Question
Four equal circles, each of radius a units, touch each other. show that area between them is $\Big(\frac{6}{7}\text{a}^2\Big)$ sq units. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
✓
Answer
Required area = [area of square - areas of quadrants of circles]Let the side = 2a unit and radius = a units
Area of square = (side side) = (2a 2a) sq. units
$=4\text{a}^2\text{sq. units}$
Area of quadrant $=\frac{1}{4}\pi\text{r}^2$
Area of 4 quadrants $=4\times\frac{1}{4}\pi\text{r}^2=\pi\text{r}^2=\frac{22}{7}\times\text{a}\times\text{a}=\frac{22}{7}\text{a}^2\text{sq. units}$
Required area $=\Big(4\text{a}^2-\frac{22}{7}\text{a}^2\Big)\text{sq. units}=\frac{6\text{a}^2}{7}$
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