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

m (Solution 1)
m (Solution 1)
Line 6: Line 6:
  
 
==Solution 1==
 
==Solution 1==
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 (-4,5). <math>y=1-x</math> and <math>y=1+x</math> intersect at (1,0). Using the [[Shoelace Theorem]] you get <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)</cmath>=<math>\frac{32}{2} = </math>\boxed{\textbf{(E)}\ 16}$.~heeeeeeheeeee
+
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 (-4,5). <math>y=1-x</math> and <math>y=1+x</math> intersect at (1,0). Using the [[Shoelace Theorem]] you get <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)</cmath>=<math>\frac{32}{2} = \boxed{\textbf{(E)}\ 16}</math>.~heeeeeeheeeee
  
 
==See Also==
 
==See Also==

Revision as of 20:53, 20 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

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

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

Invalid username
Login to AoPS