2019 AMC 8 Problems/Problem 21

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 Solution by Pi Academy

https://youtu.be/VrNy1cWn8rA?si=RxsHRW6r2ZE6v45h

~ mondays_makemetweakout

Video Solutions

https://youtu.be/DUhONk6cPy4 by Marshmellow

https://youtu.be/IgpayYB48C4?si=i9Nnwi2KpuHkL82l&t=6214 by Math-X

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

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

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

https://youtu.be/1RLoLeJjWko by Education, the Study of Everything

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

https://youtu.be/-5C6ACg5Hl0 by savannahsolver

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

https://youtu.be/j3QSD5eDpzU?t=2607 by OmegaLearn ~ pi_is_3.14

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

https://youtu.be/Xm4ZGND9WoY by The Power of Logic ~Hayabusa1

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

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