2018 AMC 10A Problems/Problem 24
Contents
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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