Difference between revisions of "2014 AMC 10A Problems/Problem 19"
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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 $x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-. | 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 $x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-. | ||
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+ | SOMEONE HELP WITH THE CODE PLEASE | ||
==Solution 2 (3D Coordinate Geometry)== | ==Solution 2 (3D Coordinate Geometry)== |
Revision as of 19:50, 29 October 2024
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 $x=\boxed{\textbf{(A) }\dfrac{3\sqrt{33}}{5}} skr-.
SOMEONE HELP WITH THE CODE PLEASE
Solution 2 (3D Coordinate Geometry)
Let's redraw the diagram, however make a 3D coordinate plane, using D as the origin.
Now we can use the distance formula in 3D, which is and plug it in for the distance of .
We get the answer as .
Continuing with solution 1, using similar triangles, we get the answer as
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Solution 3
The diagonal of the base of the cube with side length is . Hence by similarity:
.
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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