5) In a right triangle $A B C$, the legs $B C$ and $C A$ measure $7 \mathrm{~cm}$ and $24 \mathrm{~cm}$ respectively. Let $H$ be the projection of $C$ onto the hypotenuse $A B$. What is the perimeter of triangle $H B C$?
(A) $\frac{262}{25} \mathrm{~cm}$,
(B) $\frac{501}{49} \mathrm{~cm}$
(C) $\frac{392}{25} \mathrm{~cm}$,
(D) $\frac{801}{49} \mathrm{~cm}$
(E) $\frac{412}{25} \mathrm{~cm}$.
5. The answer is $\mathbf{( C )}$.

According to the Pythagorean theorem, the hypotenuse $A B$ of triangle $A B C$ measures $\sqrt{7^{2}+24^{2}} \mathrm{~cm}=$ $\sqrt{49+576} \mathrm{~cm}=\sqrt{625} \mathrm{~cm}=25 \mathrm{~cm}$. Therefore, the perimeter of triangle $A B C$ is $(24+7+$ $25) \mathrm{cm}=56 \mathrm{~cm}$. Triangle $H B C$ is also a right triangle and is similar to $A B C$. Therefore, the ratio between the perimeter of $H B C$ and that of $A B C$ is equal to the ratio of the measures of their respective hypotenuses, which is $\frac{7}{25}$. Therefore, the perimeter of $H B C$ is $56 \frac{7}{25} \mathrm{~cm}=\frac{392}{25} \mathrm{~cm}$.