Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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== Problem == | == Problem == | ||
− | == Solution == | + | 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> \textbf{(A) }\ \frac {\sqrt {2}}{2}\qquad \textbf{(B) }\ \frac {3}{4}\qquad \textbf{(C) }\ \frac {\sqrt {3}}{2}\qquad \textbf{(D) }\ 1\qquad \textbf{(E) }\ \sqrt {2}</math> | ||
+ | |||
+ | == Solutions == | ||
+ | ===Solution 1 (trig) === | ||
+ | The area of a triangle can be given by <math>\frac12 ab \sin C</math>. <math>MC=1</math> because it is the midpoint of a side, and <math>CD=2</math> because it is the same length as <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) \sin | ||
+ | 120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}</math>. | ||
+ | Note: Even if you don't know the value of <math>\sin 120^\circ</math>, you can use the fact that <math>\sin x = \sin (180^\circ - x)</math>, so <math>\sin 120^\circ = \sin 60^\circ</math>. | ||
+ | You can easily calculate <math>\sin 60^\circ</math> to be <math>\frac{\sqrt3}{2}</math> using equilateral triangles. | ||
+ | |||
+ | ~Minor Edits by doulai1 | ||
+ | |||
+ | ===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 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{\textbf{(C) }\frac{\sqrt{3}}{2}}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Draw a line from <math>M</math> to the midpoint of <math>\overline{BC}</math>. Call the midpoint of <math>\overline{BC}</math> <math>P</math>. This is an equilateral triangle, since the two segments <math>\overline{PC}</math> and <math>\overline{MC}</math> are identical, and <math>\angle C</math> is <math>60^{\circ}</math>. Using the [[Pythagorean Theorem]], point <math>M</math> to <math>\overline{BC}</math> is <math>\dfrac{\sqrt{3}}{2}</math>. Also, the length of <math>\overline{CD}</math> is 2, since <math>C</math> is the midpoint of <math>\overline{BD}</math>. So, our final equation is <math>\frac{\sqrt{3}}{2}\times2\over2</math>, which just leaves us with <math>\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}</math>. | ||
+ | |||
+ | ===Solution 4 === | ||
+ | Drop a vertical down from <math>M</math> to <math>BC</math>. Let us call the point of intersection <math>X</math> and the midpoint of <math>BC</math>, <math>Y</math>. We can observe that <math>\triangle AYC</math> and <math>\bigtriangleup MXC</math> are similar. By the [[Pythagorean theorem]], <math>AY</math> is <math>\sqrt3</math>. | ||
+ | |||
+ | Since <math>AC:MC=2:1,</math> we find <math>MX=\frac{\sqrt3}{2}.</math> Because <math>C</math> is the midpoint of <math>BD,</math> and <math>BC=2,</math> <math>CD=2.</math> Using the area formula, <math>\frac{CD*MX}{2}=\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}.</math> | ||
+ | |||
+ | ~ sdk652 | ||
+ | |||
+ | ===Solution 5 === | ||
+ | Think of <math>\triangle ABC</math> and <math>\triangle MCD</math> being independent. Now to find area's we just solve for ratios between the triangles that we can plug in the value of <math>2</math> (for a side of <math>\triangle ABC</math>) for. Looking at the information, we see that <math>C</math> is the midpoint of <math>\overline{BD}</math>, and this means that it bisects <math>\overline{BD}</math> which results in <math>BC=CD</math>. Now for the height, we can see that <math>M</math> is the midpoint of <math>\overline{AC}</math> which means that <math>AM=CM</math>, and in turn means that the height of <math>\triangle MCD</math> is half of that of <math>\triangle ABC</math>, and now plugging the ratios of the bases being the same while the height is half of the other triangle, we end up with the area of <math>\triangle MCD</math> being half of that of <math>\triangle ABC</math>. Now all that's left is to find the area of <math>\triangle ABC</math>, and for that, we plug in <math>2</math> which leads us to the answer of <math>3</math>, but since we need to divide by two, our final answer is <math>\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}.</math> | ||
+ | |||
+ | ~<math>\LaTeX</math> help by vadava_lx | ||
+ | |||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2005|ab=B|num-b=13|num-a=15}} | |
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:15, 23 June 2024
Contents
[hide]Problem
Equilateral has side length , is the midpoint of , and is the midpoint of . What is the area of ?
Solutions
Solution 1 (trig)
The area of a triangle can be given by . because it is the midpoint of a side, and because it is the same length as . Each angle of an equilateral triangle is so . The area is . Note: Even if you don't know the value of , you can use the fact that , so . You can easily calculate to be using equilateral triangles.
~Minor Edits by doulai1
Solution 2
In order to calculate the area of , we can use the formula , where is the base. We already know that , so the formula 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 .
Solution 3
Draw a line from to the midpoint of . Call the midpoint of . This is an equilateral triangle, since the two segments and are identical, and is . Using the Pythagorean Theorem, point to is . Also, the length of is 2, since is the midpoint of . So, our final equation is , which just leaves us with .
Solution 4
Drop a vertical down from to . Let us call the point of intersection and the midpoint of , . We can observe that and are similar. By the Pythagorean theorem, is .
Since we find Because is the midpoint of and Using the area formula,
~ sdk652
Solution 5
Think of and being independent. Now to find area's we just solve for ratios between the triangles that we can plug in the value of (for a side of ) for. Looking at the information, we see that is the midpoint of , and this means that it bisects which results in . Now for the height, we can see that is the midpoint of which means that , and in turn means that the height of is half of that of , and now plugging the ratios of the bases being the same while the height is half of the other triangle, we end up with the area of being half of that of . Now all that's left is to find the area of , and for that, we plug in which leads us to the answer of , but since we need to divide by two, our final answer is
~ help by vadava_lx
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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