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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