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 $\left(\frac{(20-4)-(-20+4)}{2}\right)$ = $\frac{32}{2} = \boxed{\textbf{(E)}\ 16}$ .~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

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2019 AMC 8 Problems/Problem 21

Contents

Problem 21

Solution 1

Solution 2

See Also