Difference between revisions of "2005 AMC 10A Problems/Problem 8"
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==Solution== | ==Solution== | ||
− | We see that side <math>BE</math>, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So: | + | We see that side <math>BE</math>, which we know is <math>1</math>, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So: |
<cmath>1^2 + (HE+1)^2=\sqrt{50}^2</cmath> | <cmath>1^2 + (HE+1)^2=\sqrt{50}^2</cmath> | ||
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<cmath>HE=6</cmath> | <cmath>HE=6</cmath> | ||
− | So, the area of the square is <math>6^2=\boxed{ | + | So, the area of the square is <math>6^2=\boxed{\textbf{(C) }36}</math> |
==See also== | ==See also== |
Latest revision as of 10:58, 13 December 2021
Problem
In the figure, the length of side of square is and . What is the area of the inner square ?
Solution
We see that side , which we know is , is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, . Then , and is one of the sides of the square whose area we want to find. So:
So, the area of the square is
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 | ||
All AMC 10 Problems and Solutions |
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