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A cylinder has radius 5 and height 8 . What is the largest distance between any two points on the cylinder?

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8 \sqrt{2}

8 \sqrt{5}

\sqrt{41}

2 \sqrt{41}

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Detailed Solution

Let AC represent the largest distance.
Question Image

Diameter of a sphere

= 2 \times 5 = 10

From Pythagoras theorem, we have,

A C^{2} = B C^{2} + A B^{2}

\Rightarrow A C^{2} = h^{2} + d^{2}

Putting the values of

h

and d we get,

\Rightarrow A C^{2} = 8^{2} + 10^{2}

\Rightarrow A C^{2} = 64 + 100

\Rightarrow A C^{2} = 164

\Rightarrow AC = \sqrt{4 \times 41} \Rightarrow AC = 2 \sqrt{41}

Therefore, the largest distance inside the given cylinder is

2 \sqrt{41}

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