Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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== Problem == | == Problem == | ||
+ | Equilateral <math> \triangle ABC</math> has side length <math>2</math>, <math> M</math> is the midpoint of <math> \overline{AC}</math>, and <math> C</math> is the midpoint of <math> \overline{BD}</math>. What is the area of <math> \triangle CDM</math>? | ||
+ | <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><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 == | ||
+ | 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}} |
Revision as of 15:33, 12 July 2011
Problem
Equilateral has side length , is the midpoint of , and is the midpoint of . What is the area of ?
Solution
In order to calculate the area of , we can use the formula , where is the base. We already know that , so the formula now becomes . We can drop verticals down from and to points and , respectively. We can see that . Now, we establish the relationship that . We are given that , and is the midpoint of , so . Because is a triangle and the ratio of the sides opposite the angles are is . Plugging those numbers in, we have . Cross-multiplying, we see that Since is the height , the area is .
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 |