1994 AHSME Problems/Problem 7


Squares $ABCD$ and $EFGH$ are congruent, $AB=10$, and $G$ is the center of square $ABCD$. The area of the region in the plane covered by these squares is [asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,5)--(12,-2)--(5,-9)--(-2,-2)--cycle); label("A", (0,0), W); label("B", (10,0), E); label("C", (10,10), NE); label("D", (0,10), NW); label("G", (5,5), N); label("F", (12,-2), E); label("E", (5,-9), S); label("H", (-2,-2), W); dot((-2,-2)); dot((5,-9)); dot((12,-2)); dot((0,0)); dot((10,0)); dot((10,10)); dot((0,10)); dot((5,5)); [/asy] $\textbf{(A)}\ 75 \qquad\textbf{(B)}\ 100 \qquad\textbf{(C)}\ 125 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 175$


The area of the entire region in the plane is the area of the figure. However, we cannot simply add the two areas of the squares. We find the area of $\triangle ABG$ and subtract this from $200$, the total area of the two squares.

Since $G$ is the center of $ABCD$, $BG$ is half of the diagonal of the square. The diagonal of $ABCD$ is $10\sqrt{2}$ so $BG=5\sqrt{2}$. Since $EFGH$ is a square, $\angle G=90^\circ$. So $\triangle ABG$ is an isosceles right triangle. Its area is $\frac{(5\sqrt{2})^2}{2}=\frac{50}{2}=25$. Therefore, the area of the region is $200-25=\boxed{\textbf{(E) }175.}$

--Solution by TheMaskedMagician

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS