Difference between revisions of "2019 AMC 10A Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | The two lines are <math>y= 2x-2</math> and <math>y = x/2+1</math>, which intersect the third line at <math>(4,6)</math> and <math>(6,4)</math>. So we have an isosceles triangle with base <math>2\sqrt{2}</math> and height <math>3\sqrt{2}</math> <math>\implies \boxed{(C) 6}</math>. | + | The two lines are <math>y= 2x-2</math> and <math>y = x/2+1</math>, which intersect the third line at <math>(4,6)</math> and <math>(6,4)</math>. So we have an isosceles triangle with base <math>2\sqrt{2}</math> and height <math>3\sqrt{2}</math> <math>\implies \boxed{(C) 6}</math>. |
==See Also== | ==See Also== | ||
Revision as of 17:56, 9 February 2019
- The following problem is from both the 2019 AMC 10A #7 and 2019 AMC 12A #5, so both problems redirect to this page.
Problem
Two lines with slopes 
and 
intersect at 
. What is the area of the triangle enclosed by these two lines and the line 
Solution
The two lines are 
and 
, which intersect the third line at 
and 
. So we have an isosceles triangle with base 
and height 
.
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AMC 10 Problems and Solutions | ||
| 2019 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
