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In,a Hemisphere igloo,an Eskimo s head just touches the roof [#permalink]
 06 Jun 2010, 19:41
Question Stats:
50% (04:30) correct
50% (00:29) wrong based on 4 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, 03:44, edited 1 time in total.
Moved to Ps Section
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I have found the answer is 33
It takes a bit time, do you have any practical way?
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hmm... i need to follow this one up... Thanks for the question.
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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 1488 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. Answer: B.
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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. Answer: B. 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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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. Answer: B. 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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat NEW!!!
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Surface area of sphere is - 4 * pi (r)^2
Shoulden't area of hemisphere be -2 * pi (r)^2 ????
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