Difference between revisions of "2012 AMC 12A Problems/Problem 10"

Revision as of 23:16, 13 February 2012

Problem

A triangle has area $30$ , one side of length $10$ , and the median to that side of length $9$ . Let $\theta$ be the acute angle formed by that side and the median. What is $\sin{\theta}$ ?

$\textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10}$

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]$

$AB$ is the side of length $10$ , and $CD$ is the median of length $9$ . The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$ . To find $\sin{\theta}$ , just use opposite over hypotenuse with the right triangle $\triangle DCE$ . This is equal to $\frac69=\boxed{\textbf{(D)}\ \frac23}$ .

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 $\triangle BCD = 15$ in the above figure. Therefore, $\frac{1}{2} \cdot 5 \cdot 9 \cdot \sin{\theta} = 15$ . Solving for $\theta$ gives $\theta = \frac{2}{3}$ . $\boxed{D}$ .

@@ Line 35: / Line 35: @@
 ===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 <math>\triangle BCD = 15</math> in the above figure. Therefore, <math>\frac{1}{2} \cdot 5 \cdot 9 \cdot \sin{\theta} = 15</math>. Solving for <math>\theta</math> gives <math>\theta = \frac{2}{3}</math>. \boxed{D}.
+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 <math>\triangle BCD = 15</math> in the above figure. Therefore, <math>\frac{1}{2} \cdot 5 \cdot 9 \cdot \sin{\theta} = 15</math>. Solving for <math>\theta</math> gives <math>\theta = \frac{2}{3}</math>. <math>\boxed{D}</math>.
 == See Also ==
 {{AMC12 box|year=2012|ab=A|num-b=9|num-a=11}}

2012 AMC 12A (Problems • Answer Key • Resources)
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