Question

a chord of a circle of radius 28cm. subtends an angle 45at the centre of the circle . the area of the minor segment .

Answer 1

given, radius of circle = 28 cm

and angle A=45°

Area of sector=A /360° πr^2

=45/360 × 22/ 7 × 28 × 28

=( 22 × 4 × 28 ) / 8

=2464/8

=308 cm^2

therefore, the required area of minor sector = 308 cm^2
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Your Answer: 308 cm^2
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Answer 2

Answer:

Area of the minor segment = 30.86cm²

Step-by-step explanation:

Given,

Radius of the circle = 28cm

To find,

The area of the minor segment

Solution

Recall the formula

Area of the minor segment =  area of the minor sector -  area of the triangle

Area of a sector = $\frac{\theta}{360} \pi r^2$, where θ is the angle made by the sector at the center of the circle.

The area of the triangle formed by the two radii of the sector and the chord is given by $\frac{1}{2}r^2sin\theta$

Here, we have

θ = 45°

r = 28cm

Area of the sector =  $\frac{\theta}{360} \pi r^2$

= $\frac{45}{360} \pi X28^2$

= $\frac{1}{8}$ ×$\frac{22}{7}$×28²

= 308

Area of the sector = 308cm²

Area of the triangle =  $\frac{1}{2}r^2sin\theta$

= $\frac{1}{2}$ ×28 ×28×sin45

= 196√2 cm²

Area of the triangle = 196√2 cm²

Area of the minor segment = 308 - 196√2

= 308 - 277.144

= 30.86cm²

∴Area of the minor segment = 30.86cm²

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