Difference between revisions of "1994 AHSME Problems/Problem 7"
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--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] | ||
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+ | == Solution 2 == | ||
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+ | `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> | ||
==See Also== | ==See Also== |
Latest revision as of 02:36, 28 May 2021
Contents
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 TheMaskedMagician
Solution 2
`Note the overlap area is exactly the area of . So the total area is
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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