how to show that the maximum rectange inscribed in a circle is a square?

Question

actually i know how to solve this questio but i don't know that why we take xsquare+y square=4a square
why we take here radius 4a squar

Answer 1

Let a rectangle inscribed in a circle have sides x & y. Let the DIAMETER of the circle be d. Then, by Pythagorean's theorem, we have:

x^2 + y^2 = d^2

But note that if we use a as the radius of the circle, which is half of the diameter, we have:

2a = d
4a^2 = d^2

So, subsituting into the other equation, we have:

x^2 + y^2 = 4a^2

Answer 2

Let a rectangle inscribed in a circle have sides x & y. Let the DIAMETER of the circle be d. Then, by Pythagorean's theorem, we have:

x^2 + y^2 = d^2

But note that if we use a as the radius of the circle, which is half of the diameter, we have:

2a = d
4a^2 = d^2

So, subsituting into the other equation, we have:

x^2 + y^2 = 4a^2

Answer 3

Imagine a circle and any shaped rectancle ABCD
Since all the angles are right angled in a rectangle,
the diagonal should be the diameter.
(because only diameter subtends the right angle at any point on circumference of circle)

so if sides are x and y for rectangle,

xsquare +y square = (diameter )square =(2a)square
(pythagoras)
As u have told ,u know what to do next .
so better luck with ur further part of the sum.

Answer 4

elementary! The diameter of the circle may well be the diagonal of the sq.. (Draw a diagram) you do now not even % pi for this: If d = diagonal of a sqaure, then component s = d/?2. So s = 30/?2 and perimeter P = 4s = 4(30/?2) = one hundred twenty/?2. the area is s² = (30/?2 )² = 900/2 = 450

Answer 5

Take the unit circle centered at the origin (x² + y² = 1). The vertices of your rectangle are (x,y), (-x,y), (x,-y) and (-x,-y).

The area of the rectangle is A = 2x * 2y = 4xy, subject to the constraint x² + y² = 1. So

A = 4*x*sqrt(1-x²). Differentiate:
dA/dx = 4sqrt(1-x²) + 4x*(-2x)/2sqrt(1-x²) which we want to be zero. So
4sqrt(1-x²) = 4x²/sqrt(1-x²)
4(1-x²) = 4x²
4 = 8x²
x² = 1/2
x=sqrt(1/2) and therefore y=sqrt(1/2) too.

Answer 6

You can maximise the area of the rectangle by using differentials.
Thats the easiest and the simplest thing to do!!
I

Answer 7

Let length and breath be "l" and "b". P=2l+2b or [P-2b]/2=l. Now:
Area=lxb or A=[P-2b]/2 xb. Differentiating w.r.t."b" : dA/db=P/2-2b.
For max. : dA/db=0. Hence P-4b=0 or P=4b.
Which implies that each side is b units. Hence it is a square.

Answer 8

tell me when u get to know it