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Geometry sum 2

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Geometry sum 2

by winnerhere » Sat Aug 15, 2009 4:39 am
There is a square field with three cows tethered at three different corners of the field. The ropes, used to tether the cows, are all of length 14√2. If the area of the square field is 784 m2, what is the area of the field that cannot be accessed by any of the three cows? (assume π = 22/7)

1. 184 sq.m

2.168 sq.m

3.161 sq.m

4.322 sq.m

5.84 sq.m
Last edited by winnerhere on Mon Aug 17, 2009 2:28 am, edited 1 time in total.
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by Morgoth » Sat Aug 15, 2009 11:45 am
side of the square = sqroot 784 = 28

3 cows tethered at 3 corners of the circle make 3 parts of the entire circle with diameter of 28 or radius of 14

area of the square tethered by 3 cows = 3(pi*r^2)/4 = [3*(22)*14^2]/(7*4)

= 3*11*14 = 462

area not accessed by any of the cows = area of square - 462 = 784 -462 = 322


OA?
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by winnerhere » Sun Aug 16, 2009 11:28 pm
OA is 84
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Re: Geometry sum 2

by tohellandback » Sun Aug 16, 2009 11:31 pm
winnerhere wrote:There is a square field with three cows tethered at three different corners of the field. The ropes, used to tether the cows, are all of length If the area of the square field is 784 m2, what is the area of the field that cannot be accessed by any of the three cows? (assume π = 22/7)

1. 184 sq.m

2.168 sq.m

3.161 sq.m

4.322 sq.m

5.84 sq.m
missed something there??
The powers of two are bloody impolite!!
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by winnerhere » Mon Aug 17, 2009 2:30 am
oops..sorry..the length is 14√2.

now the post is rightly edited.
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by tohellandback » Mon Aug 17, 2009 2:51 am
please check the image..the area in black is the area we need to find.
The other half of the square is grazed completely by the cows.
half of the area of square=784//2=392
area of each reason A and B is pi * (14√2)^2/8

so area of regions A and B=2* 22/7 * 2* 14 *14 * 1/8= 308

Area of the black region=392-308=84
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The powers of two are bloody impolite!!
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by winnerhere » Mon Aug 17, 2009 2:56 am
Thank u so much buddy!

:)
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