Difference between revisions of "2019 AMC 8 Problems/Problem 21"
m (→Solution 1) |
m (→Solution 1) |
||
| Line 6: | Line 6: | ||
==Solution 1== | ==Solution 1== | ||
| − | You need to first find the coordinates where the graphs intersect. <math>y=5</math>, and <math>y=x+1</math> intersect at (4,5). <math>y=5</math>, and <math>y=1-x</math> intersect at (-4,5). <math>y=1-x</math> and <math>y=1+x</math> intersect at (1,0). Using the [[Shoelace Theorem]] you get <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)</cmath>=<math>\frac{32}{2} = | + | You need to first find the coordinates where the graphs intersect. <math>y=5</math>, and <math>y=x+1</math> intersect at (4,5). <math>y=5</math>, and <math>y=1-x</math> intersect at (-4,5). <math>y=1-x</math> and <math>y=1+x</math> intersect at (1,0). Using the [[Shoelace Theorem]] you get <cmath>\left(\frac{(20-4)-(-20+4)}{2}\right)</cmath>=<math>\frac{32}{2} = \boxed{\textbf{(E)}\ 16}</math>.~heeeeeeheeeee |
==See Also== | ==See Also== | ||
Revision as of 20:53, 20 November 2019
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 (4,5). , and intersect at (-4,5). and intersect at (1,0). Using the Shoelace Theorem you get ![\[\left(\frac{(20-4)-(-20+4)}{2}\right)\]](http://latex.artofproblemsolving.com/8/1/9/81958394559aa60a28897e9a3df25fce2ce74b15.png)
=
.~heeeeeeheeeee
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. 
