Difference between revisions of "2018 AMC 10A Problems/Problem 24"
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By angle bisector theorem, <math>BG=\frac{5a}{6}</math>. By similar triangles, <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>, where <math>h</math> is the length of the altitude to <math>BC</math>. Then <math>\frac{ah}{2}=120</math> and we wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}</math>. (trumpeter) | By angle bisector theorem, <math>BG=\frac{5a}{6}</math>. By similar triangles, <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>, where <math>h</math> is the length of the altitude to <math>BC</math>. Then <math>\frac{ah}{2}=120</math> and we wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}</math>. (trumpeter) | ||
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+ | == (More basic) Solution 2 (bad) == | ||
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+ | <math>\overline{DE}</math> is midway from <math>A</math> to <math>\overline{BC}</math>, and <math>DE = \frac{BC}{2}</math>. Therefore, <math>DE</math> is a quarter of the area, which is <math>30</math>. Using the angle bisector theorem in the same fashion as the previous problem, we get that one segment is <math>5</math> times the other. We want the larger piece, as described by the problem. Because the heights are identical, one area is <math>5</math> times the other, and <math>\frac{5}{6} \cdot 90 = \boxed{75}</math>. | ||
==See Also== | ==See Also== |
Revision as of 19:56, 8 February 2018
Problem
Triangle with and has area . Let be the midpoint of , and let be the midpoint of . The angle bisector of intersects and at and , respectively. What is the area of quadrilateral ?
Solution
By angle bisector theorem, . By similar triangles, , and the height of this trapezoid is , where is the length of the altitude to . Then and we wish to compute . (trumpeter)
(More basic) Solution 2 (bad)
is midway from to , and . Therefore, is a quarter of the area, which is . Using the angle bisector theorem in the same fashion as the previous problem, we get that one segment is times the other. We want the larger piece, as described by the problem. Because the heights are identical, one area is times the other, and .
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
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