Difference between revisions of "2020 AMC 10A Problems/Problem 12"

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(Solution)
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== Solution ==
 
== Solution ==
 
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Since quadrilateral <math>UVCM</math> has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. However, also note that <math>\triangle AUV</math> is <math>\frac 14</math> the area of triangle <math>AMC</math> by similarity, so <math>[UVCM]=\frac 34\cdot [AMC].</math> Thus,
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<cmath>\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]</cmath>
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<cmath>72=\frac 34\cdot [AMC]</cmath>
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<cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 23:12, 31 January 2020

Problem

Triangle $AMC$ is isoceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$

$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$

Solution

Since quadrilateral $UVCM$ has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. However, also note that $\triangle AUV$ is $\frac 14$ the area of triangle $AMC$ by similarity, so $[UVCM]=\frac 34\cdot [AMC].$ Thus, \[\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]\] \[72=\frac 34\cdot [AMC]\] \[[AMC]=96\rightarrow \boxed{\textbf{(C)}}.\]

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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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