Difference between revisions of "2020 AMC 10A Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | + | Since quadrilateral <math>UVCM</math> has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. However, also note that <math>\triangle AUV</math> is <math>\frac 14</math> the area of triangle <math>AMC</math> by similarity, so <math>[UVCM]=\frac 34\cdot [AMC].</math> Thus, | |
| + | <cmath>\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]</cmath> | ||
| + | <cmath>72=\frac 34\cdot [AMC]</cmath> | ||
| + | <cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath> | ||
==See Also== | ==See Also== | ||
Revision as of 22:12, 31 January 2020
Problem
Triangle 
is isoceles with 
. Medians 
and 
are perpendicular to each other, and 
. What is the area of 
Solution
Since quadrilateral 
has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. However, also note that 
is 
the area of triangle by similarity, so ![$[UVCM]=\frac 34\cdot [AMC].$](http://latex.artofproblemsolving.com/e/3/5/e354a04f6ddd58b67ae5cff873290204d44bfebf.png)
Thus,
![\[\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]\]](http://latex.artofproblemsolving.com/f/6/7/f67a43e7668067b662267a0d02d6a8a70ce12c37.png)
![\[72=\frac 34\cdot [AMC]\]](http://latex.artofproblemsolving.com/1/b/7/1b71cfc5fd3bd61542b19f5cad4ad6e8eada6754.png)
![\[[AMC]=96\rightarrow \boxed{\textbf{(C)}}.\]](http://latex.artofproblemsolving.com/9/d/5/9d51a9121e32738d01cf813e1f2392b35c767203.png)
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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