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

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== Problem ==
 
== Problem ==
  
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Triangle <math>AMC</math> is isoceles with <math>AM = AC</math>. Medians <math>\overline{MV}</math> and <math>\overline{CU}</math> are perpendicular to each other, and <math>MV=CU=12</math>. What is the area of <math>\triangle AMC?</math>
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[asy]
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draw((-4,0)--(4,0)--(0,12)--cycle);
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draw((-2,6)--(4,0));
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draw((2,6)--(-4,0));
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draw((-2,6)--(2,6));
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label("M", (-4,0), W);
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label("C", (4,0), E);
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label("A", (0, 12), N);
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label("V", (2, 6), NE);
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label("U", (-2, 6), NW);
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draw(rightanglemark((-2,6),(0,4),(-4,0),30));
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[/asy]
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<math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 22:24, 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 $\triangle AMC?$ [asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); draw(rightanglemark((-2,6),(0,4),(-4,0),30)); [/asy]

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

Solution

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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