A chord with a length of 12 12 runs from pi/12 π12 to pi/6 π6 radians on a circle. What is the area of the circle?

2 Answers
Mar 7, 2016

Area of a circle is
S = (36pi)/sin^2(pi/24)=(72pi)/(1-sqrt((2+sqrt(3))/4))S=36πsin2(π24)=72π12+34

Explanation:

enter image source here

Picture above reflects the conditions set in the problem. All angles (enlarged for better understanding) are in radians counting from the horizontal X-axis OXOX counterclockwise.
AB=12AB=12
/_XOA=pi/12XOA=π12
/_XOB=pi/6XOB=π6
OA=OB=rOA=OB=r

We have to find a radius of a circle in order to determine its area.
We know that chord ABAB has length 1212 and an angle between radiuses OAOA and OBOB (where OO is a center of a circle) is
alpha=/_AOB = pi/6 - pi/12 = pi/12α=AOB=π6π12=π12

Construct an altitude OHOH of a triangle Delta AOB from vertex O to side AB. Since Delta AOB is isosceles, OH is a median and an angle bisector:
AH=HB=(AB)/2=6
/_AOH=/_BOH=(/_AOB)/2=pi/24

Consider a right triangle Delta AOH.
We know that cathetus AH=6 and angle /_AOH=pi/24.
Therefore, hypotenuse OA, which is a radius of our circle r, equals to
r=OA=(AH)/sin(/_AOH)=6/sin(pi/24)

Knowing radius, we can find an area:
S = pi*r^2 = (36pi)/sin^2(pi/24)

Let's express this without trigonometric functions.

Since
sin^2(phi) = (1-cos(2phi))/2
we can express the area as follows:
S = (72pi)/(1-cos(pi/12))

Another trigonometric identity:
cos^2(phi) = (1+cos(2phi))/2
cos(phi) = sqrt[(1+cos(2phi))/2]

Therefore,
cos(pi/12) = sqrt[(1+cos(pi/6))/2] =
= sqrt[(1+sqrt(3)/2)/2] = sqrt((2+sqrt(3))/4)

Now we can represent the area of a circle as
S = (72pi)/(1-sqrt((2+sqrt(3))/4))

Mar 10, 2016

Another approach same result

Explanation:

enter image source here
The chord AB of length 12 in the above figure runs frompi/12 to pi/6 in the circle of radius r and center O, taken as origin .
/_AOX=pi/12 and /_BOX=pi/6
So polar coordinate of A =(r,pi/12) and that of B =(r,pi/6)
Applying distance formula for polar coordinate
the length of the chord AB,12=sqrt(r^2+r^2-2*r^2*cos(/_BOX-/_AOX)

=>12^2=r^2+r^2-2*r^2*cos(pi/6-pi/12)
=>144=2r^2(1-cos(pi/12))
=>r^2=144/(2(1-cos(pi/12))
=>r^2=cancel144^72/(cancel2(1-cos(pi/12))
=>r^2=72/(1-cos(pi/12))
=>r^2=72/(1-sqrt(1/2(1+cos(2*pi/12))
=>r^2=72/(1-sqrt(1/2(1+cos(pi/6))
=>r^2=72/(1-sqrt(1/2(1+sqrt3/2)

So area of the circle
=pi*r^2
=(72pi)/(1-sqrt(1/2(1+sqrt3/2)
=(72pi)/(1-sqrt((2+sqrt3)/4)