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Area of circles inside a semi circle help!

aa.png Hi, I have worked out that the radius the semi circle to be 1.128 and the area of the black circle is 4 but don't get how to work out area of the grey circles.

Thanks

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Original post by coconut64
aa.png Hi, I have worked out that the radius the semi circle to be 1.128 and the area of the black circle is 4 but don't get how to work out area of the grey circles.

Thanks


Split the diagram in half, isolate one circle. Use symmetry and radius of the semicircle to figure it out.
Reply 2
Original post by RDKGames
Split the diagram in half, isolate one circle. Use symmetry and radius of the semicircle to figure it out.


Can you give me some more hints for the first step because I don't see how the radius can help here...

Thanks
Original post by coconut64
Can you give me some more hints for the first step because I don't see how the radius can help here...

Thanks


Can you see how the rr is related to the 2π\frac{2}{\sqrt{\pi}}? You might be able to spot it by looking at the horizontal radius.
dfsdfsdf.PNG
Reply 4
Original post by coconut64
aa.png Hi, I have worked out that the radius the semi circle to be 1.128 and the area of the black circle is 4 but don't get how to work out area of the grey circles.

Thanks

Original post by coconut64
aa.png Hi, I have worked out that the radius the semi circle to be 1.128 and the area of the black circle is 4 but don't get how to work out area of the grey circles.

Thanks


How can the black circle have an area of 4, when it is inside a semicircle of area 2?

It's a weird universe.
(edited 7 years ago)
Original post by ghostwalker
How can the black circle have an area of 4, when it is inside a semicircle of area 2?

It's a weird universe.


Reply 7
Original post by RDKGames
Can you see how the rr is related to the 2π\frac{2}{\sqrt{\pi}}? You might be able to spot it by looking at the horizontal radius.
dfsdfsdf.PNG
how would it be 2/root pi? I don't see you you can obtain that I just know it as 1.128 thanks
Original post by coconut64
how would it be 2/root pi? I don't see you you can obtain that I just know it as 1.128 thanks


We're told area of the semicircle is 2. So 12πR2=2\frac{1}{2}\pi R^2=2 and rearrange for RR which is the radius of the semicircle.

2π=1.128\frac{2}{\sqrt{ \pi }} = 1.128 to 3 decimal places so you got it rounded while I'm showing the exact value.
(edited 7 years ago)
Reply 9
Original post by RDKGames
We're told area of the semicircle is 2. So 12πR2=2\frac{1}{2}\pi R^2=2 and rearrange for RR which is the radius of the semicircle.

2π=1.128\frac{2}{\sqrt{ \pi }} = 1.128 to 3 decimal places so you got it rounded while I'm showing the exact value.


Thanks. But I still don't get how r can be worked out using the radius and the line of symmetry as I can only use it to find the area of the sector....
Original post by coconut64
Thanks. But I still don't get how r can be worked out using the radius and the line of symmetry as I can only use it to find the area of the sector....


r=12Rr=\frac{1}{2}R

You can prove this if you're being formal about it too.
Reply 11
Original post by RDKGames
r=12Rr=\frac{1}{2}R

You can prove this if you're being formal about it too.


Can I just ask is this A-level stuff or beyond because i have never learnt that but got given this question as a 'challenge'
Original post by coconut64
Can I just ask is this A-level stuff or beyond because i have never learnt that but got given this question as a 'challenge'


Basic GCSE stuff.
Reply 13
Original post by RDKGames
Basic GCSE stuff.


I am really surprised then cuz I didn't remember learning that. I am confused now, so surely the radius of the black circle is also 0.564 but it is not looking right tho

Thanks
Reply 14
Original post by RDKGames
r=12Rr=\frac{1}{2}R

You can prove this if you're being formal about it too.

What are rr and RR ?
Reply 15
Original post by notnek
What are rr and RR ?


He guided me that r is the radius inside the circle and R is the radius of the semi-circle
Reply 16
Original post by coconut64
He guided me that r is the radius inside the circle and R is the radius of the semi-circle

Is rr the radius of the black circle?

Your original question was about the grey circles which is why I wasn't sure about the variables.
Reply 17
Original post by notnek
Is rr the radius of the black circle?

Your original question was about the grey circles which is why I wasn't sure about the variables.


I just followed his guidance and found the two grey circles to be 1 cm2 but if r is 0.564, it just means that the area of the big circle is also one then? I am confused
Reply 18
Original post by coconut64
I just followed his guidance and found the two grey circles to be 1 cm2 but if r is 0.564, it just means that the area of the big circle is also one then? I am confused

I think you should start again and work out the area of the black circle, which is the easier area to work out.

In your OP you said the area of the black circle is 4 but how can it be larger than the area of the semicircle?

Have another go and post your working if you still get 4. You shouldn't need to use a calculator so don't evaluate π\pi.

It is actually possible to find the black area without involving π\pi.
Original post by coconut64
I just followed his guidance and found the two grey circles to be 1 cm2 but if r is 0.564, it just means that the area of the big circle is also one then? I am confused


I have just looked over my working and realised I gave you nonsense as far as the grey part is concerned, got mixed up with it as I have done it a few days ago aha.

Allow me to clarify;

dsafsdf.PNG

R=2πR=\frac{2}{\sqrt{\pi}}

and you can work out rr with trigonometry therefore you can find the area of the smaller circle.

Apologies.

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