Circle Geometry (1 Viewer)

[DJel]

Hi,

I can prove part one easily, however the second part of this question has me perplexed.

http://img237.imageshack.us/img237/8262/circleuo8.png


Any help would be greatly appreciated.

 

[kony]

let PX = a
and let SX = b

from (i), PX/AP = SX/BS
a/5=b/3
a=5b/3 --------------------- (1)

also, AX + XB = 16.
AX represents the hypotenuse of APX and XB represents the hypotenuse of XSB.

root(25+a²) + root(9+b²) = 16
subbing in result (1),

root(25+25b²/9) + root(9+b²) = 16
5/3.root(9+b²) + root(9+b²) = 16
root(9+b²) = 6

from this, a and b are found.

now we use a bit of trig to find what angleSBX and angleQAX are, in order to find the length of arcPQ and arcRS. the result required follows.

 

[DJel]

Hi,

thanks for the help, I have found that angles PAQ and RBS are 2pi/3 (as angles SBX and QAX = pi/3) therefore arc RS = 2pi and arc PQ = 10pi/3. Now, I think I am confused about how the string is being placed around the circle.

I am starting at R to Q then around the circumference back to Q then 2/3 of the circumference back up to P. From P to S then around the circumference back to S then 2/3 of the circumference back up to R.

I think I am doing it wrong, could someone please explain how the string is going around the two circles?

DJel.

 

[lyounamu]

Hi,

thanks for the help, I have found that angles PAQ and RBS are 2pi/3 (as angles SBX and QAX = pi/3) therefore arc RS = 2pi and arc PQ = 10pi/3. Now, I think I am confused about how the string is being placed around the circle.

I am starting at R to Q then around the circumference back to Q then 2/3 of the circumference back up to P. From P to S then around the circumference back to S then 2/3 of the circumference back up to R.

I think I am doing it wrong, could someone please explain how the string is going around the two circles?

DJel.

1. P to S
2. S to R around the big arc
3. R to Q
4. Q to P around the big arc

He already found you SX and PX. So times each one by 2 (PX = QX since tangents from external point are equal) and add them together. And then add the pratial circumferences of both circles and add everything together.

Like this: 2SX + 2PX +RS (big arc) + PQ (big arc)

Following from Kony's working out, PX = (5 x square root of 27)/3 and SX = square root of 27

Arc PQ = rl = 5 . 4pi/3
Arc RS = rl = 3. 4pi/3

Total distance = 2 x (5 x square root of 27)/3 + 2 x square root of 27 + 20pi/3 + 12pi/3
= 61.2231345....

 

[DJel]

Thanks guys.