Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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===Solution 2=== | ===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>. | ||
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+ | ===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 60°. Using the Pythagorean Theorem, point <math>M</math> to <math>\overline{BC} is \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>\dfrac{\sqrt{3}}{2}\times2\over2</math>, which just leaves us with <math>\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}</math>. | ||
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== 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 18:14, 12 February 2017
Problem
Equilateral has side length
,
is the midpoint of
, and
is the midpoint of
. What is the area of
?
Solution
Solution 1
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
.
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 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
.
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 60°. Using the Pythagorean Theorem, point
to
. Also, the length of
is 2, since
is the midpoint of
. So, our final equation is
, which just leaves us with
.
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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