2019 AMC 8 Problems/Problem 21

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

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
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All AJHSME/AMC 8 Problems and Solutions

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