A rectangular box is 10 inches wide, 10 inches long, and 5

Question

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\sqrt{2} 
(E)  10\sqrt{3}

Bunuel · Accepted Answer

The longest distance will be the diagonal of a rectangular box. Look at the diagram below: https://gmatclub.com/forum/download/file.php?id=16743 Square of the diagonal of the face (base) is d^2=a^2+b^2 and the square of the diagonal of a rectangular box is D^2=d^2+c^2=(a^2+b^2)+c^2 --> D=\sqrt{a^2+b^2+c^2} . Applying this to our question, we get: D=\sqrt{10^2+10^2+5^2}=15 . Answer: A. Similar question to practice: a-rectangular-box-has-dimensions-of-8-feet-8-feet-and-z-128483.html

JeffTargetTestPrep · Answer

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

Dienekes · Answer

One can also apply the 3-D or Deluxe Pythagorean theorem directly, which is D^2 = L^2 + W^2 + H^2, to get the value directly.

LogicGuru1 · Answer

The greatest distance between any two points in a CUBE/CUBOID is given by the formula 
 d=\sqrt{l^2+b^2+h^2}

d=\sqrt{100+100+25} ;  {given===> l=10; b=10; h=5}

d=\sqrt{225}= 15

Answer is A

BrentGMATPrepNow · Answer

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

Answer: A

Cheers,
Brent

dhruvmm · Answer

This is such an easy question... but the question stem states - straight line between any two points "on" the box. And that means face diagonal, an i missing something? Wouldn't we use the space diagonal formula is the question stated "in" the box perhaps?

Abhishek009 · Answer

= \sqrt{10^2+10^2+5^2}

= \sqrt{225}

= 15 , Answer must be (A)

Check the highlighted part the straight line is the diagonal inside the box...

dhruvmm · Answer

Hi, that's my exact point of confusion. The highlighted part says "between any two points ON the box?"
And you mentioned - "straight line is the diagonal INSIDE the box."

ON is not the same as IN - is it? minor error in the verbose but makes a world of a difference imo

ArnauG · Answer

Using the Pythagorean theorem, we can find the length of the diagonal. Let's denote the length, width, and height of the box as L, W, and H, respectively.

The length of the diagonal (D) can be calculated as:

D = √(L^2 + W^2 + H^2) = √(10^2 + 10^2 + 5^2) = √(100 + 100 + 25) = √225 = 15

Therefore, the greatest possible straight-line distance between any two points on the box is 15 inches, which corresponds to option (A).

bumpbot · Answer

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A rectangular box is 10 inches wide, 10 inches long, and 5

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