# Difference between revisions of "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 (1,0). Using the Shoelace Theorem you get $$\left(\frac{(20-4)-(-20+4)}{2}\right)$$=$\frac{32}{2}=$\boxed{\textbf{(E)}\ 16}\$.

## See Also

 2019 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 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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