Three circles each of radius \[3.5\] cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Home » Three circles each of radius cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Three circles each of radius

$3.5$

cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Three circles each of radius

$3.5$

cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Solution:

Given that the three circles are drawn such that each of them touches the other two.

Now, by joining the centers of the three circles,

We get, AB = BC = CA =

$2$

(radius) =

$7$

cm

Therefore, we can say that triangle ABC is an equilateral triangle with each side

$7$

cm.

∴ Area of the triangle of side a =

$(\sqrt{3}/4)\times {{a}^{2}}$

=

$(\sqrt{3}/4)\times {{7}^{2}}$

=

$(49/4)\sqrt{3}$

$c{{m}^{2}}$

=

$21.2176$

$c{{m}^{2}}$

Now, we know that Central angle of each sector =

$\phi ={{60}^{\circ }}(60\pi /180)$

=

$\pi /3$

radians

Thus, area of each sector =

$(1/2){{r}^{2}}\theta$

=

$(1/2)\times {{(3.5)}^{2}}\times (\pi /3)$

=

$12.25\times (22/(7\times 6))$

=

$6.4167$

$c{{m}^{2}}$

Therefore, Total area of three sectors =

$3\times 6.4167$

=

$19.25$

$c{{m}^{2}}$

so, the Area enclosed between three circles = Area of triangle ABC – Area of the three sectors

=

$21.2176-19.25$

=

$1.9676$

$c{{m}^{2}}$

Therefore, the required area enclosed between these circles is

$1.9676$

$c{{m}^{2}}$

i

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