Four equal circles each of radius $a$ units touch one another. the area enclosed between them in square units, is (take $π = \frac{22}{7}$ )

3 a^{2}

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\frac{6 a^{2}}{7}

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\frac{41 a^{2}}{7}

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\frac{a^{2}}{7}

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Solution

The correct option is C $\frac{6 a^{2}}{7}$
Radius of circle $= a$
Side of square $= a + a = 2 a$
Required area $=$ Area of square $- 4 \times$ Area of quadrant
$= (2 a)^{2} - 4 \times \frac{1}{4} π r^{2}$
$= 4 a^{2} - \frac{22}{7} a^{2}$
$= \frac{28 - 22}{7} a^{2}$
$= \frac{6}{7} a^{2}$

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Four equal circles each of radius a units touch one another. the area enclosed between them in square units, is (take π=227)

The correct option is C 6a27Radius of circle =aSide of square =a+a=2aRequired area = Area of square −4× Area of quadrant =(2a)2−4×14πr2 =4a2−227a2 =28−227a2 =67a2

Four equal circles each of radius $a$ units touch one another. the area enclosed between them in square units, is (take $π = \frac{22}{7}$ )

The correct option is C $\frac{6 a^{2}}{7}$
Radius of circle $= a$
Side of square $= a + a = 2 a$
Required area $=$ Area of square $- 4 \times$ Area of quadrant
$= (2 a)^{2} - 4 \times \frac{1}{4} π r^{2}$
$= 4 a^{2} - \frac{22}{7} a^{2}$
$= \frac{28 - 22}{7} a^{2}$
$= \frac{6}{7} a^{2}$