Difference between revisions of "2016 AMC 10B Problems/Problem 9"

(Problem)
(Solution)
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==Solution==
 
==Solution==
 
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Let the point in the first quadrant be <math>(a, a^2)</math>.
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Then, the area of the triangle is <math>\frac{2a\cdot a^2}{2}=a^3</math>.
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Solving the equation <math>a^3=64</math> for <math>a</math>, we get <math>a=4</math>, so <math>BC=2a=8</math>.
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So the answer  is <math>(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 13:55, 21 February 2016

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 $BC$?

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

Solution

Let the point in the first quadrant be $(a, a^2)$. Then, the area of the triangle is $\frac{2a\cdot a^2}{2}=a^3$. Solving the equation $a^3=64$ for $a$, we get $a=4$, so $BC=2a=8$. So the answer is $(C) 8$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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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