2024 AMC 8 Problems/Problem 11
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 4
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution (easy to digest) by Power Solve
- 7 Video Solution by NiuniuMaths (Easy to understand!)
- 8 Video Solution 3 by SpreadTheMathLove
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 Video Solution by Interstigation
- 11 See Also
Problem
The coordinates of are
,
, and
, with
. The area of
is 12. What is the value of
?
Solution 1
The triangle has base which means its height satisfies
This means that
so the answer is
Solution 2
By the Shoelace Theorem, has area
. From the problem, this is equal to
. We now solve for y.
OR
OR
OR
However, since, as stated in the problem, , our only valid solution is
.
~ cxsmi
Solution 4
As in the figure, the triangle is determined by the vectors and
. Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that
. Expanding the determinants, we find that
or
. Solving each equation individually, we find that
or
. However, the problem states that
, so the only valid solution is
.
~ cxsmi (again!)
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315
~Math-X
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=2UIVXOB4f0o
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=RRTxlduaDs8
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=-64aBL-lEVg
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1063
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.