Difference between revisions of "2014 AMC 10A Problems/Problem 19"
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By Pythagorean Theorem in three dimensions, the distance <math>XY</math> is <math>\sqrt{4^2+4^2+10^2}=2\sqrt{33}</math>. | By Pythagorean Theorem in three dimensions, the distance <math>XY</math> is <math>\sqrt{4^2+4^2+10^2}=2\sqrt{33}</math>. | ||
− | Let the length of the segment <math>XY</math> that is inside the cube with side length <math>3</math> be <math>x</math>. By similar triangles, <math>\dfrac{x}{3}=\dfrac{2\sqrt{33}}{10}</math>, giving <math>x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}}</math>. | + | Let the length of the segment <math>XY</math> that is inside the cube with side length <math>3</math> be <math>x</math>. By similar triangles, <math>\dfrac{x}{3}=\dfrac{2\sqrt{33}}{10}</math>, giving <math>x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 03:54, 2 February 2021
Contents
[hide]Problem
Four cubes with edge lengths ,
,
, and
are stacked as shown. What is the length of the portion of
contained in the cube with edge length
?
Solution
By Pythagorean Theorem in three dimensions, the distance is
.
Let the length of the segment that is inside the cube with side length
be
. By similar triangles,
, giving
.
Video Solution
~IceMatrix
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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