Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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== Solution == | == Solution == | ||
===Solution 1=== | ===Solution 1=== | ||
− | The area of a | + | The area of a triangle 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=== | ===Solution 2=== |
Revision as of 21:31, 26 November 2013
Contents
[hide]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 twice the length of
. 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
.
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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