $A M=18$. Find the sides of the trapezoid if its perimeter is 112, and the area is 672.
Solution: $M, E, K, F$ are the points of tangency of the circle with sides $A B$, $B C, C D, A D$ respectively. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-12.jpg?height=443&width=948&top_left_y=287&top_left_x=951)

$$
A M=A F=a=18, B M=B E=x, C E=C K=y, D K=D F=z
$$

$P_{A B C D}=2(18+x+y+z)=112, x+y+z=38$.

$$
\begin{aligned}
& S_{A B C D}=(18+x+y+z) r_{G n .}= \\
& (18+x+y+z) h / 2=672
\end{aligned}
$$

Then $28 h=672, h=24$. Let $B P \perp A D, C N \perp A D$. In triangle $A B P$ by the Pythagorean theorem we have: $h^{2}+(a-x)^{2}=(a+x)^{2}, h^{2}=4 a x, x=8, y+z=30$. In triangle $C D N$ by the Pythagorean theorem we have: $h^{2}+(z-y)^{2}=(z+y)^{2}, h^{2}=4 y z, \quad y z=144$. We find $y, z$ by solving the equation $t^{2}-30 t+144=0, t_{1}=24, t_{2}=6$. We have two solutions: 1) $y=6, z=24,2$ ) $y=24, z=6$. Finally, we get 1) $A B=26, B C=14, C D=30, A D=42$, 2) $A B=26$, $B C=32, C D=30, A D=24$.

Answer: 1) $A B=26, B C=14, C D=30, A D=42$, 2) $A B=26, B C=32, C D=30, A D=24$.

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-12.jpg?height=286&width=487&top_left_y=1799&top_left_x=248)
1)

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-12.jpg?height=282&width=688&top_left_y=1798&top_left_x=1118)