19. Two identical rectangles with sides of length $3 \mathrm{~cm}$ and $9 \mathrm{~cm}$ are overlapping as in the diagram. What is the area of the overlap of the two rectangles?
A $12 \mathrm{~cm}^{2}$
B $13.5 \mathrm{~cm}^{2}$
C $14 \mathrm{~cm}^{2}$
D $15 \mathrm{~cm}^{2}$
E $16 \mathrm{~cm}^{2}$
SOLUTION
D

First we prove that the four white triangles are congruent. Note that they each have a right angle. Also angles $H G A$ and $F G E$ are equal (vertically opposite). The quadrilateral $A C E G$ is a parallelogram since each of its sides come from the rectangles. Hence angles $F G E$ and $G A C$ are equal (corresponding), and angles $G A C$ and $E C D$ are equal (corresponding). Therefore each triangle has a right-angle, and an angle equal to $H G A$, and therefore they each have the same angles. Moreover, they each have a corresponding side of length $3 \mathrm{~cm}$. By labelling triangle $A B C$ with lengths $3, x$ and $y$, Pythagoras' Theorem gives $x^{2}+3^{2}=y^{2}[1]$. The triangles $A B C$ and $C D E$ are congruent, so $C D=C B=x$. Since $A D=9$, we can see $x+y=9[2]$.
Equation [2] rearranges to $x=9-y$, so $x^{2}=(9-y)^{2}=81-$ $18 y+y^{2}$. Substituting this into [1] gives $81-18 y+y^{2}+3^{2}=y^{2}$, so $18 y=90$ and $y=5$. Thus $x=9-5=4$. The area of each white triangle is $\frac{1}{2} \times 3 x=\frac{1}{2} \times 3 \times 4=6 \mathrm{~cm}^{2}$. Thus the overlap is $3 \times 9-2 \times 6=15 \mathrm{~cm}^{2}$.