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

## 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, $$\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]$$ $$72=\frac 34\cdot [AMC]$$ $$[AMC]=96\rightarrow \boxed{\textbf{(C)}}.$$

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