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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