2019 AMC 8 Problems/Problem 21

Revision as of 20:52, 20 November 2019 by Saksham27 (talk | contribs) (Solution 1)

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 (1,0). Using the Shoelace Theorem you get \[\left(\frac{(20-4)-(-20+4)}{2}\right)\]=$\frac{32}{2} =$\boxed{\textbf{(E)}\ 16}$.~heeeeeeheeeee

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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