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Area of circles inside a semi circle help! Watch

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Size:  34.9 KB 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)
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Size:  34.9 KB 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.
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    (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
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    (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 r is related to the \frac{2}{\sqrt{\pi}}? You might be able to spot it by looking at the horizontal radius.
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    (Original post by coconut64)
    Name:  aa.png
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Size:  34.9 KB 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)
    Name:  aa.png
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Size:  34.9 KB 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.
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    (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.
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    (Original post by RDKGames)
    Can you see how the r is related to the \frac{2}{\sqrt{\pi}}? You might be able to spot it by looking at the horizontal radius.
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    how would it be 2/root pi? I don't see you you can obtain that I just know it as 1.128 thanks
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    (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 \frac{1}{2}\pi R^2=2 and rearrange for R which is the radius of the semicircle.

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

    \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....
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    (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=\frac{1}{2}R

    You can prove this if you're being formal about it too.
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    (Original post by RDKGames)
    r=\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'
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    (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.
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    (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

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    (Original post by RDKGames)
    r=\frac{1}{2}R

    You can prove this if you're being formal about it too.
    What are r and R ?
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    (Original post by notnek)
    What are r and R ?
    He guided me that r is the radius inside the circle and R is the radius of the semi-circle
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    (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 r the radius of the black circle?

    Your original question was about the grey circles which is why I wasn't sure about the variables.
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    (Original post by notnek)
    Is r 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
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    (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 \pi.

    It is actually possible to find the black area without involving \pi.
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    (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;

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    R=\frac{2}{\sqrt{\pi}}

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

    Apologies.
 
 
 
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