Difference between revisions of "2005 AMC 10A Problems/Problem 8"
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<cmath>HE=6</cmath> | <cmath>HE=6</cmath> | ||
So, the area of the square is <math>6^2=\boxed{36} \Rightarrow \mathrm{(C)}</math>. | So, the area of the square is <math>6^2=\boxed{36} \Rightarrow \mathrm{(C)}</math>. | ||
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+ | CHECK OUT Video Solution: https://youtu.be/z6T9Zm6Jvk0 | ||
==See also== | ==See also== |
Revision as of 19:41, 30 October 2020
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 1, 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 .
CHECK OUT Video Solution: https://youtu.be/z6T9Zm6Jvk0
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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