. Let $ABCD$ be an isosceles trapezoid, with the larger base $[AB]$. Suppose the diagonals intersect at a point $O$ such that $\frac{OA}{OC}=2$. Given that the area of triangle $BOC$ is 10, what is the area of trapezoid $ABCD$?
. Let $H$ be the foot of the altitude from $C$ in triangle $B O C$, and $K$ be the foot of the altitude from $A$ in triangle $A O D$.

![](https://cdn.mathpix.com/cropped/2024_05_10_3c9ab831f1301fbf5b93g-163.jpg?height=337&width=874&top_left_y=300&top_left_x=595)

The area of triangles $B O C$ and $A D O$ is equal to $10=C H \times O B=A K \times D O$. The area of triangle $D O C$ is $C H \times O D=C H \times O B / 2=5$. The area of triangle $A O B$ is $A K \times O B=A K \times 2 \times O D=20$. Finally, the area of trapezoid $A B C D$ is $2 \times 10+5+20=45$.