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

(Solution 1)
(Problem 21)
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==Problem 21==
 
==Problem 21==
What is the area of the triangle formed by the lines <math>y=5</math>, <math>y=1+x</math>, and <math>7=1-x</math>?
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What is the area of the triangle formed by the lines <math>y=5</math>, <math>y=1+x</math>, and <math>y=1-x</math>?
  
 
<math>\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16</math>
 
<math>\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16</math>

Revision as of 17:02, 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}$.

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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