Question

6. A chord of a circle radius 15cm substends an angle of 60° at the center. Find the areas of the corresponding minor and major segments of the circle .

$( \: use \: \pi = 3.14 \: and \: \sqrt{3 \: } = 1.73)$
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Answer 1

Step-by-step explanation:

[Please see the attached image for the diagram]

Given that,

Radius (r) = 15 cm

and Angle (θ) = 60°

Now,

Area of the minor segment APB = Area of sector OAPB - Area of ΔOAB

Now,

Area of sector OAPB =

$\frac{60}{360}*\pi*15^{2}$       [∵Area of a Sector = (θ/360°) × π × r²]

$=\frac{1}{6}*3.14*15*15\\=\frac{1}{2}*3.14*5*15\\=117.75cm^{2}$

Finding Area of ΔAOB:

ΔAOB is isosceles as two sides are equal.  (OA = OB = Radius of the Circle)

∴ ∠A = ∠B

Sum of all interior angles of a triangle = 180°

∴ ∠A + ∠B + ∠C = 180°

⇒ 2 ∠A = 180° - 60°

⇒ ∠A = 120°/2

⇒ ∠A = 60°

ΔAOB is equilateral as ∠A = ∠B = ∠C = 60°

∴ OA = OB = AB = 15 cm

Therefore, Area of the equilateral Δ AOB = √3/4 × (OA)² = √3/4 × 15²  

                                              = (225√3)/4 cm² = 97.3 cm²

Area of the minor segment APB = Area of sector OAPB - Area of ΔOAB

                                                     = (117.75 - 97.3) cm²

                                                     = 20.45 cm²

∴ Area of the Minor Segment APB = 20.45 cm²

Finding Area of the Circle:

Area of the Circle = π × r²

                             = 3.14 * 15² cm²

                             = 706.5 cm²

∴ Area of Major Segment = Area of the circle - Area of the Minor Segment

                                           = (706.5 - 20.45) cm²

                                           = 686.05 cm²

∴Area of the Major Segment is 686.05 cm²

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