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Updated on: 01 Oct 2023, 02:14
Originally posted by Sajjad1994 on 10 Feb 2021, 23:30.
Last edited by BottomJee on 01 Oct 2023, 02:14, edited 2 times in total.
Last edited by BottomJee on 01 Oct 2023, 02:14, edited 2 times in total.
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Diameter: \(\frac{6}{\sqrt{π}}\)
Height: 1
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:07) correct
29%
(01:58)
wrong
based on 248
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Consider a right circular cylinder for which the following quantities are all numerically equal: the volume in cubic centimeters, nine-times the height in centimeters, the area of the circular base in square centimeters.
In the table, select values for Diameter and Height that satisfy the conditions above. Select only one answer in each column.
In the table, select values for Diameter and Height that satisfy the conditions above. Select only one answer in each column.
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| Diameter | Height | ------ |
| \(\frac{3}{\sqrt{π}}\) | ||
| \(\frac{6}{\sqrt{π}}\) | ||
| 1 | ||
| 9 | ||
| \(9\sqrt{π}\) |
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Official Answer
Diameter: \(\frac{6}{\sqrt{π}}\)
Height: 1
rocky620
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15 Feb 2021, 05:52
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Let r & h be the radius and height of the cylinder
Volume = 9h = Area of the Base
\(\pi*r^2*h = 9h = \pi*r^2\)
\(\pi*r^2*h = \pi*r^2\)
Height, h=1
\(9h = \pi*r^2\)
\(9 = \pi*r^2\)
\(\frac{9}{\pi} =r^2\)
\(\frac{3}{\sqrt{\pi}} =r\)
Diameter = \(\frac{6}{\sqrt{\pi}}\)
Volume = 9h = Area of the Base
\(\pi*r^2*h = 9h = \pi*r^2\)
\(\pi*r^2*h = \pi*r^2\)
Height, h=1
\(9h = \pi*r^2\)
\(9 = \pi*r^2\)
\(\frac{9}{\pi} =r^2\)
\(\frac{3}{\sqrt{\pi}} =r\)
Diameter = \(\frac{6}{\sqrt{\pi}}\)
17 Aug 2024, 12:54
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Geometry is removed in the latest edition right? such qs are not expected? or are they?
