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Difference between revisions of "1994 AHSME Problems/Problem 7"

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Since <math>G</math> is the center of <math>ABCD</math>, <math>BG</math> is half of the diagonal of the square. The diagonal of <math>ABCD</math> is <math>10\sqrt{2}</math> so <math>BG=5\sqrt{2}</math>. Since <math>EFGH</math> is a square, <math>\angle G=90^\circ</math>. So <math>\triangle ABG</math> is an isosceles right triangle. Its area is <math>\frac{(5\sqrt{2})^2}{2}=\frac{50}{2}=25</math>. Therefore, the area of the region is <math>200-25=\boxed{\textbf{(E) }175.}</math>
 
Since <math>G</math> is the center of <math>ABCD</math>, <math>BG</math> is half of the diagonal of the square. The diagonal of <math>ABCD</math> is <math>10\sqrt{2}</math> so <math>BG=5\sqrt{2}</math>. Since <math>EFGH</math> is a square, <math>\angle G=90^\circ</math>. So <math>\triangle ABG</math> is an isosceles right triangle. Its area is <math>\frac{(5\sqrt{2})^2}{2}=\frac{50}{2}=25</math>. Therefore, the area of the region is <math>200-25=\boxed{\textbf{(E) }175.}</math>
  
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 +
 
 +
== Solution 2 ==
 +
 
 +
`Note the overlap area <math>\triangle ABG</math> is exactly <math>1/4</math> the area of <math>ABCD</math>.  So the total area is <math>100 + 100 - 100/4 = 175</math>
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==See Also==
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{{AHSME box|year=1994|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 01:36, 28 May 2021

Problem

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] $\textbf{(A)}\ 75 \qquad\textbf{(B)}\ 100 \qquad\textbf{(C)}\ 125 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 175$

Solution

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

Solution 2

`Note the overlap area $\triangle ABG$ is exactly $1/4$ the area of $ABCD$. So the total area is $100 + 100 - 100/4 = 175$

See Also

1994 AHSME (Problems • Answer Key • Resources)
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

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