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A circular coin is placed on a table. Four equal larger coins are placed around the first coin, so that each touches the coin and also two of the larger coins if the radius of the first coin is r and the of the others is 5 cm, find r

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5 (\sqrt{2} + 1)

5 (\sqrt{2} - 4)

5 (\sqrt{2} - 6)

5 (\sqrt{2} - 1)

answer is D.

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Detailed Solution

The image for the given problem is provided below,
Question Image

Here, the measure of AC,

AC = 5 + 5

AC = 10 AB

AC = 5 + r

cm
Also, the measure of BC,

BC = 5 + r

cm
From Pythagoras theorem,

A C^{2} = A B^{2} + B C^{2}

10^{2} = (5 + r)^{2} + (5 + r)^{2}

10^{2} = 2 (5 + r)^{2}

5 \sqrt{2} = 5 + r

r = 5 (\sqrt{2} - 1) cm

Therefore, the inner circle's radius =

5 (\sqrt{2} - 1) cm .

The right answer is 4.

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