2018 AMC 10A Problems/Problem 24
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 .
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
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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