Latest Elitmus Aptitude Question SOLUTION: if a ractanlgle area is equal to 2*area of a circle (radius r) if rectangle is inscribed in a circle so what will be the perimeter of rectangle

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(#M40151289) ELITMUS Numerical Ability question geometry Keep an EYE

if a ractanlgle area is equal to 2*area of a circle
(radius r) if rectangle is inscribed in a circle so what will be the perimeter of rectangle

Asked In Elitmus umesh (9 years ago)

Solved bns Read Solution (5)

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bns  (9 years ago) Best Solution

Impossible question...
rectangle is inscribed in a circle and ractanlgle area is equal to 2*area of a circle.
how could it be?

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pratham patil  (9 years ago)

4r*sqrt(1+pi)

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Ramdas Singathiya  (9 years ago)

let L=length of rectangle
B=height of rectangle
r= radius of circle
now given => L*B= 2*(pi*R*R).........................(1)

when rectangle is inscribed in circle (Draw to paper), we will get that
=> R^2= (L/2)^2 + (B/2)^2..............................(using pythogoras theorem)
=> (R*R)= (L*L)/4 +(B*B)/4
=> (R*R)*4 =(L*L) +(B*B)................................(2)

we have to find 2*(L+B) = 2*sqrt[ (L+B)^2]
=> 2*sqrt[(L*L)+(B*B)+2*L*B]............................ using (a+b)^2 = a^2 +b^2 +2*a*b formula
now use equ. (1) and (2)

=> 2* sqrt[4*(R*R)+ 2*(2*pi*R*R)]
=> 2*2*R*sqrt [1+pi]................................... here we taken common 4*R*R from sqrt
=> 4*R * sqrt[1+pi].. !!!!!!!!!!!!!!!!!!!!!!!!!!!!! :)

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ANSHU KUMAR  (9 years ago)

Ans : sqrt(8pi+8)*r
sol : r= sqrt(l*l + b*b) / 2

2*pi*r*r = l*b
put the value of r in above equation ... and by solving we get : 2(l+b) = sqrt(l*l+b*b) *r

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Digesh Singh Bisht  (1 year ago)

@Ramdas is a legend . He can answer to wrong question also.

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