Difference between revisions of "2016 AMC 12B Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | Plotting points <math>B</math> and <math>C</math> on the graph shows that they are at <math>\left( -x,x^2\right)</math> and <math>\left( x,x^2\right)</math>, which is isosceles. By setting up the triangle area formula you get: <math>64=\frac{1}{2}*2x*x^2 = | + | By: Albert471 |
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+ | Plotting points <math>B</math> and <math>C</math> on the graph shows that they are at <math>\left( -x,x^2\right)</math> and <math>\left( x,x^2\right)</math>, which is isosceles. By setting up the triangle area formula you get: <math>64=\frac{1}{2}*2x*x^2 = 64=x^3</math> Making x=4, and the length of <math>BC</math> is <math>2x</math>, so the answer is <math>\boxed{\textbf{(C)}\ 8}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=5|num-a=7}} | {{AMC12 box|year=2016|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:47, 4 August 2017
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
By: Albert471
Plotting points and
on the graph shows that they are at
and
, which is isosceles. By setting up the triangle area formula you get:
Making x=4, and the length of
is
, so the answer is
.
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 12 Problems and Solutions |
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