Difference between revisions of "2016 AMC 10B Problems/Problem 9"
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<math>\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16</math> | <math>\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16</math> | ||
| − | + | ==Solution== | |
| + | <asy>import graph;size(7cm,IgnoreAspect); | ||
| + | real f(real x) {return x*x;} | ||
| + | draw((0,0)--(4,16)--(-4,16)--cycle,blue); | ||
| + | draw(graph(f,-5,5,operator ..),gray); | ||
| + | xaxis("$x$");yaxis("$y$",-1); | ||
| + | label("$y=x^2$",(4.5,20.25),E); | ||
| + | draw((4.2,0)--(4.2,16),Arrows); | ||
| + | label("$r^2$",(4.2,0)--(4.2,16),E); | ||
| + | draw((0,17)--(4,17),Arrows); | ||
| + | label("$r$",(0,17)--(4,17),N); | ||
| + | </asy> | ||
| + | The area of the triangle is <math>\frac{(2r)(r^2)}{2} = r^3</math>, so <math>r^3=64\implies r=4</math>, giving a total distance across the top of <math>8</math>, which is answer <math>\textbf{(C)}</math>. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 19:46, 23 December 2022
Contents
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
![[asy]import graph;size(7cm,IgnoreAspect); real f(real x) {return x*x;} draw((0,0)--(4,16)--(-4,16)--cycle,blue); draw(graph(f,-5,5,operator ..),gray); xaxis("$x$");yaxis("$y$",-1); label("$y=x^2$",(4.5,20.25),E); draw((4.2,0)--(4.2,16),Arrows); label("$r^2$",(4.2,0)--(4.2,16),E); draw((0,17)--(4,17),Arrows); label("$r$",(0,17)--(4,17),N); [/asy]](http://latex.artofproblemsolving.com/f/f/1/ff1ff1c3607f4680df88baa2e03b81634391f450.png)
The area of the triangle is 
, so 
, giving a total distance across the top of 
, which is answer 
.
Video Solution
~savannahsolver
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. 
