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Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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draw(C--D--M--cycle);</asy><math> \textrm{(A)}\ \frac {\sqrt {2}}{2}\qquad \textrm{(B)}\ \frac {3}{4}\qquad \textrm{(C)}\ \frac {\sqrt {3}}{2}\qquad \textrm{(D)}\ 1\qquad \textrm{(E)}\ \sqrt {2}</math>
 
draw(C--D--M--cycle);</asy><math> \textrm{(A)}\ \frac {\sqrt {2}}{2}\qquad \textrm{(B)}\ \frac {3}{4}\qquad \textrm{(C)}\ \frac {\sqrt {3}}{2}\qquad \textrm{(D)}\ 1\qquad \textrm{(E)}\ \sqrt {2}</math>
 
== Solution ==
 
== Solution ==
 +
===Solution 1===
 +
The area of a circle can be given by <math>\frac12 ab \text{sin} C</math>. <math>MC=1</math> because it is the midpoint of a side, and <math>CD=2</math> because it is twice the length of <math>BC</math>. Each angle of an equilateral triangle is <math>60^\circ</math> so <math>\angle MCD = 120^\circ</math>. The area is <math>\frac12 (1)(2) \text{sin} 120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}</math>.
 +
 +
===Solution 2===
 
In order to calculate the area of <math>\triangle CDM</math>, we can use the formula <math>A=\dfrac{1}{2}bh</math>, where <math>\overline{CD}</math> is the base. We already know that <math>\overline{CD}=2</math>, so the formula now becomes <math>A=h</math>. We can drop verticals down from <math>A</math> and <math>M</math> to points <math>E</math> and <math>F</math>, respectively. We can see that <math>\triangle AEC \sim \triangle MFC</math>. Now, we establish the relationship that <math>\dfrac{AE}{MF}=\dfrac{AC}{MC}</math>. We are given that <math>\overline{AC}=2</math>, and <math>M</math> is the midpoint of <math>\overline{AC}</math>, so <math>\overline{MC}=1</math>. Because <math>\triangle AEB</math> is a <math>30-60-90</math> triangle and the ratio of the sides opposite the angles are <math>1-\sqrt{3}-2</math> <math>\overline{AE}</math> is <math>\sqrt{3}</math>. Plugging those numbers in, we have <math>\dfrac{\sqrt{3}}{MF}=\dfrac{2}{1}</math>. Cross-multiplying, we see that <math>2\times\overline{MF}=\sqrt{3}\times1\implies \overline{MF}=\dfrac{\sqrt{3}}{2}</math> Since <math>\overline{MF}</math> is the height <math>\triangle CDM</math>, the area is <math>\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}</math>.
 
In order to calculate the area of <math>\triangle CDM</math>, we can use the formula <math>A=\dfrac{1}{2}bh</math>, where <math>\overline{CD}</math> is the base. We already know that <math>\overline{CD}=2</math>, so the formula now becomes <math>A=h</math>. We can drop verticals down from <math>A</math> and <math>M</math> to points <math>E</math> and <math>F</math>, respectively. We can see that <math>\triangle AEC \sim \triangle MFC</math>. Now, we establish the relationship that <math>\dfrac{AE}{MF}=\dfrac{AC}{MC}</math>. We are given that <math>\overline{AC}=2</math>, and <math>M</math> is the midpoint of <math>\overline{AC}</math>, so <math>\overline{MC}=1</math>. Because <math>\triangle AEB</math> is a <math>30-60-90</math> triangle and the ratio of the sides opposite the angles are <math>1-\sqrt{3}-2</math> <math>\overline{AE}</math> is <math>\sqrt{3}</math>. Plugging those numbers in, we have <math>\dfrac{\sqrt{3}}{MF}=\dfrac{2}{1}</math>. Cross-multiplying, we see that <math>2\times\overline{MF}=\sqrt{3}\times1\implies \overline{MF}=\dfrac{\sqrt{3}}{2}</math> Since <math>\overline{MF}</math> is the height <math>\triangle CDM</math>, the area is <math>\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}</math>.
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2005|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2005|ab=B|num-b=13|num-a=15}}

Revision as of 15:50, 29 December 2012

Problem

Equilateral $\triangle ABC$ has side length $2$, $M$ is the midpoint of $\overline{AC}$, and $C$ is the midpoint of $\overline{BD}$. What is the area of $\triangle CDM$? [asy]defaultpen(linewidth(.8pt)+fontsize(8pt)); pair B = (0,0); pair A = 2*dir(60); pair C = (2,0); pair D = (4,0); pair M = midpoint(A--C); label("$A$",A,NW);label("$B$",B,SW);label("$C$",C, SE);label("$M$",M,NE);label("$D$",D,SE); draw(A--B--C--cycle); draw(C--D--M--cycle);[/asy]$\textrm{(A)}\ \frac {\sqrt {2}}{2}\qquad \textrm{(B)}\ \frac {3}{4}\qquad \textrm{(C)}\ \frac {\sqrt {3}}{2}\qquad \textrm{(D)}\ 1\qquad \textrm{(E)}\ \sqrt {2}$

Solution

Solution 1

The area of a circle can be given by $\frac12 ab \text{sin} C$. $MC=1$ because it is the midpoint of a side, and $CD=2$ because it is twice the length of $BC$. Each angle of an equilateral triangle is $60^\circ$ so $\angle MCD = 120^\circ$. The area is $\frac12 (1)(2) \text{sin} 120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}$.

Solution 2

In order to calculate the area of $\triangle CDM$, we can use the formula $A=\dfrac{1}{2}bh$, where $\overline{CD}$ is the base. We already know that $\overline{CD}=2$, so the formula now becomes $A=h$. We can drop verticals down from $A$ and $M$ to points $E$ and $F$, respectively. We can see that $\triangle AEC \sim \triangle MFC$. Now, we establish the relationship that $\dfrac{AE}{MF}=\dfrac{AC}{MC}$. We are given that $\overline{AC}=2$, and $M$ is the midpoint of $\overline{AC}$, so $\overline{MC}=1$. Because $\triangle AEB$ is a $30-60-90$ triangle and the ratio of the sides opposite the angles are $1-\sqrt{3}-2$ $\overline{AE}$ is $\sqrt{3}$. Plugging those numbers in, we have $\dfrac{\sqrt{3}}{MF}=\dfrac{2}{1}$. Cross-multiplying, we see that $2\times\overline{MF}=\sqrt{3}\times1\implies \overline{MF}=\dfrac{\sqrt{3}}{2}$ Since $\overline{MF}$ is the height $\triangle CDM$, the area is $\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}$.

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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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