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03 Apr 2020, 05:55
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In a circle with centre 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 what is the length of OC?
A. \(\sqrt{\frac{\pi}{4}}\)
B. \(\sqrt{\frac{\pi}{6}}\)
C. \(\sqrt{\frac{\pi}{3\sqrt{3}}}\)
D. \(\sqrt{\frac{\pi}{4\sqrt{3}}}\)
E. \(\sqrt{\frac{\pi}{5\sqrt{3}}}\)
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03 Apr 2020, 06:40
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Area of sector =60/360πr^2
Area of Sector=1/6π=R
Area of Triangle OCD is r/2
∆OCD=1/12π
Also, the area of triangle ocd is √3/4x^2(where x is length of OC)
1/12π=√3/4x^2
X=√(π/3√3)
OA is C
Posted from my mobile device
Area of Sector=1/6π=R
Area of Triangle OCD is r/2
∆OCD=1/12π
Also, the area of triangle ocd is √3/4x^2(where x is length of OC)
1/12π=√3/4x^2
X=√(π/3√3)
OA is C
Posted from my mobile device
03 Apr 2020, 07:12
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Area of region R = \(\pi r^2 * \frac{360}{6}\) = \(\frac{\pi }{6}\) , since r = 1
Now, \(\frac{\pi }{6} * \frac{1}{2}\) = △ OCD
or, △ OCD = \(\frac{\pi }{12}\) .. equation (i)
Now, since OC=OD, so ∠OCD=∠ODC.
Hence, ∠OCD=∠ODC=∠COD=60 deg
Therefore, OC=CD=OD
Now we can equate from equation (i),
△ OCD = \(\frac{\pi }{12}\),
or, \(\frac{\sqrt{3}}{4} * x^2\)= \(\frac{\pi }{12}\)
therefore, x = \(\sqrt{\frac{\pi }{ 3 \sqrt3}}\)
Answer: C
Now, \(\frac{\pi }{6} * \frac{1}{2}\) = △ OCD
or, △ OCD = \(\frac{\pi }{12}\) .. equation (i)
Now, since OC=OD, so ∠OCD=∠ODC.
Hence, ∠OCD=∠ODC=∠COD=60 deg
Therefore, OC=CD=OD
Now we can equate from equation (i),
△ OCD = \(\frac{\pi }{12}\),
or, \(\frac{\sqrt{3}}{4} * x^2\)= \(\frac{\pi }{12}\)
therefore, x = \(\sqrt{\frac{\pi }{ 3 \sqrt3}}\)
Answer: C
