Math, asked by vivekkushwah776, 1 year ago

who can prove the area of the circle?​

Answers

Answered by anuththarabashini52
0

Answer:

A=πr2

markasbrainest

Answered by rohit175622
1

Step-by-step explanation:

proof by having you look at the following figures:

Circles inscribed inside a square and inside an octagon

Did you notice that as we went from 4 sides to 8 sides, the area outside the circle, but inside the square shrank considerably? This is illustrated below in brown

Circles inscribed inside a square and inside an octagon

Now if we keep increasing the number of sides to a very very big number ( call the resulting polygon an n-gon), this space will be so small, it will look like the circle is the same as the n-gon with a lot of sides

Keep this fact in mind since we will refer back to it later!

Now, consider the following figure. Hang on! we are halfway of completing the proof of the area of a circle :

Circles inscribed inside a square and inside an octagon

The area of triangle AOB is 1/2 ( base × height) = 1/2 (s × r)

However, we can make 8 such triangles inside the octagon as show below:

Circles inscribed inside a square and inside an octagon

This means that the area of the entire octagon is 8 ×( 1/2 (s × r)) = 1/2 r × 8s

Notice that 8s is equal to the perimeter of the octagon

As stated before, if we increase the number of sides to infinity or a very big number, the resulting n-gon ( The regular polygon which number of sides is a big number) will will almost look like a circle

This means that the perimeter of the octagon will almost be the same as the perimeter of the circle

As a result, the closer the perimeter of the polygon is to the circle, the closer the area of the polygon is to the area of the circle

It is reasonable then to replace 8s by 2 × pi × r, which is the perimeter of the circle, to calculate the area of the polygon or the circle when the number of sides is very big

Doing so we get:

Area of circle or polygon equal = 1/2 r × 2 × pi × r = pi × r2

Proof of the area of the circle

Thanks

I think this may help you

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