2024 AMC 8 Problems/Problem 11
Contents
Problem
The coordinates of 
are 
, 
, and 
, with 
. The area of is 12. What is the value of 
?
![[asy] draw((3,11)--(11,7)--(5,7)--(3,11)); dot((5,7)); label("$A(5,7)$",(5,7),S); dot((11,7)); label("$B(11,7)$",(11,7),S); dot((3,11)); label("$C(3,y)$",(3,11),NW); [/asy]](http://latex.artofproblemsolving.com/3/a/1/3a1150ef27a4bba82b5f1aa665dc3e1dbe92d9f8.png)
Solution 1
The triangle has base 
which means its height satisfies
![\[\dfrac{6h}{2}=3h=12.\]](http://latex.artofproblemsolving.com/e/4/4/e448c160a210aadf166e977d28c622077d4e0ddf.png)
This means that 
so the answer is 
Solution 2
![[asy] size(10cm); draw((5,7)--(11,7)--(3,11)--cycle); draw((3,11)--(3,7)--(5,7),red); draw((3,7.5)--(3.5,7.5)--(3.5,7)); label("$A(5,7)$", (5,7),S); label("$B(11,7)$", (11,7),S); label("$C(3,y)$", (3,11),W); label("$D(3,7)$", (3,7),SW); [/asy]](http://latex.artofproblemsolving.com/6/c/3/6c3d5e8cefbc0744689ea5ca762f90f5209701e9.png)
Label point 
as the point at which 
. We now have ![$[\triangle ABC] = [\triangle BCD] - [\triangle ACD]$](http://latex.artofproblemsolving.com/f/3/0/f308cc469812756561fc3e1e8503d82e6b1c9e6a.png)
, where the brackets denote areas. On the righthand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of 
are 
and 
. The two side lengths of 
are and 
Now,
![\[[\triangle ABC] = 12 = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2 = 3(y-7)\]](http://latex.artofproblemsolving.com/f/7/d/f7dcf7c67e31a3176ece6003c89c42b7d320ba2d.png)
Dividing by 
gives 
so 
-Benedict T (countmath1)
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=2UIVXOB4f0o
Video Solution by Math-X (First understand the problem!!!)
~Math-X
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=RRTxlduaDs8
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. 
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.
