Difference between revisions of "2020 AMC 10A Problems/Problem 12"
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==Solution 2 (CHEATING)== | ==Solution 2 (CHEATING)== | ||
Draw a to-scale diagram with your graph paper and straightedge. Measure the height and approximate the area. | Draw a to-scale diagram with your graph paper and straightedge. Measure the height and approximate the area. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/ZGwAasE32Y4 | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
Revision as of 22:16, 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. Also note that 
has 
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)
Solution 2 (CHEATING)
Draw a to-scale diagram with your graph paper and straightedge. Measure the height and approximate the area.
Video Solution
~IceMatrix
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. 
