Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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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>. | 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>. | ||
− | ===Solution 4 | + | ===Solution 4 === |
Drop a vertical down from <math>M</math> to <math>BC</math>. WLOG, let us call the point of intersection <math>X</math> and the midpoint of <math>BC</math>, <math>Y</math>. We can observe that <math>\triangle AYC</math> and <math>\triangle MXC</math> are similar. By the Pythagorean theorem, <math>AY</math> is <math>\sqrt3</math>. Since <math>AC:MC=2:1,</math> we find <math>MX=\frac{\sqrt3}{2}.</math> Because <math>C</math> is the midpoint of <math>BD,</math> and <math>BC=2,</math> <math>CD=2.</math> Using the area formula, <math>\frac{CD*MX}{2}=\frac{\sqrt3}{2},</math> <math>\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}.</math> | Drop a vertical down from <math>M</math> to <math>BC</math>. WLOG, let us call the point of intersection <math>X</math> and the midpoint of <math>BC</math>, <math>Y</math>. We can observe that <math>\triangle AYC</math> and <math>\triangle MXC</math> are similar. By the Pythagorean theorem, <math>AY</math> is <math>\sqrt3</math>. Since <math>AC:MC=2:1,</math> we find <math>MX=\frac{\sqrt3}{2}.</math> Because <math>C</math> is the midpoint of <math>BD,</math> and <math>BC=2,</math> <math>CD=2.</math> Using the area formula, <math>\frac{CD*MX}{2}=\frac{\sqrt3}{2},</math> <math>\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}.</math> | ||
Revision as of 14:34, 1 January 2020
Contents
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 .
Solution 4
Drop a vertical down from to . WLOG, 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 | |
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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