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Show that the rectangle of maximum perimeter which can be incribed in a circle of radius a is a square of side a √2.

Show that the rectangle of maximum perimeter which can be incribed in a circle of radius a is a square of side  a√2.

Grade:12

1 Answers

Swapnil Saxena
102 Points
9 years ago

Let the radius of the circle to be R.

When a rectangle is inscribed in a circle , the diameter of the circle will act as the diagonal for the rectangle.

So assume that the sides of the rectangle to be a, b

applying pythagoras theorm

Then a2 + b2 = 4R2  ===> b = (4R2 - a2 )1/2

The Perimeter of the rectangle is 2(a+b) = 2 (a + (4R2 - a2 )1/2)

differetiating the equattion with respect to a and put it equal to 0

= 2(1- (2a)/2(4R2 - a2 )1/2)=0

= 2(1- (a)/(4R2 - a2 )1/2)=0

= 1= (a)/(4R2 - a2 )1/2

= a=(4R2 - a2 )1/2

= a2= (4R2 - a2 )

= 2a2 = 4R2

= a = R(2)1/2

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