Difference between revisions of "2020 AMC 10A Problems/Problem 12"
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== Problem == | == Problem == | ||
+ | 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> | ||
== Solution == | == Solution == |
Revision as of 22:24, 31 January 2020
Problem
Triangle is isoceles with . Medians and are perpendicular to each other, and . What is the area of [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]
Solution
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
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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