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

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==Solution 1==
 
==Solution 1==
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You need to first find the coordinates where the graphs intersect. <math>y=5</math>, and <math>y=x+1</math> intersect at (4,5). <math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>. <math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(1,0)</math>. Using the [[Shoelace Theorem]] you get <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)</cmath>=<math>\frac{32}{2} </math>=<math> \boxed{\textbf{(E)}\ 16}</math>.~heeeeeeheeeee
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You need to first find the coordinates where the graphs intersect. <math>y=5</math>, and <math>y=x+1</math> intersect at <math>(4,5)</math>. <math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>. <math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(1,0)</math>. Using the [[Shoelace Theorem]] you get <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)</cmath>=<math>\frac{32}{2} </math>=<math> \boxed{\textbf{(E)}\ 16}</math>.~heeeeeeheeeee
  
 
==Solution 2==
 
==Solution 2==

Revision as of 11:50, 21 November 2019

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

See Also

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

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