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A
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Difficulty:
65%
(hard)
Question Stats:
17% (00:06) correct
83%
(01:48)
wrong
based on 12
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A right circular cone of radius 9 cm and height 12 cm is cut parallel to the base , to form two pieces such that the height of smaller cone is half of the height of the frustum. What is the curved surface area, in square cm , of the frustum ?
1) 60π
2) 72π
3) 90π
4) 104π
5) 120π
Sent from my A0001 using GMAT Club Forum mobile app
1) 60π
2) 72π
3) 90π
4) 104π
5) 120π
Sent from my A0001 using GMAT Club Forum mobile app
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12 Mar 2018, 23:51
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Not a GMAT question. One does not need to know the curved surface area of a frustum of a cone. Still:
We need to find he curved surface area of the lower part of the cone, so the area of the surface without the bottom and top. It's area = \(\pi*h*(R + r)\).
Since the height is cut in the ratio 1:2, then the height of the frustum is \(h_1= 8\) cm and the height of the smaller, upper cone is \(h_2= 4\) cm. Next, from similar triangles property: \(\frac{r}{R} =\frac{h_2}{H}\) (where H is the height of the big cone and h2 is the height of the smaller, upper cone). r/9 = 4/12 --> r = 3.
Therefore, the area of the frustum is \(\pi*h_1*(R + r)=\pi*8(9 + 3)=96\pi\).
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