Question

three equal circles,each of radii 6cm touch one another.find the area enclosed between them. use root 3=1.73

Answer 1

you can put π=3.142 . i hope you have understand

Answer 2

Solution:

If we can join the radius point on each circle it leads a triangle by enclosing the circle (It is in the below  attachment), since the side of each triangle are same it would be the equilateral triangle. To find the area enclosed between them we must exclude the area of equilateral triangle from that sector.

Area of equilateral Triangle is $\bold=\frac{\sqrt{3}}{4} \times a^{2}}$. Where, ‘a’ stands for the side of equilateral triangle. In this case the value of a,  

= 6+6 = 12 cm

Area of equilateral Triangle $=\frac{\sqrt{3}}{4} \times 12^{2}$

$=36 \times \sqrt{3} \mathrm{cm}^{2}$

Area of sector would be $=\pi \times r^{2} \times \frac{\theta}{360}$

$= \pi \times r^{2} \times \frac{60}{360}$

$= \pi \times 6 \times 6 \times \frac{60}{360}$

$\mathrm{H}=6 \times \pi \mathrm{cm}^{2}$

The value of for the three sectors $=3 \times 6 \times \pi$

                                       $=18 \times \pi \mathrm{cm}^{2}$

The required value of area enclosed by them,  

$=(36 \times \sqrt{3})-(18 \times \pi) \mathrm{cm}^{2}$

$=8.532 {cm}^{2}$