[The ratio of the areas of triangles with a common base or common height $]$ Complex Trapezoids (other)

The area of trapezoid $A B C D$ with bases $A D$ and $B C (A D > B C)$ is 48, and the area of triangle $A O B$, where $O-$ is the point of intersection of the diagonals of the trapezoid, is 9. Find the ratio of the bases of the trapezoid $A D: B C$.
$S_{\triangle \mathrm{AOD}}=\frac{D O}{O B} \cdot S_{\triangle \mathrm{AOB}}$

## Solution

Notice that triangles $A B D$ and $A C D$ are equal in area, as they share the same base and have equal heights. Therefore, triangles $C O D$ and $A O B$ are also equal in area.

Let $\frac{A D}{B C}=x>1$. From the similarity of triangles $A O D$ and $C O B$, it follows that $\frac{A O}{O C}=\frac{D O}{O B}=\frac{A D}{B C}=x$. Then,

$$
S_{\triangle \mathrm{AOD}}=\frac{D O}{O B} \cdot S_{\triangle \mathrm{AOB}}=9 x, \quad S_{\triangle \mathrm{BOC}}=\frac{C O}{O A} \cdot S_{\triangle \mathrm{AOB}}=\frac{9}{x}
$$

and since

$$
S_{\triangle \mathrm{AOD}}+S_{\triangle \mathrm{BOC}}=S_{\mathrm{ABCD}}-S_{\triangle \mathrm{AOB}}-S_{\triangle \mathrm{COD}}=48-2 \cdot 9=30
$$

we obtain the equation $9 x+\frac{9}{x}=30$, the larger root of which $(x=3)$ satisfies the condition $x>1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-35.jpg?height=352&width=734&top_left_y=676&top_left_x=665)

## Answer

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