[ Pythagorean Theorem (direct and inverse).]

## [ Perimeter of a triangle $\quad$]

The perimeter of a right triangle $ABC (\angle C=90^{\circ})$ is 72, and the difference between the median $CK$ and the altitude $CM$ is 7.

Find the area of triangle $ABC$.
Let $A C=b, B C=a, C K=x$. Then $A B=2 x, C M=x-7$. According to the condition $a+b=72-2 x$.

Since $2 S_{A B C}=a b=2 x(x-7),(a+b)^{2}-\left(a^{2}+b^{2}\right)=2 a b$, then $4(36-x)^{2}-4 x^{2}=4 x(x-7)$. From this, $(36-x)^{2}-x^{2}$ $=x^{2}-7 x, x^{2}+65 x-1296=0, \quad x=\frac{-65+\sqrt{4225+5184}}{2}=\frac{-65+97}{2}=16$.

Therefore, $S_{A B C}=x(x-7)=16 \cdot 9=144$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-05.jpg?height=369&width=788&top_left_y=696&top_left_x=635)

## Answer

144. 

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