Difference between revisions of "2019 AMC 8 Problems/Problem 21"
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==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>. |
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==Video Solutions== | ==Video Solutions== |
Revision as of 19:36, 24 October 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.
, and intersect at ,
, and intersect at ,
and intersect at .
Using the Shoelace Theorem we get: So our answer is .
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.