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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>
  
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==Solution==
+
shit
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<asy>import graph;size(7cm,IgnoreAspect);
 
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real f(real x) {return x*x;}
 
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draw((0,0)--(4,16)--(-4,16)--cycle,blue);
 
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draw(graph(f,-5,5,operator ..),gray);
 
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xaxis("$x$");yaxis("$y$",-1);
 
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label("$y=x^2$",(4.5,20.25),E);
 
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draw((4.2,0)--(4.2,16),Arrows);
 
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label("$r^2$",(4.2,0)--(4.2,16),E);
 
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draw((0,17)--(4,17),Arrows);
 
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label("$r$",(0,17)--(4,17),N);
 
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</asy>
 
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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 17:09, 22 November 2022

Problem

All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$

shit

Video Solution

https://youtu.be/1pi0eiD3jHc

~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

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