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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
