Question:
In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is
- \({\left( {\frac{\pi }{{3\sqrt 3 }}} \right)^{\frac{1}{2}}}\)
- \({\left( {\frac{\pi }{4}} \right)^{\frac{1}{2}}}\)
- \({\left( {\frac{\pi }{6}} \right)^{\frac{1}{2}}}\)
- \({\left( {\frac{\pi }{{4\sqrt 3 }}} \right)^{\frac{1}{2}}}\)
Correct Answer: 1
It is given that radius of the circle = 1 cm
Chord AB subtends an angle of 60° on the centre of the given circle. R be the region bounded by the radii OA, OB and the arc AB.
Therefore, R = $\frac{60°}{360°}$×Area of the circle = $\frac{1}{6}$×$\pi×(1)^2$ = $\frac{\pi}{6}$ sq. cm
It is given that OC = OD and area of triangle OCD is half that of R. Let OC = OD = x.
Area of triangle COD = $\frac{1}{2}×OC×OD×sin60°$
$\frac{\pi}{6×2}$ = $\frac{1}{2}×x×x×\frac{\sqrt{3}}{2}$
$\Rightarrow$ $x^2 = \frac{\pi}{3\sqrt{3}}$
$\Rightarrow$ $x$ = $(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}$ cm.
It is given that radius of the circle = 1 cm
Chord AB subtends an angle of 60° on the centre of the given circle. R be the region bounded by the radii OA, OB and the arc AB.
Therefore, R = $\frac{60°}{360°}$×Area of the circle = $\frac{1}{6}$×$\pi×(1)^2$ = $\frac{\pi}{6}$ sq. cm
It is given that OC = OD and area of triangle OCD is half that of R. Let OC = OD = x.
Area of triangle COD = $\frac{1}{2}×OC×OD×sin60°$
$\frac{\pi}{6×2}$ = $\frac{1}{2}×x×x×\frac{\sqrt{3}}{2}$
$\Rightarrow$ $x^2 = \frac{\pi}{3\sqrt{3}}$
$\Rightarrow$ $x$ = $(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}$ cm.
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