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

Revision as of 07:42, 21 December 2022

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 $(-4,5)$ , and 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 $x$ 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 $(0, 1)$ 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/-5C6ACg5Hl0

~savannahsolver

https://www.youtube.com/watch?v=SBGcumVOroI&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=23

Video Solutions

https://www.youtube.com/watch?v=x4cF3o3Fzj8&t=123s

2019 AMC 8 (Problems • Answer Key • Resources)
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Difference between revisions of "2019 AMC 8 Problems/Problem 21"

Revision as of 07:42, 21 December 2022

Contents

Problem 21

Solution 1

Solution 2

Solution 3

Video Solutions

Video Solutions

See Also

@@ Line 31: / Line 31: @@
 ==Video Solutions==
-https://www.youtube.com/watch?v=mz3DY1rc5ao - Happytwin
+https://www.youtube.com/watch?v=mz3DY1rc5ao
+- Happytwin
 Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY
@@ Line 37: / Line 39: @@
 https://www.youtube.com/watch?v=9nlX9VCisQc
-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/-5C6ACg5Hl0