Latest Elitmus Aptitude Question SOLUTION: the circle O having a diameter of 2cm, has a square inscribed in it.each side of the square is then taken as a diameter to form 4 smaller circles O'.find the total area of all four

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(#M40008868) ELITMUS Logical Reasoning question Keep an EYE

the circle O having a diameter of 2cm, has a square inscribed in it.each side of the square is then taken as a diameter to form 4 smaller circles O'.find the total area of all four O' circles which is outside the cirle O.
a)2
b)pi-2
c)2-pi/4
d)2-pi/2

Asked In Elitmus amrit bhagat (12 years ago)

Solved Sourabh Sar Read Solution (10)

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Read Solution (10) :

Sourabh Sar  (12 years ago)

ans is (a)2.

Well area of cirle O=pi.(Plzz dnt ask How!!!)

Nw the diameter of circle is also the diagonal of the square.
Hence each side of square will be sqrt(2).

=>Area of square=2

since each side of square is also the diameter of other 4 circles.
Hence summation of area of 4 circles=2*pi...........(1)

If u hav drwan its fig u'll find that to obtain the required ans u hav to
subtract the area of 4 semi-circles formed on the side of the square from the each of the small
portion outside the square.

to get that area of small portion =area of circle O-area of square
=pi-2.......(2)

this small portion has to be substracted from the four semi-circles.
Hence, area of 4 semi-circles=2*pi/2= pi......[from (1)]

required ans=total area of 4 semi-circles - area of small portion(from (2))
=pi-(pi-2)
=2.
Kudos..:-))

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arvind  (10 years ago)

The first thing I did was to find the side of the square inscribed in the circle,
as the diameter of the circle will be the diagonal of the square, this will form two 45 45 90 right triangles with a hypotenuse of 2.
Since s = ssqrt(2), 2 = s(sqrt(2).
divide 2 by sqrt(2) = 2/sqrt(2)
Multiply the numerator and denominator by sqrt(2) to obtain 2(sqrt(2)/2 for each side of the square, or sqrt(2)
Now, these are being used as diameters for four more circles, so we determine the radius of each one of the four to be sqrt(2)/2
So this value squared times pi will be the area of each circle, or 2/4 = 1/2 times pi = 1/2 pi for each circle times four circles results in 2 pi as the area for the four circles combined.


The area of the original circle is pi r^2, radius being 1 so the area of the original circle will come to 1pi.
So, since the outer circles obviously cover the original circle completely, the area outside the original should be 2pi minus pi, or one pi.

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t.venkat rao  (12 years ago)

First consider the 4 semicircles which are formed by the sides of the sqare and remove the 4 extrnal parts thart r frmed out side the sqare and included in the circle so that the required part can be obtained therfore area of 4 semi circles =pi and area of 4 external parts=pi-2 result=pi-(pi-2)=2.

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Aditya   (10 years ago)

AO = OB = 1cm (radii of circle O)

Area of quarter part of circle O = (1/4) pi 12 = pi/4

AB is the side of square inscribed in circle O hence AB act as a diameter to the circle O”

AB = √2 (diameter of Circle O”)

Area of circle O’’ = pi (1/√2)2 = pi /2

Half Area of Circle O’’ = pi/4

Area of triangle (AOB) inside circle O” =1/2 x 1 x1 = 1/2

Area of two sectors (P and Q) of circle O”

= Half Area of Circle O’’- Area of triangle AOB

= (pi/4) - 1/2

Required Area = Area of circle O” - (Quarter area of circle O + area of sectors P & Q)

= pi/2 - (pi/4 + pi/4 - 1/2 ) =1/2

Therefore overall area for 4 such circles = 4 x 1/2= 2

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Ashu  (12 years ago)

a)2

2 sq cm is the total area of all four O' circles which is outside the cirle O.

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Ankit Agrahari  (12 years ago)

well accrding to me ...it is pi-2
Area of O = pi
Area of 4 circles = 2*pi
Area of Sq. = 2 (side=1.414)
so area of region outside the circle O will be 2*pi-pi-2=pi-2

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Vivek Pathane  (9 years ago)

First draw a cicle with o, then a square ABCD inside it.
now, AB becomes sqrt(2) since diameter of circle is also the diagonal of square.
next find the area of the sector AOB.
with a little calculation yu will get titha as 90deg.
area of sector AOB=(titha/360)*pi*r^2=(90/360)*pi*1^2=pi/4
Now,area of the segment formed by AB=Area of sector AOB - Area of tri. AOB
= pi/4-(0.5*r^2*sin titha)
=pi/4-(0.5*1^2*sin90)
=(pi/4)-(1/2)

Now consider the outer circle O' with diameter sqrt(2)
half area of this circle is pi*(1/sqrt(2))^2/2=pi/4

nw area of part of the circle which falls outside the circle o is
= half area of circle o' - area of segment
=(pi/4)-[(pi/4) -(1/2)]
=1/2

this was for one outer circle,
for 4 outer circles= 4*(1/2)=2

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amrit bhagat  (12 years ago)

explian the ans plz

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kundan prasad  (8 years ago)

Ans: a)2

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  (5 years ago)

Can anyone draw the figure and explain ?

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