Difference between revisions of "2024 AMC 8 Problems/Problem 11"
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-Benedict T (countmath1) | -Benedict T (countmath1) | ||
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| + | ==Video Solution by Math-X (First understand the problem!!!)== | ||
| + | https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315 | ||
==Video Solution (easy to digest) by Power Solve== | ==Video Solution (easy to digest) by Power Solve== | ||
https://www.youtube.com/watch?v=2UIVXOB4f0o | https://www.youtube.com/watch?v=2UIVXOB4f0o | ||
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| − | |||
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~Math-X | ~Math-X | ||
Revision as of 07:58, 31 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by Math-X (First understand the problem!!!)
- 5 Video Solution (easy to digest) by Power Solve
- 6 Video Solution by NiuniuMaths (Easy to understand!)
- 7 Video Solution 3 by SpreadTheMathLove
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 9 See Also
- 10 See Also
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 by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=2UIVXOB4f0o
~Math-X
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
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.
