Difference between revisions of "1994 AHSME Problems/Problem 7"
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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 <math>\triangle ABG</math> and subtract this from <math>200</math>, the total area of the two squares. | 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 <math>\triangle ABG</math> and subtract this from <math>200</math>, the total area of the two squares. | ||
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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> | ||
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician |
Revision as of 16:23, 28 June 2014
Problem
Squares and
are congruent,
, and
is the center of square
. The area of the region in the plane covered by these squares is
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 and subtract this from
, the total area of the two squares.
Since is the center of
,
is half of the diagonal of the square. The diagonal of
is
so
. Since
is a square,
. So
is an isosceles right triangle. Its area is
. Therefore, the area of the region is
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician