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>, | + | 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=5.</math> |
+ | <math>y=5</math>, and <math>y=x+1</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>, | + | 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. | ||
− | <math>y=1-x</math> and <math>y=1+x</math> intersect at <math>(0,1)</math>. | + | 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>. | + | 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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+ | 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== |
Revision as of 13:32, 7 November 2020
Problem 21
What is the area of the triangle formed by the lines , , and ?
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 and get Plugging this into the equation, , and intersect at , we call this line x.
Doing the same thing, we get Thus also. , and intersect at , we call this line y.
It's apparent the only solution to is Thus, and intersect at , we call this line z.
Using the Shoelace Theorem we get: So our answer is
We might also see that the lines and are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by 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 so
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 which is equal to .
Video Solutions
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 (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.