Question

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Answer 1

$\bold{\huge{\underline{Question-}}}$

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.


$\bold{\huge{\underline{Answer-}}}$According to the question we know that each angle are equilateral triangle is 60°
and, area of circle is πr² arc of radius 6 cm




$\bold{\huge{\underline{Solution-}}}$




$\bold{\huge{Step-1}}$


Here we know each angle are equilateral triangle is 60° so,

$\bold{Area\:of\:sector\:OCDE}$.......(1)


$= \frac{60°}{360°} \times \pi {r}^{2} \\ \\ \\ = \frac{1}{6} \times \pi \times 6 \times 6 \\ \\ \\ = \boxed{ \frac{132}{7} {cm}^{2} }$






$\bold{\huge{Step-2}}$



$\bold{Area\:of \:equilateral\: triangle\: BAO}$.........(2)



$= \frac{ \sqrt{3} }{4} \times 12 \times 12 \\ \\ \\ = \boxed{ \sqrt[36]{3} {cm}^{2} }$







$\bold{\huge{Step-3}}$



$\bold{Area\:of\:circle}$ ..............(3)


$= \frac{22}{7} \times 6 \times 6 \\ \\ \\ = \boxed{ \frac{792}{7} {cm}^{2} }$






$\bold{\huge{Step-4}}$



$\bold{Area\:of\:shaded\: region}$



= Now, from the (3) + (2) - (1)



$= \frac{792}{7} + \sqrt[36]{3} - \frac{132}{7} \\ \\ \\ = \boxed{ \sqrt[36]{3} + \frac{660}{7} {cm}^{2} }$

Answer 2

❥QUESTION

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

❥Answer

It is given that OAB is an equilateral triangle having each angle as 60°

Area of the sector is common in both.

Radius of the circle = 6 cm.

Side of the triangle = 12 cm.

Area of the equilateral triangle = (√3/4) (OA)2= (√3/40×122 = 36√3 cm2

Area of the circle = πR2 = (22/7)×62 = 792/7 cm2

Area of the sector making angle 60° = (60°/360°) ×πr2 cm2

= (1/6)×(22/7)× 62 cm2 = 132/7 cm2

Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector

= 36√3 cm2 +792/7 cm2-132/7 cm2

= (36√3+660/7) cm2