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

2019 AMC 8 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2019 AMC 8 Problems/Problem 21

Contents

Problem 21

Solution 1

Solution 2

Video explaining solution

See Also