8. Given the right triangle $A B C$ with a right angle at $A$, we construct a square $B C D E$ on the hypotenuse (with $D, E$ on the opposite side of $A$ relative to $B C$). Knowing that the areas of triangles $A B E$ and $A C D$ are $6 \mathrm{~m}^{2}$ and $27 \mathrm{~m}^{2}$ respectively, what is the area of triangle $A B C$?
(A) $3 \sqrt{2} m^{2}$
(B) $6 \mathrm{~m}^{2}$
(C) $12 \mathrm{~m}^{2}$
(D) $9 \sqrt{2} \mathrm{~m}^{2}$
(E) $18 \mathrm{~m}^{2}$
8. The answer is $(\mathbf{D})$.

FIRST SOLUTION

We show that the product of the two given areas is equal to the square of the area of the triangle $A B C$ we are looking for. Let $P$ be the foot of the altitude from vertex $A$ of triangle $A C D$, $Q$ the foot of the altitude from vertex $A$ of triangle $A B E$, and $H$ the foot of the altitude from vertex $A$ of triangle $A B C$. The quadrilateral $A P C H$ is, by construction, a rectangle (the angles at $P$ and $H$ are right angles because they are formed by altitudes, the angle at $C$ is complementary to a right angle) so $A P$ is congruent to $C H$; similarly, $A Q$ is congruent to $B H$. We therefore know that $S(A C D)=\frac{1}{2} C D \cdot C H$ and $S(A B E)=\frac{1}{2} B E \cdot B H$; but, since $C D=B E=C B$ and since $C H \cdot B H=A H$ by the theorem of Euclid, the product of the two areas is

$$
\frac{C D \cdot C H}{2} \frac{B E \cdot B H}{2}=\frac{C B^{2} \cdot C H \cdot B H}{4}=\left(\frac{C B \cdot A H}{2}\right)^{2}=S(A B C)^{2} .
$$

Therefore, the area of $A B C$ is $\sqrt{27 m^{2} \cdot 6 m^{2}}=9 \sqrt{2} m^{2}$.
![](https://cdn.mathpix.com/cropped/2024_04_17_8e3f264af5d5452f36c1g-10.jpg?height=620&width=1406&top_left_y=427&top_left_x=354)

## SECOND SOLUTION

Draw the square $A X Y Z$ such that $B$ is on side $Z A$, $C$ is on side $A X$, $D$ is on side $X Y$, and $E$ is on side $Y Z$. Note that the triangles $A B C$, $X C D$, $D Y E$, and $E Z B$ are congruent: they have congruent angles and the same hypotenuse. On the other hand, $D X$ (which is congruent to $A C$) is the altitude relative to $A C$ in triangle $A C D$, so the area of triangle $A C D$ is $A C^{2} / 2$. Similarly, since $C Z$ and $A B$ are congruent, the area of triangle $A B E$ is $A B^{2} / 2$. Therefore, for the area $S$ of triangle $A B C$ we have

$$
S=\frac{A B \cdot A C}{2}=\sqrt{\frac{A B^{2}}{2} \cdot \frac{A C^{2}}{2}}=\sqrt{27 m^{2} \cdot 6 m^{2}}=9 \sqrt{2} \mathrm{~m}^{2} .
$$