2012 AMC 12A Problems/Problem 10
Problem
A triangle has area 
, one side of length 
, and the median to that side of length 
. Let 
be the acute angle formed by that side and the median. What is 
?
Solution
Solution 1
![[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0); draw(A--B--C--cycle); draw(D--C); draw(E--C); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,SW); label("$E$",E,S); label("$9$",(D--C),NW); label("$6$",(E--C)); label("$10$",(A--B),SE); [/asy]](http://latex.artofproblemsolving.com/6/f/0/6f0ff42fa4415e92913383af798361d060587c2f.png)
is the side of length , and 
is the median of length . The altitude of 
to is 
because the 0.5(altitude)(base)=Area of the triangle. is 
. To find , just use opposite over hypotenuse with the right triangle 
. This is equal to 
.
Solution 2
It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of 
in the above figure. Expressing the area in terms of , 
. Solving for gives 
. 
.
See Also
| 2012 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
