IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

ankit.fbd09

Prove that area of circle is 22/7 r*r

Or pi r(square)

prasad_aspirant

i m not that big genius

puneet

Let a regular polygon of n sides be inscribed in a circle of radius r, and let s be the length of each side.

Let us divide the polygon into n isosceles triangles, and let us denote the area of each triangle by A_T. Then the sum of those n triangles will approximate the area of the circle.

Now the area of each triangle is half the base s times the height h.

A_T = ½sh . . . . . . . . . (1)

We will now express both s and h in terms of the radius r, and then substitute those expressions in line (1).

As we saw above, each chord s subtends a central angle of

360°
n

.

And the side of each isosceles triangle is the radius r. Therefore,

h
r

= cos ½

360°
n

= cos

180°
n

or,

h = r cos

180°
n

. . . . . . . . (2)

Also,

s/2
r

= sin

180°
n

so that

s = 2r sin

180°
n

. . . . . . . . (3)

Upon substituting lines (3) and (2) into line (1), we have

A_T = ½sh

=

½· 2r sin

180°
n

· r cos

180°
n

A_T = ½sh

=

r² sin

180°
n

· cos

180°
n

This is the area of one of the triangles. The area A of the entire circle is approximated by all n triangles:

A = nA_T

=

n· r² sin

180°
n

· cos

180°
n

But we have seen that

n sin

180°
n

?.

Therefore, finally,

A

?r² cos

180°
n

. . . . . . . . (4)

Suppose now that the number of sides n is an enormously large number -- more than the number of stars in the entire universe! Then

the polygon will "exhaust" the circle. The central angle

180°
n

(line 4)

will be indistinguishable from 0°. We will have

A = ?r² cos 0°.

But cos 0° = 1. Therefore,

A = ?r².

Or, on replacing r² with (

D
2

)² =

D²
4

,

A

=

?
4

D².

This is what we wanted to prove.

elessar_iitkgp

Lets prove this by integration. Consider a thin circular area element of radius x inside the given circle and concentric with the given circle. The width of this element be dx.

If this element is straightened out, its approximately a rectangular strip of length
2

x and width dx.
Area of the strip dA = 2

x dx
A = ₀

^r 2

x dx =

r² /2

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