Question

AB is a chord of a circle with centre O and radius 4cm. AB is length 4cm and divides circle into two segments.

Answer 1

Given data: Radius of the circle with center‘O’, r = 4 cm = OA = OB Length of the chord AB = 4 cm OAB is an equilateral triangle and angle AOB = 60° + ? Angle subtended at centre ? = 60° Area of the segment ( shaded region ) = ( area of sector) - ( area of triangleAOB ) = \frac{\theta }{360}\times\prod r^{2} - \frac{\sqrt{3}}{4}\left ( side\right )^{2} = \frac{ 60 }{ 360 }\times \prod 4 ^{2} - \frac{\sqrt{ 3 }}{ 4 }\left ( 4\right )^{2} On solving the above equation, we get, = 58.67 – 6.92 = 51.75 cm2 Therefore, the required area of the segment is 51.75 cm2 

Answer 2

given... AB is the diameter and PQ is the chord of the circle with centre O and AB bisects PQ at R also   PR = RQ = 8 cm and RB = 4 cm

we know that the line from the centre of the circle which bisects the chord is perpendicular to the chord

⇒ AB ⊥ PQ

Let the radius of the circle be r

⇒ OP = OB = r

⇒ OR + RB = r

⇒ OR = r – RB = r-4

Now in right ∆ ORP

OR²+PR²=OP²

⇒ (r – 4)²+8²=R²

⇒r²+4²-2*4*r+64=r²

⇒16-8r+64=0

⇒8r=80

⇒r=80/8

⇒r=10

∴RADIUS 10

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