Difference between revisions of "2019 AMC 8 Problems/Problem 21"
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==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, 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 | ||
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+ | Video explaining solution: | ||
+ | https://www.youtube.com/watch?v=9nlX9VCisQc | ||
==See Also== | ==See Also== |
Revision as of 13:54, 14 April 2020
Contents
Problem 21
What is the area of the triangle formed by the lines , , and ?
Solution 1
You need to first find the coordinates where the graphs intersect. , and intersect at . , and intersect at . and intersect at . Using the Shoelace Theorem you get ==.~heeeeeeheeeee
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 which is equal to . ~SmileKat32
Video explaining solution: https://www.youtube.com/watch?v=9nlX9VCisQc
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.