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 = \(\frac{ θ}{360}\) ∙ πr^{2}
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 = \(\frac{60^{\circ}}{360^{\circ}}\) × πr^{2} [Since θ = 60]
= \(\frac{1}{6}\) × πr^{2}
= \(\frac{1}{6}\) × \(\frac{22}{7}\) × 28^{2} m^{2}.
= \(\frac{1}{6}\) × \(\frac{22}{7}\) × 784 m^{2}.
= \(\frac{17248}{42}\) m^{2}.
= \(\frac{1232}{3}\) m^{2}.
= 410\(\frac{2}{3}\) m^{2}.
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.
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