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

Revision as of 19:51, 18 August 2020

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

~heeeeeeheeeee ~more edits by BakedPotato66

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

Video explaining solution

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

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

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

@@ Line 6: / Line 6: @@
 ==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 <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>(0,1)</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
+First we need to 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>(0,1)</math>.
+Using the [[Shoelace Theorem]] we get: <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}</cmath> <math>=</math> So our answer is <math>\boxed{\textbf{(E)}\ 16}</math>.
+~heeeeeeheeeee
+~more edits by BakedPotato66
 ==Solution 2==

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