Geometry Toughies

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by rtfact » Mon Nov 17, 2008 3:46 am
1. C

from (1) we can't determine the circumference.
from (2) we don't know where the cord is located in the circle - the chord can be the diameter or smaller than the diameter.

from (1+2): we can determine that the chord is on a central angle of 60 degrees 360/6. The hight from the center of the circle (mark as: point k) to the chord divides the chord in to two straight triangles with a base of 3 (half of the chord's length). The angles of each triangle are 30:60:90, hence the hypotenuse equals 2*3=6=Radius.

2. B
There are three options to create a cylinder with a maximum volume, because there are 3 bases possible.

diameter=8, R=4, H=12, Volume=pi*R^2*H=pi*192
diameter=10, R=5, H=8, Volume=pi*R^2*H=pi*200
diameter=8, R=4, H=10, Volume=pi*R^2*H=pi*160

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by tomato1 » Mon Nov 17, 2008 4:03 am
1. C

Check th attachment for explanation
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by gmat009 » Wed Nov 19, 2008 12:11 pm
rtfact wrote: 2. B
There are three options to create a cylinder with a maximum volume, because there are 3 bases possible.

diameter=8, R=4, H=12, Volume=pi*R^2*H=pi*192
diameter=10, R=5, H=8, Volume=pi*R^2*H=pi*200
diameter=8, R=4, H=10, Volume=pi*R^2*H=pi*160
Can someone explain me why we cann't take Diameter=12 and H=10
Vol = pi*(6*6)*10=360pi which is maximum volume

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by mals24 » Wed Nov 19, 2008 12:34 pm
@gmat009

We cannot have the diameter as 12 and height as 10 because in this case the cylinder will not fit in the box.
The cylinder would have fit if the dimensions were diameter = 12 and height also 12.


Image

However if the length is 12 and the width 10 then the cylinder will easily fit. So the maximum radius should be 5 and the maximum height will be 8.

Hope its clear