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The volume of the rectangular solid above is 720. If AF = 15, which of
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05 Dec 2022, 02:55
05 Dec 2022, 02:55
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The volume of the rectangular solid above is 720. If AF = 15, which of the following is closest to the distance from C to F ?
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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: The volume of the rectangular solid above is 720. If AF = 15, which of
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07 Dec 2022, 07:42
07 Dec 2022, 07:42
1
Expert Reply
Volume = l*w*h
\(720=6*15*X\)
\(X=8\)
Distance from A to E
\(8^2 + 15^2 = 17^2\)
Distance C to F
\(17^2 + 6^2 =\)
\(289 + 36 = 325\)
325 is approximately \(18^2\)
Answer: E
\(720=6*15*X\)
\(X=8\)
Distance from A to E
\(8^2 + 15^2 = 17^2\)
Distance C to F
\(17^2 + 6^2 =\)
\(289 + 36 = 325\)
325 is approximately \(18^2\)
Answer: E
Re: The volume of the rectangular solid above is 720. If AF = 15, which of
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09 Dec 2022, 04:00
09 Dec 2022, 04:00
Expert Reply
OE
The distance from C to F is the diagonal of the box, so use the Super Pythagorean theorem: a^2 + b^2 + c^2 = d^2, where a, b, and c are the sides of the box and d is the diagonal. You have the length and height of the box, so use the volume formula to find the width: 720 = (15)(6)w, so the width is 8. Now plug your numbers into the formula: 15^2 + 8^2 + 6^2 = d^2, so 325 = d^2, and d = 18.03.
The distance from C to F is the diagonal of the box, so use the Super Pythagorean theorem: a^2 + b^2 + c^2 = d^2, where a, b, and c are the sides of the box and d is the diagonal. You have the length and height of the box, so use the volume formula to find the width: 720 = (15)(6)w, so the width is 8. Now plug your numbers into the formula: 15^2 + 8^2 + 6^2 = d^2, so 325 = d^2, and d = 18.03.
