Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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===Solution 3=== | ===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 | + | 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}</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>\dfrac{\sqrt{3}}{2}\times2\over2</math>, which just leaves us with <math>\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}</math>. |
== See Also == | == See Also == |
Revision as of 23:36, 28 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
is
. 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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