Difference between revisions of "2018 AMC 10A Problems/Problem 24"
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== Solution 3 == | == Solution 3 == | ||
The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> by Law of Sines. So the area of <math>\bigtriangleup ABG</math> is <math>100</math>. Since <math>\overline{DE}</math> is the midsegment of <math>\bigtriangleup ABC</math>, so <math>\overline{DF}</math> is the midsegment of <math>\bigtriangleup ABG</math> . So the area of <math>\bigtriangleup ACG</math> to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math> , so the area of <math>\bigtriangleup ACG</math> is <math>25</math>, by similar triangles. Therefore the area of quad <math>FDBG</math> is <math>100-25=\boxed{75}</math> | The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> by Law of Sines. So the area of <math>\bigtriangleup ABG</math> is <math>100</math>. Since <math>\overline{DE}</math> is the midsegment of <math>\bigtriangleup ABC</math>, so <math>\overline{DF}</math> is the midsegment of <math>\bigtriangleup ABG</math> . So the area of <math>\bigtriangleup ACG</math> to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math> , so the area of <math>\bigtriangleup ACG</math> is <math>25</math>, by similar triangles. Therefore the area of quad <math>FDBG</math> is <math>100-25=\boxed{75}</math> | ||
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==Solution 4 == | ==Solution 4 == |
Revision as of 23:20, 27 January 2019
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 1
Let , , , and the length of the perpendicular to through be . By angle bisector theorem, we have that where . Therefore substituting we have that . By similar triangles, we have that , and the height of this trapezoid is . Then, we have that . We wish to compute , and we have that it is by substituting. (rachanamadhu)
Solution 2
For this problem, we have because of SAS 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 .
Solution 3
The area of to the area of is by Law of Sines. So the area of is . Since is the midsegment of , so is the midsegment of . So the area of to the area of is , so the area of is , by similar triangles. Therefore the area of quad is
Solution 4
The area of quadrilateral is the area of minus the area of . Notice, , so , and since , the area of . Given that the area of is , using on side yields . Using the Angle Bisector Theorem, , so the height of . Therefore our answer is -Solution by ktong
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.