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>. | 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>. | ||
− | == | + | == Solution 2 == |
<math>\overline{DE}</math> is midway from <math>A</math> to <math>\overline{BC}</math>, and <math>DE = \frac{BC}{2}</math>. Therefore, <math>\bigtriangleup ADE</math> is a quarter of the area of <math>\bigtriangleup ABC</math>, which is <math>30</math>. Subsequently, we can compute the area of quadrilateral <math>BDEC</math> to be <math>120 - 90 = 30</math>. Using the angle bisector theorem in the same fashion as the previous problem, we get that <math>\overline{BG}</math> is <math>5</math> times the length of <math>\overline{GC}</math>. 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>. | <math>\overline{DE}</math> is midway from <math>A</math> to <math>\overline{BC}</math>, and <math>DE = \frac{BC}{2}</math>. Therefore, <math>\bigtriangleup ADE</math> is a quarter of the area of <math>\bigtriangleup ABC</math>, which is <math>30</math>. Subsequently, we can compute the area of quadrilateral <math>BDEC</math> to be <math>120 - 90 = 30</math>. Using the angle bisector theorem in the same fashion as the previous problem, we get that <math>\overline{BG}</math> is <math>5</math> times the length of <math>\overline{GC}</math>. 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>. |
Revision as of 21:54, 8 February 2018
Contents
[hide]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 .
Solution 2
is midway from to , and . Therefore, is a quarter of the area of , which is . Subsequently, we can compute the area of quadrilateral to be . Using the angle bisector theorem in the same fashion as the previous problem, we get that is times the length of . 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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