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# PS - Polygons

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In,a Hemisphere igloo,an Eskimo s head just touches the roof [#permalink]  06 Jun 2010, 18:41
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Difficulty:

5% (low)

Question Stats:

57% (02:45) correct 42% (01:20) wrong based on 7 sessions
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data

Last edited by RaviChandra on 18 Nov 2010, 02:44, edited 1 time in total.
Moved to Ps Section
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Re: PS - Polygons [#permalink]  07 Jun 2010, 03:35
I have found the answer is 33

It takes a bit time, do you have any practical way?
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Re: PS - Polygons [#permalink]  17 Nov 2010, 22:37
hmm... i need to follow this one up...

Thanks for the question.
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Re: PS - Polygons [#permalink]  18 Nov 2010, 05:32
Expert's post
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data

Look at the diagram below:
Attachment:

AngleSemicircle.gif [ 3.75 KiB | Viewed 1964 times ]
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so R=65. As the child can play over an area of 9,856 square units then the radius of this playing are is: playing \ area=\pi{r^2}=9,856 --> r^2=\frac{9,856}{\pi} --> r\approx{56}. Thus the child's height will be H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33.

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Re: PS - Polygons [#permalink]  20 Nov 2010, 14:31
Bunuel wrote:
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data

Look at the diagram below:
Attachment:
AngleSemicircle.gif
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so R=65. As the child can play over an area of 9,856 square units then the radius of this playing are is: playing \ area=\pi{r^2}=9,856 --> r^2=\frac{9,856}{\pi} --> r\approx{56}. Thus the child's height will be H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33.

Hi Bunnel ,
Got 2 doubts here
1) Aint the area should be playing \ area=\pi{r^2}/2 since its an hemisphere ?
2) While we are just concentrating on the right half should the area that of the quarter ?
M confused or may be i'm thinking in the wrong direction , Please explain .
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This time , its my time .

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Re: PS - Polygons [#permalink]  21 Nov 2010, 00:52
Expert's post
girisshhh84 wrote:
Bunuel wrote:
lnarayanan wrote:
In,a Hemisphere igloo,an Eskimo’s head just touches the roof when he stands erect at the centre of the floor, but his son can play over an area of 9856 square units without stooping.If the Eskimo’s height is 65 units,What is his son’s height?
A) 25 units , B) 33 units , C) 35 units , D) 37 units ,E) Insufficient data

Look at the diagram below:
Attachment:
AngleSemicircle.gif
Now, the RADIUS of the igloo equals to the hight of the Eskimo, so R=65. As the child can play over an area of 9,856 square units then the radius of this playing are is: playing \ area=\pi{r^2}=9,856 --> r^2=\frac{9,856}{\pi} --> r\approx{56}. Thus the child's height will be H=\sqrt{R^2-r^2}=\sqrt{65^2-56^2}=33.

Hi Bunnel ,
Got 2 doubts here
1) Aint the area should be playing \ area=\pi{r^2}/2 since its an hemisphere ?
2) While we are just concentrating on the right half should the area that of the quarter ?
M confused or may be i'm thinking in the wrong direction , Please explain .

I think you are just confused with the diagram:

Hemisphere is half of a sphere and the diagram gives the cross section of it. But the base of a hemisphere (the base of an igloo) is still a circle, so the playing area of the child is a circle limited by his height.
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Re: PS - Polygons [#permalink]  21 Nov 2010, 13:15
Surface area of sphere is - 4 * pi (r)^2

Shoulden't area of hemisphere be -2 * pi (r)^2 ????
Re: PS - Polygons   [#permalink] 21 Nov 2010, 13:15
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# PS - Polygons

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