2005 AMC 10A Problems/Problem 25
Contents
Problem
In we have , , and . Points and are on and respectively, with and . What is the ratio of the area of triangle to the area of the quadrilateral ?
Solution 1
We have
But , so
Note: If it is hard to understand why , you can use the fact that the area of a triangle equals . If angle , we have that .
Video Solution
CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00
Solution 2 (no trig)
We can let .
Since , .
So, .
This means that .
Thus,
-Conantwiz2023
Solution 3 (trig)
Using this formula:
Since the area of is equal to the area of minus the area of ,
.
Therefore, the desired ratio is
Note: was not used in this problem.
Solution 4
Let be on such that then we have Since we have Thus and Finally, after some calculations, .
~ Nafer
~ LaTeX changes by tkfun
Solution 5
Let the area of triangle ABC be denoted by [ABC] and the area of quadrilateral ABCD be denoted by [ABCD].
Let the area of be . and share a height, and the ratio of their bases are , so the area of is .
Similarly, and share a height, and the ratio of their bases is , so the ratio of . Therefore, The ratio which is answer choice .
~JH. L
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg