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

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First we need to find the coordinates where the graphs intersect.  
 
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>,  
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We want the points x and y to be the same. Thus, we set <math>5=x+1,</math> and get <math>x=4.</math> Plugging this into the equation, <math>y=1-x,</math>
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<math>y=5</math>, and <math>y=1+x</math> intersect at <math>(4,5)</math>, we call this line x.
  
<math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>,
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Doing the same thing, we get <math>x=-4.</math> Thus <math>y=5</math> also.
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<math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>, we call this line y.
  
<math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>.  
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It's apparent the only solution to <math>1-x=1+x</math> is <math>0.</math> Thus, <math>y=1.</math>
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<math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>, we call this line z.
  
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>.
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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
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We might also see that the lines <math>y</math> and <math>z</math> are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by <math>-1</math> to get the other. As the base is horizontal, this is a isosceles triangle with base 8, as the intersection points have distance 8. The height is <math>5-1=4,</math> so <math>\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}</math>
  
~more edits by BakedPotato66
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Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!
  
 
==Solution 2==
 
==Solution 2==
Graphing the lines, using the intersection points we found in Solution 1, 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 <math>\frac{4\cdot8}{2}</math> which is equal to <math>\boxed{\textbf{(E)}\ 16}</math>.  
+
Graphing the lines, using the intersection points we found in Solution 1, 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 <math>\frac{4\cdot8}{2}</math> which is equal to <math>\boxed{\textbf{(E)}\ 16}</math>.
  
~SmileKat32
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==Solution 3==
  
~more edits by BakedPotato69
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<math>y = x + 1</math> and <math>y = -x + 1</math> have <math>y</math>-intercepts at <math>(1, 0)</math> and slopes of <math>1</math> and <math>-1</math>, respectively. Since the product of these slopes is <math>-1</math>, the two lines are perpendicular. From <math>y = 5</math>, we see that <math>(-4, 5)</math> and <math>(4, 5)</math> are the other two intersection points, and they are <math>8</math> units apart. By symmetry, this triangle is a <math>45-45-90</math> triangle, so the legs are <math>4\sqrt{2}</math> each and the area is <math>\frac{(4\sqrt{2})^2}{2} = \boxed{\textbf{(E)}\ 16}</math>.
 +
==Video Solutions==
 +
https://www.youtube.com/watch?v=mz3DY1rc5ao - Happytwin
 +
 
 +
Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY
  
==Video Solutions==
 
 
https://www.youtube.com/watch?v=9nlX9VCisQc
 
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
  
https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx
+
https://youtu.be/RvtOX17DemY
  
https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25
+
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 19:48, 16 January 2021

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.

We want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$ $y=5$, and $y=1+x$ intersect at $(4,5)$, we call this line x.

Doing the same thing, we get $x=-4.$ Thus $y=5$ also. $y=5$, and $y=1-x$ intersect at $(-4,5)$, we call this line y.

It's apparent the only solution to $1-x=1+x$ is $0.$ Thus, $y=1.$ $y=1-x$ and $y=1+x$ intersect at $(0,1)$, we call this line z.

Using the Shoelace Theorem we get: \[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\] $=$ So our answer is $\boxed{\textbf{(E)}\ 16.}$

We might also see that the lines $y$ and $z$ are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by $-1$ to get the other. As the base is horizontal, this is a isosceles triangle with base 8, as the intersection points have distance 8. The height is $5-1=4,$ so $\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}$

Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!

Solution 2

Graphing the lines, using the intersection points we found in Solution 1, 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 $\boxed{\textbf{(E)}\ 16}$.

Solution 3

$y = x + 1$ and $y = -x + 1$ have $y$-intercepts at $(1, 0)$ and slopes of $1$ and $-1$, respectively. Since the product of these slopes is $-1$, the two lines are perpendicular. From $y = 5$, we see that $(-4, 5)$ and $(4, 5)$ are the other two intersection points, and they are $8$ units apart. By symmetry, this triangle is a $45-45-90$ triangle, so the legs are $4\sqrt{2}$ each and the area is $\frac{(4\sqrt{2})^2}{2} = \boxed{\textbf{(E)}\ 16}$.

Video Solutions

https://www.youtube.com/watch?v=mz3DY1rc5ao - Happytwin

Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY

https://www.youtube.com/watch?v=9nlX9VCisQc

https://www.youtube.com/watch?v=Z27G0xy5AgA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=3 ~ MathEx

https://youtu.be/RvtOX17DemY

~savannahsolver

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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