Question

Find the area of shaded region wheres circle of radius 6cm has been drawn with vertex o at equilateral triangle oab of side 12

Answer 1

❏SOLUTION:-

• Area of the circle with centre O and radius 6 cm

$\sf =\pi r²=\pi (6)²cm²$

$\sf =3.14×36=113.04cm²$⠀⠀⠀⠀

⠀⠀

•Area of sector OLQP such that $\bf\angle LOP=60⁰$

⠀⠀⠀⠀⠀⠀⠀⠀|$❏ \sf\triangle OAB$ is an equilateral triangle -Each angle of $\sf \triangle OAB=60⁰$|

⠀⠀⠀

$\sf = \pi r {}^{2} \times \dfrac{ \theta}{360 {}^{0} }$

⠀⠀⠀

$\sf = 3.14 \times (6) {}^{2} \times \dfrac{ 60 {}^{0} }{360 {}^{0} }$

⠀⠀⠀

$\sf = 3.14 \times 36 \times \dfrac{1}{6} cm {}^{2}$

⠀⠀⠀

$= \sf 3.14 \times 6cm {}^{2}$

⠀⠀⠀

$= \sf18.84 cm {}^{2}$

❏Area of an equilateral triangle OAB with side 12cm

$\sf = \dfrac{ \sqrt{3} }{4} \times (side) {}^{2} cm {}^{2}$

⠀⠀⠀

$\sf= \dfrac{ \sqrt{3} }{4} \times (12) {}^{2} cm {}^{2}$

⠀⠀⠀

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

⠀⠀⠀

$\sf = 1.79 \times 36cm {}^{2}$

⠀⠀⠀

$= \sf 62.28cm {}^{2}$

⠀⠀⠀

❏Area of shaded region

⠀⠀

=Area of circle +area of equilateral triangle-2×Area of sector OLQP

⠀⠀⠀⠀

=(113.04+62.18-2×18.84)cm²

⠀⠀⠀

=(175.32-37.68)cm²

⠀⠀⠀

=137.64 cm²