Difference between revisions of "2016 AMC 10B Problems/Problem 9"
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Then, the area of the triangle is <math>\frac{2a\cdot a^2}{2}=a^3</math>. | Then, the area of the triangle is <math>\frac{2a\cdot a^2}{2}=a^3</math>. | ||
Solving the equation <math>a^3=64</math> for <math>a</math>, we get <math>a=4</math>, so <math>BC=2a=8</math>. | Solving the equation <math>a^3=64</math> for <math>a</math>, we get <math>a=4</math>, so <math>BC=2a=8</math>. | ||
| − | So the answer is <math>\boxed{\textbf{(C)} 8}</math>. | + | So the answer is <math>\boxed{\textbf{(C)}\ 8}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2016|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:59, 21 February 2016
Problem
All three vertices of 
lie on the parabola defined by 
, with 
at the origin and 
parallel to the 
-axis. The area of the triangle is 
. What is the length of 
?
Solution
Let the point in the first quadrant be 
.
Then, the area of the triangle is 
.
Solving the equation 
for 
, we get 
, so 
.
So the answer is 
.
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
