Areas and perimeters of sectors (degrees)

Like we saw in our chapters describing circumferences and areas of circles we now have the following rules.  

What if we don't have an entire circle?  

Well, half a circle would have half the area or half the circumference. One quarter of a circle would have a quarter of the area, or a quarter of the circumference.  In fact all we need to know is what fraction the sector is of a whole circle.  For this all we need to know is the angle of the sector.

Looking at the quarter circle, the angle of the sector is $90$90°. The fraction of the circle is $\frac{90}{360}=\frac{1}{4}$90360​=14​. 

More generally, If the angle of the sector is $\theta$θ, then the fraction of the circle is represented by

$fraction=\frac{\theta}{360}$fraction=θ360​ (due to there being $360$360° in a circle).

 

Example

Question: Find the area and circumference of a sector with central angle of $126$126° and radius of $7$7cm. Evaluate to $2$2 decimal places.  

Think: What fraction is this sector of a whole circle?  What are the rules for circumference and area?

Do:  This sector is $\frac{126}{360}=0.35$126360​=0.35 of a circle.

Circumference of a whole circle is $C=2\pi r$C=2πr, so the perimeter of the sector is

$0.35\times2\pi r$0.35×2πr	$=$=	$0.35\times2\pi\times7$0.35×2π×7
	$=$=	$0.35\times14\pi$0.35×14π
	$=$=	$4.9\times\pi$4.9×π
	$=$=	$15.39$15.39 cm (rounded to $2$2 decimal places)

Area of a circle is $A=\pi r^2$A=πr2, so the area of the sector is

$0.35\times\pi r^2$0.35×πr2	$=$=	$0.35\pi\times7^2$0.35π×72
	$=$=	$17.15\pi$17.15π
	$=$=	$53.88$53.88 cm2  (rounded to 2 decimal places)

 

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3