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

(Created page with "== Problem == A triangle has area <math>30</math>, one side of length <math>10</math>, and the median to that side of length <math>9</math>. Let <math>\theta</math> be the acut...")
 
(Solution)
Line 6: Line 6:
  
 
== Solution ==
 
== Solution ==
 +
 +
==Solution 1==
 +
 
<center><asy>
 
<center><asy>
 
unitsize(5mm);
 
unitsize(5mm);
Line 29: Line 32:
  
 
<math>AB</math> is the side of length <math>10</math>, and <math>CD</math> is the median of length <math>9</math>. The altitude of <math>C</math> to <math>AB</math> is <math>6</math> because the 0.5(altitude)(base)=Area of the triangle. <math>\theta</math> is <math>\angle CDE</math>. To find <math>\sin{\theta}</math>, just use opposite over hypotenuse with the right triangle <math>\triangle DCE</math>. This is equal to <math>\frac69=\boxed{\textbf{(D)}\ \frac23}</math>.
 
<math>AB</math> is the side of length <math>10</math>, and <math>CD</math> is the median of length <math>9</math>. The altitude of <math>C</math> to <math>AB</math> is <math>6</math> because the 0.5(altitude)(base)=Area of the triangle. <math>\theta</math> is <math>\angle CDE</math>. To find <math>\sin{\theta}</math>, just use opposite over hypotenuse with the right triangle <math>\triangle DCE</math>. This is equal to <math>\frac69=\boxed{\textbf{(D)}\ \frac23}</math>.
 +
 +
===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}.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2012|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2012|ab=A|num-b=9|num-a=11}}

Revision as of 00:16, 14 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}.

See Also

2012 AMC 12A (ProblemsAnswer KeyResources)
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