Circular Measure - Additional Mathematics

Circular Measure
Cambridge O Level Additional Mathematics 4037  

Circular measure is one of the easiest chapters when it comes to understanding the basic concepts.
Let's first look at these concepts.

The first thing that you need to know is the relation between degrees and radians.

180 ° = π rads

The smaller part of the circle is called the minor sector the larger part is the major sector.

Finding the Arc length 

The arc length is the length of arc along AB. It is represented by s. r is the radius of circle and θ is subtended by the sector (of which the area is to be found).

 s = rθ  (theta must be in radians)

e.g. If r = 3 cm 

       θ = π/3

s = rθ 

= 3(π/3)

= π

= 3.14cm            

Finding the Area of a sector

To find the area of a sector, we use the following formula:

A = 1/2 * r^2 * θ ( theta must be in radians)

Another formula for the area of a sector is
A = 1/2 * r * s
where s is the length of the arc.

e.g. r = 3cm and θ = 1.2rad
A = 1/2 * 3 * 3 * 1.2
    = 5.4 cm^2



Finding the Area of a triangle and Area of a segment 

Area of the triangle:
A = 1/2 * r * r * sin θ



E.g.  r = 3cm and θ = 30° = π/6 rads
A = 1/2 * 3 * 3 * 0.5

    = 2.25cm                [Either convert 30° to radians, set calculator to rads and then find sin(π/6) or set calculator to ° and find sin(30). The answer will be same].


The shaded part of the sector is called the segment. The remaining part of the sector forms an isosceles triangle. The line separating the segment and the isosceles triangle is called chord. To find the area of the Segment, we first find out the area of the sector and then subtract the area of the triangle from it.

E.g. r = 3cm and θ = 30°
A = (1/2 * r^2 * θ) - (1/2 * r^2 *sinθ)
    = (1/2 * 3^2 *  π/6) - (1/2 * 3^2 * 0.5)
    = 0.106cm^2




Another topic that you should know for this chapter is Similarities and the rules for similar triangles. You should also know the Sine and Cosine rule and be able to recognize the angles formed between parallel lines (corresponding, alternate, allied).

* Also note that a right angle is formed at tangents.

Although the basic formulas from this chapter are pretty simple, the questions from the question papers can sometimes look complicated. However, no matter how complex the question is, you will always notice that when you break the question into parts, the answers always lie in the basic concepts. Either it is the length of an arc or area of a sector/segment or it is the area of a triangle that you have to find.


So let's solve a few questions now.
Q. The diagram shows two intersecting circles of radius 6cm and 4cm with centres 7cm apart. Find the perimeter and the area of the shaded region common to both circles.

First lets find out the perimeter of the shaded region. 
If you look at the diagram carefully, you will notice that the perimeter is the sum of the arc length of the larger circle and the arc length of the sector of the smaller circle.


So lets first find out the arc length.
s = rθ,  we know the r of both the circles but we do not know θ of either.


Now, the radius of the larger sector is 6cm, smaller is 4cm and the line joining the centres is 7cm. So, these 3 lines are forming a triangle.
Let the angle for the larger triangle be θ1 and the smaller be θ2.
Therefore, taking (θ1/2) and applying the Cosine rule:
 4^2 = 6^2 + 7^2 - {2 * 6 * 7 * cos (θ1/2)}
16 = 85 - 84cos(θ1/2)
84cos(θ1/2) = 85 - 16
cos(θ1/2) = 69/84
(θ1/2) = 0.607 rads
θ1 = 1.214 rads

θ2/2 = π - 0.607 - π/2= 0.964
θ2 = 1.928 rads 

Therefore, s1= 6*1.214 = 7.284cm
s2 = 4*1.928 = 7.712cm 
perimeter = 7.284 + 7.712 = 14.996cm

A1 = (0.5 * 36 * 1.214) - (0.5 * 36 * sin(1.214)) = 4.985 cm^2
A2 = (0.5 * 16 * 1.928) - (0.5 * 16 * sin(1.928)) = 7.929 cm^2
Total area = 12.91 cm^2


Q.





First let's write all the angles.
Let's take L as the symbol for angle.
LCOD = π/6 rad (alternate angle)
AC is a straight line.
LCOD + LAOD = π
LAOD = 2.617 rads
LBOC = 2.617 rads

length of arc AB = 10 * π/6 = 5.236cm
length of arc CD = 20 * π/6 = 10.472cm
Using the Cosine rule:
AD^2 = 10^2 + 20^2 - 2*10*20*cos 2.617
AD = 29.09 cm = BC
Perimeter = 29.09 +29.09 + 5.236 + 10.47 = 73.9cm

 
This is rather easier.
Area of sector AOB = 0.5 * 100 * π/6 = 26.18cm^2
Area of sector COD = 0.5 * 400 * π/6 = 104.7cm^2
Area of triangle AOD = Area of triangle BOC = 0.5 * 10 * 20 * sin 2.617 = 50.1cm^2
Total Area = 231.1cm^2