Difference between revisions of "2024 AMC 8 Problems/Problem 11"
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<cmath>\dfrac{6h}{2}=3h=12.</cmath> | <cmath>\dfrac{6h}{2}=3h=12.</cmath> | ||
This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math> | This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math> | ||
| + | |||
| + | ==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> | ||
| + | Label point <math>D(3,7)</math> as the point at which <math>CD\perp DA</math>. We now have <math>[\triangle ABC] = [\triangle BCD] - [\triangle ACD]</math>, 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 <math>\triangle ACD</math> are <math>y-7</math> and <math>5-3=2</math>. The two side lengths of <math>\triangle BCD</math> are <math>y-7</math> and <math>11-3 = 8.</math> Now, | ||
| + | |||
| + | <cmath>[\triangle ABC] = 12 = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2 = 3(y-7)</cmath> | ||
| + | |||
| + | Dividing by <math>3</math> gives <math>y -7 = 4,</math> so <math>y = \boxed{\textbf{(D)\ 11}}.</math> | ||
| + | |||
| + | -Benedict T (countmath1) | ||
==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== | ||
Revision as of 15:04, 26 January 2024
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 by Math-X (First understand the problem!!!)
~Math-X
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=2UIVXOB4f0o
