2006 Cyprus MO/Lyceum/Problem 14

Problem

2006 CyMO-14.PNG

The rectangle $AB\Gamma \Delta$ is a small garden divided to the rectangle $AZE\Delta$ and to the square $ZB\Gamma E$, so that $AE=2\sqrt{5}\ \text{m}$ and the shaded area of the triangle $\Delta BE$ is $4\ \text{m}^2$. The area of the whole garden is

$\mathrm{(A)}\ 24\ \text{m}^2\qquad\mathrm{(B)}\ 20\ \text{m}^2\qquad\mathrm{(C)}\ 16\ \text{m}^2\qquad\mathrm{(D)}\ 32\ \text{m}^2\qquad\mathrm{(E)}\ 10\sqrt{5}\ \text{m}^2$

Solution

The area of triangle $\Delta BE$ is equal to the area of triangle $\Delta AE$, so the area of rectangle $\Delta AZE$ is $4*2=8$. Let $AZ=x$ and $XE=y$. $xy=8$, and $x^2+y^2=20$. Thus

$(x+y)^2=x^2+y^2+2xy=36\Rightarrow x+y=6$. Thus $x=2$ and $y=4$. So we have $[A\Delta \Gamma B]=8+4^2=24\Rightarrow \mathrm{(A)}$.

Note: The answer theoretically can be 12, since we are not given that $AZ<ZE$. If $AZ=4$ and $ZE=2$, we have a 2*2 square and a 4*2 rectangle, with a diagonal of $2\sqrt{5}$. But 12 is not one of the answers included.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 13
Followed by
Problem 15
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