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Greatest straight distance between points

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by GmatGreen » Sat Mar 01, 2014 10:35 am
A rectangular box is 10 inches wide, 10 inches long, and 5 inches high, what is the greatest possible (straight-line) distance, in inches, between any two points on the box?

A) 15
B) 20
C) 25
D) 10r2
E) 10r3
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Source: — Problem Solving |

by Patrick_GMATFix » Sat Mar 01, 2014 10:42 am
There is a straightforward formula for this: length^2 + width^2 + height^2 = diagonal^2 where diagonal is the longest straight line between two points in a box/cube. I go through the question in detail in the full solution below (taken from the GMATFix App).

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by Brent@GMATPrepNow » Sat Mar 01, 2014 12:54 pm
GmatGreen wrote:A rectangular box is 10 inches wide, 10 inches long, and 5 inches high, what is the greatest possible (straight-line) distance, in inches, between any two points on the box?

A) 15
B) 20
C) 25
D) 10√2
E) 10√3
The greatest distance will be when the two points are in opposite corners.

In these instances, we have a nice rule that says:
If x, y, and z are the three measurements of a box, then the distance between two points in OPPOSITE CORNERS equals √(x² + y² + z²)

So, for your question, the distance = √(10² + 10² + 5²) = √225 = 15 = A

If you're interested, we have a free video that explains the above rule (via a practice question): https://www.gmatprepnow.com/module/gmat-geometry?id=869

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Abhishek009 » Mon Mar 03, 2014 9:00 am
GmatGreen wrote:A rectangular box is 10 inches wide, 10 inches long, and 5 inches high, what is the greatest possible (straight-line) distance, in inches, between any two points on the box?

A) 15
B) 20
C) 25
D) 10r2
E) 10r3
The greatest straight line which can be drawn inside a cuboid is the Diagonal.

Formula of Diagonal of a Cuboid is √(x² + y² + z²)

√(10² + 10² + 5²)

√(100 + 100 + 25)

√225

15

Hence Answer is [spoiler](A)[/spoiler]
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Cuboid.jpg
Abhishek
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by theCodeToGMAT » Mon Mar 03, 2014 10:28 pm
Inclined Diagnol = sqrt(10^2 + 10^2 + 5^2)

= sqrt(225) = 15

[spoiler]{A}[/spoiler]
R A H U L
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by Scott@TargetTestPrep » Sat Jul 18, 2015 7:00 pm
GmatGreen wrote:A rectangular box is 10 inches wide, 10 inches long, and 5 inches high, what is the greatest possible (straight-line) distance, in inches, between any two points on the box?

A) 15
B) 20
C) 25
D) 10r2
E) 10r3
Solution:

To solve this problem we must remember that given any rectangular solid, the longest line segment that can be drawn within the solid will be one that goes from a corner of the solid, through the center of the solid, to the opposite corner, or in other words, the space diagonal of the solid.

The space diagonal can be calculated using the extended Pythagorean theorem:

diagonal^2 = length^2 + width^2 + height^2

Using the values from the given information we have:

d^2 = 10^2 + 10^2 + 5^2

d^2 = 100 + 100 + 25

d^2 = 225

√d^2 = √225

d = 15

Answer: A

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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by nikhilgmat31 » Wed Jul 29, 2015 5:25 am
Hope all questions in GMAT are like this.
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