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

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==Solution 1==
 
==Solution 1==
You need to first find the coordinates where the graphs intersect. <math>y=5</math>, and <math>y=x+1</math> intersect at <math>(4,5)</math>. <math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>. <math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>. Using the [[Shoelace Theorem]] you get <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)</cmath>=<math>\frac{32}{2} </math>=<math> \boxed{\textbf{(E)}\ 16}</math>.~heeeeeeheeeee
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First we need to find the coordinates where the graphs intersect.  
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<math>y=5</math>, and <math>y=x+1</math> intersect at <math>(4,5)</math>,
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<math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>,
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<math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>.  
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Using the [[Shoelace Theorem]] we get: <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}</cmath> <math>=</math> So our answer is <math>\boxed{\textbf{(E)}\ 16}</math>.
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~heeeeeeheeeee
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~more edits by BakedPotato66
  
 
==Solution 2==
 
==Solution 2==

Revision as of 19:51, 18 August 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

First we need to 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 we get: \[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\] $=$ So our answer is $\boxed{\textbf{(E)}\ 16}$.

~heeeeeeheeeee ~more edits by BakedPotato66

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

https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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

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