JIPMAT 2022 (QA) - If the radius of each of four outer circles is r, then the radius of the innermost circle is <img src="https://balti.afterboards.in/KJbLKQT21zndr44" width=200px> | PYQs + Solutions

JIPMAT 2022

Geometry

Circles

Medium

If the radius of each of four outer circles is $r$ , then the radius of the innermost circle is

\frac{\sqrt{2}}{\sqrt{2+1}} r

\frac{1}{\sqrt{2}} r

(\sqrt{2}-1) r

\sqrt{2 r}

Correct Option: 3

1) In this figure, the four outer circles of radius

r

are arranged around a square touching the innermost circle.

2) Consider half of the square's diagonal:

\newline

- It connects the center of the innermost circle to the corner of the square

\newline

- Also equals the radius we're looking for plus the side of the square

3) Let radius of innermost circle be

x

\newline

Side of square =

2x

(since small circle touches sides of square)

4) Due to symmetry, centers of large circles form a square of side

2r

5) From Pythagorean theorem:

\newline

(r + x)^2 = r^2 + r^2

\newline

r^2 + 2rx + x^2 = 2r^2

\newline

x^2 + 2rx - r^2 = 0

6) Solving quadratic equation:

\newline

x = \frac{-2r + \sqrt{4r^2 + 4r^2}}{2}

\newline

x = \frac{-2r + 2r\sqrt{2}}{2}

\newline

x = r(\sqrt{2} - 1)

Therefore, radius of innermost circle is

(\sqrt{2} - 1)r

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