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A cylinder has radius 5 and height 8 . What is the largest distance be
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10 Jan 2025, 02:14
10 Jan 2025, 02:14
Expert Reply
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Question Stats:
60% (02:01) correct
40% (01:40) wrong based on 5 sessions
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A cylinder has radius 5 and height 8 . What is the largest distance between any two points on the cylinder?
(A) $\(8 \sqrt{2}\)$
(B) $\(8 \sqrt{5}\)$
(C) $\(\sqrt{41}\)$
(D) $\(2 \sqrt{41}\)$
(E) \(13\)
(A) $\(8 \sqrt{2}\)$
(B) $\(8 \sqrt{5}\)$
(C) $\(\sqrt{41}\)$
(D) $\(2 \sqrt{41}\)$
(E) \(13\)
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Re: A cylinder has radius 5 and height 8 . What is the largest distance be
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18 Feb 2025, 04:11
18 Feb 2025, 04:11
Expert Reply
OFFICIAL EXPLANATION
The largest distance between two points on the cylinder is its diagonal's length.
GRE CYLINDER.png [ 30.98 KiB | Viewed 420 times ]
In a cylinder having radius ' $\(r\)$ ' \& height ' $\(h\)$ ' the diagonals length is $\(\sqrt{(2 r)^2+h^2}\)$
So, the diagonal's length of cylinder having radius 5 and height 8 would be $\(\sqrt{(2 \times 5)^2+8^2}=\sqrt{100+64}=\sqrt{164}=2 \sqrt{41}\)$
Hence the answer is (D).
The largest distance between two points on the cylinder is its diagonal's length.
Attachment:
GRE CYLINDER.png [ 30.98 KiB | Viewed 420 times ]
In a cylinder having radius ' $\(r\)$ ' \& height ' $\(h\)$ ' the diagonals length is $\(\sqrt{(2 r)^2+h^2}\)$
So, the diagonal's length of cylinder having radius 5 and height 8 would be $\(\sqrt{(2 \times 5)^2+8^2}=\sqrt{100+64}=\sqrt{164}=2 \sqrt{41}\)$
Hence the answer is (D).
gmatclubot
Re: A cylinder has radius 5 and height 8 . What is the largest distance be [#permalink]
18 Feb 2025, 04:11
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