Difference between revisions of "2019 AMC 8 Problems/Problem 21"

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(Solution 2)
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Graphing the lines, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get <math>\frac{4\cdot8}{2}</math> which is equal to <math>\boxed{\textbf{(E)}\ 16}</math>. ~SmileKat32
 
Graphing the lines, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get <math>\frac{4\cdot8}{2}</math> which is equal to <math>\boxed{\textbf{(E)}\ 16}</math>. ~SmileKat32
  
Video explaining solution:
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==Video explaining solution==
 
https://www.youtube.com/watch?v=9nlX9VCisQc
 
https://www.youtube.com/watch?v=9nlX9VCisQc
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 +
https://www.youtube.com/watch?v=mz3DY1rc5ao
  
 
==See Also==
 
==See Also==

Revision as of 15:09, 6 May 2020

Problem 21

What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$?

$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$

Solution 1

You need to first find the coordinates where the graphs intersect. $y=5$, and $y=x+1$ intersect at $(4,5)$. $y=5$, and $y=1-x$ intersect at $(-4,5)$. $y=1-x$ and $y=1+x$ intersect at $(0,1)$. Using the Shoelace Theorem you get \[\left(\frac{(20-4)-(-20+4)}{2}\right)\]=$\frac{32}{2}$=$\boxed{\textbf{(E)}\ 16}$.~heeeeeeheeeee

Solution 2

Graphing the lines, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get $\frac{4\cdot8}{2}$ which is equal to $\boxed{\textbf{(E)}\ 16}$. ~SmileKat32

Video explaining solution

https://www.youtube.com/watch?v=9nlX9VCisQc

https://www.youtube.com/watch?v=mz3DY1rc5ao

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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