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

(Fixed up problem)
(Solution 1)
Line 6: Line 6:
  
 
==Solution 1==
 
==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 <cmath>\left(\frac{(-20+4)-(20-4}{2}\right)</cmath>=<math>\frac{32}{2}=</math>\boxed{\textbf{(E)}\ 16}$.
  
 
==See Also==
 
==See Also==

Revision as of 16:40, 20 November 2019

Problem 21

What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $7=1-x$?

$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$

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}$.

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