Author Topic: Maths  (Read 3883 times)

nid404

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Maths
« on: October 01, 2009, 03:48:32 pm »
Check the diagram below

the quadrilateral is a square

Find the area of the shaded region

note-the four shapes touching the vertices are not triangles.....i couldn't draw it perfectly


thanks :)

Offline Ghost Of Highbury

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Re: Maths
« Reply #1 on: October 01, 2009, 06:01:29 pm »
kya question hai be?

ekdum jhakaas~~

:P..will try
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Offline Ghost Of Highbury

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Re: Maths
« Reply #2 on: October 01, 2009, 06:02:56 pm »
but what are the four things touching the verteices of the square?

are they a sector ???
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nid404

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Re: Maths
« Reply #3 on: October 01, 2009, 06:04:59 pm »
I don't think u can call it a sector

wait i'll try and make the drawing better

nid404

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Re: Maths
« Reply #4 on: October 01, 2009, 06:19:15 pm »
here i think this will be much clear

oh the sides are of 1 unit....i think i mentioned it before

please help me answer it

Offline astarmathsandphysics

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Re: Maths
« Reply #5 on: October 01, 2009, 06:39:06 pm »
Call the triangle staring from a corner x, a tiangle on a side z and the one in the middle y
The whole area =4x+y+4z=1
Put a compass on the corner of the square and form part (1/4) of a circle. The area of that part is 3x+y+2z=\frac {\pi}{4}
TDo this again but on the opposite side of the square to obtain another quarter of a ficle. The difference beween these two quarters and the area of the square istwo triangles and the middle from diagonal to diagonal ie 2x+y=\frac{\pi}{2} -1

Solve these equations simultaneously.

Will try and finish it when I get home in a few hours
« Last Edit: October 16, 2009, 08:42:31 am by astarmathsandphysics »

nid404

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Re: Maths
« Reply #6 on: October 01, 2009, 06:45:44 pm »
I don't know the answer....post your method anyway

@astar-they r not triangles is what i am specifically told by the teacher.....

Offline astarmathsandphysics

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Re: Maths
« Reply #7 on: October 01, 2009, 06:53:40 pm »
I am not using any properties of triangle So method should work.

nid404

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Re: Maths
« Reply #8 on: October 01, 2009, 07:03:59 pm »
okay then...
when you get back home, could you do it with the diagram please??

Offline Ghost Of Highbury

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Re: Maths
« Reply #9 on: October 01, 2009, 07:23:32 pm »
refer to the diagram attached..
.
first find the area of CBD...pir2/4 = 3.14/4 = 0.785

now, the area in peach is = y; the green area = x

total area of square = 1
square comprises of 4 peach shaded region + 4 green shaded region + 1 z
= 4y + 4x + z =1 (1)
CBD = 0.785
therefore, the area of the remaining part = 1 - 0.785 = 0.215

this area comprises of 2 peach shaded region + 1 green shaded region = 2y + x = 0.215

rearrange => y=(0.215 - x)/2     
put this value of 'y' in the (1) equation
7x + 2z = 1.785 (keep this equation in mind)

----------------------------------
now consider the sector CBD
area = 0.785
it comprises of 3 green + 2 peach + 1 z

3x + 2y + z = 0.785
substitute 'y' value here

y = (0.215 - x)/2

so that gives us...
2x + z = 0.57
--
now for the final sim. equations

7x + 2z = 1.785
2x + z   = 0.57  (*2)

--------------------
  7x + 2z = 1.785   (subtract them)
  4x + 2z = 1.14
 -----------------
   3x = 0.645
   x = 0.215

------------
substitute in any equation to find the value of z

lets use 2x + z = 0.57


(2*0.215) + z = 0.57
z = 0.57 - 0.43

z = 0.14

done!

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Offline Ghost Of Highbury

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Re: Maths
« Reply #10 on: October 01, 2009, 07:50:53 pm »
is this an A/AS level question????????
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nid404

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Re: Maths
« Reply #11 on: October 02, 2009, 05:23:13 am »
thanks.....

I don't know the answer....it's just a challenging question that our teacher gave us(IIT)

I did another method and got 0.136.....so I guess that is right too

Offline Ghost Of Highbury

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Re: Maths
« Reply #12 on: October 02, 2009, 05:26:49 am »
ur welcome...
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Offline Ghost Of Highbury

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Re: Maths
« Reply #13 on: October 02, 2009, 12:03:15 pm »
u asked ur ma'am?
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nid404

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Re: Maths
« Reply #14 on: October 02, 2009, 12:03:44 pm »
i will know only on monday