Question

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

$48$

$72$

$96$

$144$

$192$

Answer: C

Concepts:median (geometry)centroid coordinate geometry

Difficulty rating: 1660

Video solution:

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Written solution:

Let the centroid be the origin. Since a centroid divides each median in a $2:1$ ratio, we may place median $MV$ horizontally with $M=(8,0)$ and $V=(-4,0)$ , and median $CU$ vertically with $C=(0,8)$ and $U=(0,-4)$ .

Because $U$ is the midpoint of $AM$ , we get $A=2U-M=(-8,-8)$ . The area of $\triangle AMC$ is $\dfrac12 |(16,8)\times(8,16)|$ $=\dfrac12(256-64)=96$ . Thus, C is the correct answer.

2020 AMC 10A Problem 12

Answer: C

Difficulty rating: 1660

Video solution:

Written solution:

Problem 12 in Other Years