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

(Video explaining solution)
m (Video Solutions)
(12 intermediate revisions by 7 users not shown)
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 <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.  
 +
 
 +
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>
 +
<math>y=5</math>, and <math>y=1+x</math> intersect at <math>(4,5)</math>, we call this line x.
 +
 
 +
Doing the same thing, we get <math>x=-4.</math> Thus <math>y=5</math> also.
 +
<math>y=5</math>, and <math>y=1-x</math> intersect at <math>(-4,5)</math>, we call this line y.
 +
 
 +
It's apparent the only solution to <math>1-x=1+x</math> is <math>0.</math> Thus, <math>y=1.</math>
 +
<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>
 +
 
 +
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>
 +
 
 +
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, 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
+
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>.
 +
 
 +
==Video Solutions==
 +
Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY
  
==Video explaining solution==
 
 
https://www.youtube.com/watch?v=9nlX9VCisQc
 
https://www.youtube.com/watch?v=9nlX9VCisQc
  
Line 17: Line 34:
  
 
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://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25
  
 
==See Also==
 
==See Also==

Revision as of 16:43, 22 November 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.

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

Video Solutions

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

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=aStuVhoD8wc- Also includes other problems from 21-25

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