How to Find the Maximum Distance Between Points on a 3D Object

Quarter Wit, Quarter WisdomHow do we find the the two farthest points on a 3D object? For example, we know that on a circle, any two points that are diametrically opposite will be the farthest from each other than from any other points on the circle. Which two points will be the farthest from each other on a square? The diagonally opposite vertices. Now here is a trickier question – which two points are farthest from each other on a rectangular solid? Again, they will be diagonally opposite, but the question is, which diagonal?

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)

There are various different diagonals in a rectangular solid. Look at the given figure:
MaxDistRectangularSolid

 

 

 

 

 

BE is a diagonal, BG is a diagonal, GE is a diagonal, and BH is a diagonal. So which two points are farthest from each other? B and E, G and E, B and G, or B and H?

The inside diagonal BH can be seen as the hypotenuse of the right triangle BEH. So both BE and EH will be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BHG. So both HG and BG will also be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BDH. So both BD and DH will also be shorter in length than BH.

Thus, we see that BH will be longer than all other diagonals, meaning B and H are the points that are the farthest from each other. Solving for the exact value of BH then should not be difficult.

In our question we know that:

l = 10 inches
w = 10 inches
h = 5 inches

Let’s consider the right triangle DHB. DH is the length, so it is 10 inches.

DB is the diagonal of the right triangle DBC. If DC = w = 10 and BC = h = 5, then we can solve for DB^2 using the Pythagorian Theorem:

DB^2 = DC^2 + BC^2
DB^2 = 10^2 + 5^2 = 125

Going back to triangle DHB, we can now say that:

BH^2 = HD^2 + DB^2
BH^2 = 10^2 + 125
BH = √(225) = 15

Thus, our answer to this question is A.

Similarly, which two points on a cylinder will be the farthest from each other? Let’s examine the following practice GMAT question to find out:

The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C?

(A) 5 * √2
(B) 5 * √3
(C) 5 * √5
(D) 10
(E) 15

Look at where the farthest points will lie – diametrically opposite from each other and also at the opposite sides of the length of the cylinder:
MaxDistanceCylinder

 

 

 

 

 

 

The diameter, the height and the distance between the points forms a right triangle. Using the given measurements, we can now solve for the distance between the two points:

Diameter^2 + Height^2 = Distance^2
10^2 + 5^2 = Distance^2
Distance = 5 * √5

Thus, our answer is C.

In both cases, if we start from one extreme point and traverse every length once, we reach the farthest point. For example, in case of the rectangular solid, say we start from H, cover length l and reach D – from D, we cover length w and reach C, and from C, we cover length h and reach B. These two are the farthest points.

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Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!