Difference between revisions of "2016 AMC 10B Problems/Problem 9"
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Fuzimiao2013 (talk | contribs) (Undo revision 182850 by Weluvrameninmumbai (talk)) (Tag: Undo) |
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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
[hide]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
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.