[ Midline of a trapezoid ] $[$ Area of a trapezoid $]$

Find the area of trapezoid $A B C D$ with side $B C=5$, if the distances from vertices $A$ and $D$ to line $B C$ are 3 and 7, respectively.
Let $M$ be the midpoint of the side $A D$; $A_{1}, M_{1}$, and $D_{1}$ be the projections of points $A, M$, and $D$ onto the line $B C$. Since $M M_{1}$ is the midline of the trapezoid $A D D_{1} A_{1}$, then $M M_{1}=1 / 2\left(A A_{1}+D D_{1}\right)=5$. According to problem $\underline{54964} S_{A B C D}=2 S_{B M C}=$ $B C \cdot M M_{1}=5 \cdot 5=25$.

## Answer

25. 

Problem