Question

# A circumcircle is drawn around an equilateral triangle. If the circum-radius r = 14 cm, find the area of the shaded region.

A

61.68 cm2

B

65.32 cm2

C

69.51 cm2

D

64.23 cm2

Solution

## The correct option is A 61.68 cm2 The shaded region can be divided into three segments.  δ ABC is an equilateral triangle.  Hence all the angles are equal = 60° Consider chord AB, the angle made by the chord at circumference = ∠ACB = 60° Now, let us consider the chord AB. Chord AB makes an angle of 60° at C, i.e. ∠ACB=60∘ Then, the angle ∠AOB=2∠ACB = 2×60∘  ( angle a chord makes at the centre is double the angle the chord makes at any point on the circumference of the circle) = 120∘ Now let us consider the sector AOB, The area of the sector AOB = Θ360∘×πr2  =  120360×227×72 =  13×22×14×2 =  6163  = 105.33cm2 Now that we know the area of the sector, we need to find the area of triangle AOB in order to find the area of the segment ABD. Draw OE   ⊥ AB ;  AE = 12 AB ( Perpendicular to a chord passing through the centre will bisects the chord  Here ∠AOE=12∠AOB = 12×120  = 60∘ By applying trigonometric identities in rt. triangle AOE we get, AE = OA sin 60 = √3r2  = 14×√32 = √3 cm EO = OA Cos 60 =r×12 = 142 = 7 cm In triangle AOB, AB = 2 AE (OE is the perpendicular bisector of AB) Therefore, AB = 2 x 7 √3 = 14 √3 cm  Area of the triangle AOB = 12 x AB x OE  = 12×14√3×7 = 7√3×7 = 49√3  = 49 x 1.73 = 84.77 cm2 Therefore, the area of the segment ABD = Area of the sector AOBD – Area of the triangle AOB = 105.33 – 84.77 = 20.56 cm2 The area of 1 segment   = 20.56cm2 Therefore the area of the total segment = 3 x area of 1 segment = 3 x 20.56 = 61.68 cm2 Therefore the total area of the shaded region is 61.68 cm2

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