16. In the acute triangle $A B C$, the sides $A B$ and $B C$ measure $17 \, \text{m}$ and $25 \, \text{m}$, and the height $B H$ measures $15 \, \text{m}$. Consider the triangle $A^{\prime} B^{\prime} C^{\prime}$, contained in $A B C$, whose sides are parallel to those of $A B C$ and are $2 \, \text{m}$ away from these. How many $\text{m}^2$ does the area of $A^{\prime} B^{\prime} C^{\prime}$ measure?

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(A) $315 / 2$
(B) $280 / 3$
(C) 140
(D) 105
(E) 126
Problem 16. The correct answer is (B).

First, we notice that triangles $ABC$ and $A'B'C'$ share the same incenter. Indeed, considering the center $I$ and the radius $r$ (measured in meters) of the inscribed circle in $ABC$, the circle with center $I$ and radius $|r-2|$ will be tangent to the three sides of $A'B'C'$. Therefore, denoting $\mathcal{A}$ and $\mathcal{A}'$, in order, the areas (in $\mathrm{m}^{2}$) of $ABC$ and $A'B'C'$, the ratio $\mathcal{A}' / \mathcal{A}$ will be equal to $\left(\frac{r-2}{r}\right)^{2}$, that is, $\mathcal{A}'=\left(\frac{r-2}{r}\right)^{2} \cdot \mathcal{A}$. To determine the values of $\mathcal{A}$ and $r$, using the Pythagorean theorem, we find that (in meters) $\overline{AH}=\sqrt{\overline{AB}^{2}-\overline{BH}^{2}}=8$ and $\overline{CH}=\sqrt{\overline{BC}^{2}-\overline{BH}^{2}}=20$, from which $\overline{AC}=28$ and $\mathcal{A}=\frac{28 \cdot 15}{2}=210$. It follows that (still in meters) $r=\frac{2 \mathcal{A}}{\overline{AB}+\overline{BC}+\overline{AC}}=\frac{2 \cdot 210}{70}=6$ and thus $\mathcal{A}'=\left(\frac{4}{6}\right)^{2} \cdot 210=\frac{280}{3}$.

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