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

(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>
 
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>
[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]
 
  
 
<math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math>
 
<math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math>

Revision as of 22:25, 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

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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