Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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== Solutions == | == Solutions == | ||
− | ===Solution 1 ( | + | ===Solution 1 (trig) === |
The area of a triangle can be given by <math>\frac12 ab \sin C</math>. <math>MC=1</math> because it is the midpoint of a side, and <math>CD=2</math> because it is the same length as <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) \sin | The area of a triangle can be given by <math>\frac12 ab \sin C</math>. <math>MC=1</math> because it is the midpoint of a side, and <math>CD=2</math> because it is the same length as <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) \sin | ||
120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}</math>. | 120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}</math>. | ||
Note: Even if you don't know the value of <math>\sin 120^\circ</math>, you can use the fact that <math>\sin x = \sin (180^\circ - x)</math>, so <math>\sin 120^\circ = \sin 60^\circ</math>. | Note: Even if you don't know the value of <math>\sin 120^\circ</math>, you can use the fact that <math>\sin x = \sin (180^\circ - x)</math>, so <math>\sin 120^\circ = \sin 60^\circ</math>. | ||
You can easily calculate <math>\sin 60^\circ</math> to be <math>\frac{\sqrt3}{2}</math> using equilateral triangles. | You can easily calculate <math>\sin 60^\circ</math> to be <math>\frac{\sqrt3}{2}</math> using equilateral triangles. | ||
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+ | ~Minor Edits by doulai1 | ||
===Solution 2=== | ===Solution 2=== |
Revision as of 15:59, 7 January 2022
Contents
Problem
Equilateral has side length , is the midpoint of , and is the midpoint of . What is the area of ?
Solutions
Solution 1 (trig)
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.
~Minor Edits by doulai1
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 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 . 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,
~ sdk652
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.