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.
From Area and Perimeter of a Sector of a Circle to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Oct 10, 24 05:15 PM
Oct 10, 24 10:06 AM
Oct 10, 24 03:19 AM
Oct 09, 24 05:16 PM
Oct 08, 24 10:53 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.