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

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https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx
 
https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx
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https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25
  
 
==See Also==
 
==See Also==

Revision as of 21:41, 13 September 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, using the intersection points we found in Solution 1, 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

~more edits by BakedPotato66

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

https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25

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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