Problem Solving

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

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