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Three coins of same measure (1 cm radius) are kept on a table in such a way that they touch each other. The area of the field formed by the coins in between is :
1. \(\left( \frac{\pi}{2} - \sqrt 3 \right)\) sq.cm.
2. \(\left( \sqrt 3- \frac{\pi}{2} \right)\) sq.cm.
3. \(\left( 2\sqrt 3- \frac{\pi}{2} \right)\) sq.cm.
4. \(\left(3 \sqrt 3- \frac{\pi}{2} \right)\) sq.cm.

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Correct Answer - Option 2 : \(\left( \sqrt 3- \frac{\pi}{2} \right)\) sq.cm.

Given:

Radius of coin = 1 cm

Formula Used:

Area of minor sector = \(\frac{A}{360} × πr^2\)

Area of equilateral triangle = \(\frac{√3}{4} × (Side)^2\)

Calculation:      

Radius of each coin = 1 cm

With all three centers, an equilateral triangle of side 2 cm is formed.

The area enclosed by coin = Area of an equilateral triangle - 3 × Area of a sector of angle 60° 

⇒ The area enclosed by coins = \(\frac{√3}{4}(2)^2 - 3 \times \frac{60}{360}\times π(1)^2\)

⇒ The area enclosed by coins = \((√3 - \frac{π}{2}) cm^2\)

∴ The area enclosed by coins is \((√3 - \frac{π}{2}) cm^2\)

The correct option is 2 i.e. \((√3 - \frac{π}{2}) cm^2\)     

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