[ [Assist

In triangle $ABC$, the height $BD$ is 11.2 and the height $AE$ is 12. Point $E$ lies on side $BC$ and $BE$ : $EC=5: 9$. Find the side $AC$.

#
The product of the base and height for the given triangle is constant.

## Solution

Let $B E=5 x, C E=9 x$. By the Pythagorean theorem from the right triangle $A E C$, we find that

$$
\begin{aligned}
A C= & \sqrt{E C^{2}+A E^{2}}= \\
& \sqrt{81 x^{2}+144}
\end{aligned}
$$

From the formula for the area of a triangle, it follows that

$$
\begin{gathered}
A C \cdot B D=B C \cdot A E, \text { or } \\
\sqrt{81 x^{2}+144} \cdot 11.2=14 x \cdot 12
\end{gathered}
$$

From this equation, we find that $x$ $=1$. Therefore,

$$
A C=\sqrt{81+144}=15
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-08.jpg?height=409&width=594&top_left_y=1802&top_left_x=46)

## Answer

15. 

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