Difference between revisions of "2019 AMC 8 Problems/Problem 21"

Revision as of 20:36, 24 October 2020

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

First we need to 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 we get: $\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}$ $=$ So our answer is $\boxed{\textbf{(E)}\ 16}$ .

Solution 2

Graphing the lines, using the intersection points we found in Solution 1, 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

~more edits by BakedPotato69

Video Solutions

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

https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25

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.

@@ Line 15: / Line 15: @@
 Using the [[Shoelace Theorem]] we get: <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}</cmath> <math>=</math> So our answer is <math>\boxed{\textbf{(E)}\ 16}</math>.
-~heeeeeeheeeee
-~more edits by BakedPotato66
 ==Solution 2==

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