Problem 1026

Solution:

We show that the product of the two given areas is equal to the square of the area of the triangle $ABC$ we are looking for. Let $P$ be the foot of the altitude from vertex $A$ of triangle $ACD$, $Q$ the foot of the altitude from vertex $A$ of triangle $ABE$, $H$ the foot of the altitude from vertex $A$ of triangle $ABC$. The quadrilateral $APCH$ 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) hence $AP$ is congruent to $CH$; likewise, $AQ$ is congruent to $BH$. We therefore know that $S(ACD)=\frac{1}{2} CD \cdot CH$ and $S(ABE)=\frac{1}{2} BE \cdot BH$; but, since $CD=BE=CB$ and since $CH \cdot BH=AH$ by Euclid's theorem, the product of the two areas equals
$$\frac{CD \cdot CH}{2} \cdot \frac{BE \cdot BH}{2} = \frac{CB^2 \cdot CH \cdot BH}{4} = \left(\frac{CB \cdot AH}{2}\right)^2 = S(ABC)^2.$$
Hence the area of $ABC$ is $\sqrt{27~m^2 \cdot 6~m^2} = 9 \sqrt{2}~m^2$.

Draw the square $AXYZ$ such that $B$ is on side $ZA$, $C$ is on side $AX$, $D$ is on side $XY$ and $E$ is on side $YZ$. Note that triangles $ABC$, $XCD$, $DYE$ and $EZB$ are congruent: they have congruent angles and the same hypotenuse. On the other hand, $DX$ (which is congruent to $AC$) is the height relative to $AC$ in triangle $ACD$, hence the area of triangle $ACD$ equals $AC^2 / 2$. Likewise, since $CZ$ and $AB$ are congruent, the area of triangle $ABE$ equals $AB^2 / 2$. Therefore for the area $S$ of triangle $ABC$ we have
$$S = \frac{AB \cdot AC}{2} = \sqrt{\frac{AB^2}{2} \cdot \frac{AC^2}{2}} = \sqrt{27~m^2 \cdot 6~m^2} = 9 \sqrt{2}~\mathrm{m}^2.$$