2019 AMC 8 Problems/Problem 21

Revision as of 19:36, 24 October 2020 by Andrewaudrey02 (talk | contribs) (Solution 2)

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 $\boxed{\textbf{(E)}\ 16}$.

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

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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

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