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 12:26, 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
Solving, gives
and
We eliminate the possibilty of
because
Thus, the side lengnth of the square, by Pythagorean Theorem, is
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.