Area and Perimeter of a Sector of a Circle

We will discuss the Area and perimeter of a sector of a circle

We know that

Therefore,

Area of a sector of a circle = \(\frac{ \theta^{\circ}}{360^{\circ}}\) × Area of the circle = \(\frac{ θ}{360}\) ∙ πr²

where r is the radius of the circle and \(\theta^{\circ}\) is the sectorial angle.

Also, we know that

Therefore,

Arc MN = \(\frac{ \theta^{\circ}}{360^{\circ}}\) × Circumference of the circle = \(\frac{ θ}{360}\) ∙ 2πr = \(\frac{πθr}{180}\)

where r is the radius of the circle and \(\theta^{\circ}\) is the sectorial angle.

Thus,

perimeter of a sector of a circle = (\(\frac{πθ}{180}\) ∙ r + 2r) = (\(\frac{πθ}{180}\) + 2)r

where r is the radius of the circle and θ° is the sectorial angle.

Problems on Area and Perimeter of a Sector of a Circle:

1. A plot of land is in the shape of a sector of a circle of radius 28 m. If the sectorial angle (central angle) is 60°, find the area and the perimeter of the plot. (Use π = \(\frac{22}{7}\).)

Solution:

Area of the plot = \(\frac{60^{\circ}}{360^{\circ}}\) ×  πr² [Since θ = 60]

                       = \(\frac{1}{6}\) ×  πr²

                       = \(\frac{1}{6}\) × \(\frac{22}{7}\) × 28² m².

                       = \(\frac{1}{6}\) × \(\frac{22}{7}\) × 784 m².

                       = \(\frac{17248}{42}\) m².

                       = \(\frac{1232}{3}\) m².

                       = 410\(\frac{2}{3}\) m².

Perimeter of the plot = (\(\frac{πθ}{180}\) + 2)r

                              = (\(\frac{22}{7}\) ∙ \(\frac{60}{180}\) + 2) 28 m

                              = (\(\frac{22}{21}\) + 2) 28 m

                              = \(\frac{64}{21}\) ∙ 28 m

                              = \(\frac{1792}{21}\) m

                              = \(\frac{256}{3}\) m

                              = 85\(\frac{1}{3}\) m.

10th Grade Math

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