Find the area of the shaded regionin the figure below, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral traingle OAB of side 12 cm as centre.
Here we can see that
Area of required figure = Area of circle+Area of triangle− Area of the sector
⇒A(circle)=`Ï€"r"^2`
⇒A(circle)=`\pi` `6^2`
⇒A(circle)=36`\pi`
⇒A(triangle)=`\frac{\sqrt{3}}{4}a^2`
⇒A(triangle)=`\frac{\sqrt{3}}{4}(12^2)`
⇒A(triangle)= `36\sqrt{3}`
⇒A(sector)=`1/2``"r"^2``\theta`
⇒A(sector)=`1/2`(`"6"^2`)`\frac{\pi}{3}`
⇒A(sector)=6`\pi`
Therefore, Required Area=36`\pi` + `36 \sqrt{3}` − 6`\pi`
⇒ Required Area=36`\pi` + `36 \sqrt{3}`
Area of required figure = Area of circle+Area of triangle− Area of the sector
⇒A(circle)=`Ï€"r"^2`
⇒A(circle)=`\pi` `6^2`
⇒A(circle)=36`\pi`
⇒A(triangle)=`\frac{\sqrt{3}}{4}a^2`
⇒A(triangle)=`\frac{\sqrt{3}}{4}(12^2)`
⇒A(triangle)= `36\sqrt{3}`
⇒A(sector)=`1/2``"r"^2``\theta`
⇒A(sector)=`1/2`(`"6"^2`)`\frac{\pi}{3}`
⇒A(sector)=6`\pi`
Therefore, Required Area=36`\pi` + `36 \sqrt{3}` − 6`\pi`
⇒ Required Area=36`\pi` + `36 \sqrt{3}`
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