Difference between revisions of "2021 Fall AMC 10B Problems/Problem 15"
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| + | Note that <math>\triangle APB \cong \triangle BQC.</math> Then, it follows that <math>\overline{PB} \cong \overline{QC}.</math> Thus, <math>QC = PB = PR + RB = 7 + 6 = 13.</math> Define <math>x</math> to be the length of side <math>CR,</math> then <math>RQ = 13-x.</math> Because <math>\overline{BR}</math> is the altitude of the triangle, we can use the property that <math>QR \cdot RC = BR^2.</math> Substituting the given lengths, we have <cmath>(13-x) \cdot x = 36.</cmath> Solving, gives <math>x = 4</math> and <math>x = 9.</math> We eliminate the possibilty of <math>x=4</math> because <math>RC > QR.</math> Thus, the side lengnth of the square, by Pythagorean Theorem, is <cmath>\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.</cmath> Thus, the area of the sqaure is <math>(\sqrt{117})^2 = 117.</math> Thus, the answer is <math>\boxed{(\textbf{D}.)}.</math> | ||
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| + | ~NH14 | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=16|num-b=14}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=16|num-b=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:26, 23 November 2021
Solution
Note that 
Then, it follows that 
Thus, 
Define 
to be the length of side 
then 
Because 
is the altitude of the triangle, we can use the property that 
Substituting the given lengths, we have ![\[(13-x) \cdot x = 36.\]](http://latex.artofproblemsolving.com/c/a/6/ca645dc62b5b9083e880a83a0e861099eeec54a2.png)
Solving, gives 
and 
We eliminate the possibilty of 
because 
Thus, the side lengnth of the square, by Pythagorean Theorem, is ![\[\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.\]](http://latex.artofproblemsolving.com/b/3/a/b3ab8ad3129aca4e4a429de69029760e4aaa7cf0.png)
Thus, the area of the sqaure is 
Thus, the answer is 
~NH14
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
