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555-605 (Medium)|   Geometry|      
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Bunuel
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A square is inscibed in a circle of radius s and circumscribed about 21 Feb 2022, 06:45
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A square is inscibed in a circle of radius s and circumscribed about a circle of radius r. Which of the following is an expression for the perimeter of the square?

(A) 8s
(B) 8r
(C) 4s + 4r
(D) 2√2s + 2√2r
(E) 2s + 2r
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B
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A square is inscibed in a circle of radius s and circumscribed about 21 Feb 2022, 15:01
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If the side of the square is a
Then the radius of outer circle = a/√2
And the radius of the inner circle = a/2

Perimeter of the square = 4a
Which is equal to 8*(a/2) or 8r which is option B.

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Pastillita
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A square is inscibed in a circle of radius s and circumscribed about 12 Mar 2022, 07:04
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kungfury42
If the side of the square is a
Then the radius of outer circle = a/√2
And the radius of the inner circle = a/2

Perimeter of the square = 4a
Which is equal to 8*(a/2) or 8r which is option B.

Posted from my mobile device
Show more


Hi there, Im a bit confused by your explanation pertaining to the radius of the outer circle. Please do correct me if I am mistaken but surely the radius of the outer circle is a*root(2) / 2? I arrived at the same answer as you, with the same logic for the radius of the inner circle but when it comes to the perimeter of the square based on the radius of the outer circle, I had a different answer to you.

My logic was that the radius of the outer circle is 's', so the diameter = 2s. We know that the diameter is a diagonal of the inscribed square, so it can be expressed as
side of square*root(2). Let "side of square" be 'a'.

Hence, a*root(2) = diameter = 2s

Thus a = 2s/root(2) so perimeter (4*a) = 8s/root(2) = (4s*root(2) )as an expression for the perimeter. Did I make a mistake somewhere or am I correct?
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A square is inscibed in a circle of radius s and circumscribed about 12 Mar 2022, 09:58
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Pastillita
kungfury42
If the side of the square is a
Then the radius of outer circle = a/√2
And the radius of the inner circle = a/2

Perimeter of the square = 4a
Which is equal to 8*(a/2) or 8r which is option B.

Posted from my mobile device


Hi there, Im a bit confused by your explanation pertaining to the radius of the outer circle. Please do correct me if I am mistaken but surely the radius of the outer circle is a*root(2) / 2? I arrived at the same answer as you, with the same logic for the radius of the inner circle but when it comes to the perimeter of the square based on the radius of the outer circle, I had a different answer to you.

My logic was that the radius of the outer circle is 's', so the diameter = 2s. We know that the diameter is a diagonal of the inscribed square, so it can be expressed as
side of square*root(2). Let "side of square" be 'a'.

Hence, a*root(2) = diameter = 2s

Thus a = 2s/root(2) so perimeter (4*a) = 8s/root(2) = (4s*root(2) )as an expression for the perimeter. Did I make a mistake somewhere or am I correct?
Show more

Hi Pastillita, I'm happy to help.

Quote:
Please do correct me if I am mistaken but surely the radius of the outer circle is a*root(2) / 2?
Show more
This is absolutely correct. And in case you were wondering, the radius of the outer circle in my answer a/√2 is the indeed same expression as yours but with the common \(\sqrt{2}\) cancelled from both Numerator and Denominator.

\(\frac{a\sqrt{2}}{2}=\frac{a\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{a}{\sqrt{2}}\)

Quote:
My logic was that the radius of the outer circle is 's', so the diameter = 2s. We know that the diameter is a diagonal of the inscribed square, so it can be expressed as
side of square*root(2). Let "side of square" be 'a'.

Hence, a*root(2) = diameter = 2s

Thus a = 2s/root(2) so perimeter (4*a) = 8s/root(2) = (4s*root(2) )as an expression for the perimeter. Did I make a mistake somewhere or am I correct?
Show more
This is a perfectly sound logic. And you're absolutely correct in calculating the perimeter of the square in terms of the outer radius as \(4s\sqrt{2}\) however this is not present in the options, and hence we will have to stick with expressing the perimeter in terms of the radius of the inner circle.

I would also like to point that you could have saved on so much of calculations by a simple observation. You have the perimeter of the square as \(4a\) and you have the radius of the outer circle as \(s=\frac{a\sqrt{2}}{2}\).

Now simply rewrite \(s=\frac{a\sqrt{2}}{2}\) as \(a=\frac{2s}{\sqrt{2}}\) and plug this value of \(a\) back in \(4a\) to obtain the perimeter of the square in terms of the outer radius \(s\)

\(4a = 4*a = 4*\frac{2s}{\sqrt{2}} = 4*\sqrt{2}s\)

Hope this was helpful.
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A square is inscibed in a circle of radius s and circumscribed about 13 Mar 2022, 10:10
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kungfury42
Pastillita
kungfury42
If the side of the square is a
Then the radius of outer circle = a/√2
And the radius of the inner circle = a/2

Perimeter of the square = 4a
Which is equal to 8*(a/2) or 8r which is option B.

Posted from my mobile device


Hi there, Im a bit confused by your explanation pertaining to the radius of the outer circle. Please do correct me if I am mistaken but surely the radius of the outer circle is a*root(2) / 2? I arrived at the same answer as you, with the same logic for the radius of the inner circle but when it comes to the perimeter of the square based on the radius of the outer circle, I had a different answer to you.

My logic was that the radius of the outer circle is 's', so the diameter = 2s. We know that the diameter is a diagonal of the inscribed square, so it can be expressed as
side of square*root(2). Let "side of square" be 'a'.

Hence, a*root(2) = diameter = 2s

Thus a = 2s/root(2) so perimeter (4*a) = 8s/root(2) = (4s*root(2) )as an expression for the perimeter. Did I make a mistake somewhere or am I correct?

Hi Pastillita, I'm happy to help.

Quote:
Please do correct me if I am mistaken but surely the radius of the outer circle is a*root(2) / 2?
This is absolutely correct. And in case you were wondering, the radius of the outer circle in my answer a/√2 is the indeed same expression as yours but with the common \(\sqrt{2}\) cancelled from both Numerator and Denominator.

\(\frac{a\sqrt{2}}{2}=\frac{a\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{a}{\sqrt{2}}\)

Quote:
My logic was that the radius of the outer circle is 's', so the diameter = 2s. We know that the diameter is a diagonal of the inscribed square, so it can be expressed as
side of square*root(2). Let "side of square" be 'a'.

Hence, a*root(2) = diameter = 2s

Thus a = 2s/root(2) so perimeter (4*a) = 8s/root(2) = (4s*root(2) )as an expression for the perimeter. Did I make a mistake somewhere or am I correct?
This is a perfectly sound logic. And you're absolutely correct in calculating the perimeter of the square in terms of the outer radius as \(4s\sqrt{2}\) however this is not present in the options, and hence we will have to stick with expressing the perimeter in terms of the radius of the inner circle.

I would also like to point that you could have saved on so much of calculations by a simple observation. You have the perimeter of the square as \(4a\) and you have the radius of the outer circle as \(s=\frac{a\sqrt{2}}{2}\).

Now simply rewrite \(s=\frac{a\sqrt{2}}{2}\) as \(a=\frac{2s}{\sqrt{2}}\) and plug this value of \(a\) back in \(4a\) to obtain the perimeter of the square in terms of the outer radius \(s\)

\(4a = 4*a = 4*\frac{2s}{\sqrt{2}} = 4*\sqrt{2}s\)

Hope this was helpful.
Show more


Thank you for taking the time to give me this detailed reply, much appreciated.
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