2005 AMC 10B Problems/Problem 14
Contents
Problem
Equilateral has side length , is the midpoint of , and is the midpoint of . What is the area of ?
Solutions
Solution 1 (simplest)
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.
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 .
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,
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See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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