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# In the figure above, a circle is inscribed in a square with side b and

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Joined: 02 Sep 2009
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In the figure above, a circle is inscribed in a square with side b and  [#permalink]

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13 Dec 2017, 23:00
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15% (low)

Question Stats:

82% (01:19) correct 18% (01:32) wrong based on 88 sessions

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In the figure above, a circle is inscribed in a square with side b and a square with side a is inscribed in the circle. What is the area oF the square with side b ?

(1) a = 4
(2) The radius of the circle is $$2\sqrt{2}$$

Attachment:

square.gif [ 3.81 KiB | Viewed 1421 times ]

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Re: In the figure above, a circle is inscribed in a square with side b and  [#permalink]

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14 Dec 2017, 00:40
2
When a circle is inscribed in that bigger square of side 'b', diameter of circle is same as the side of the square. So diameter of the given circle is 'b'. And when a square is inscribed inside a circle, then the diagonal of that square is same as the diameter of the circle. So the diagonal of the smaller square is 'b'. But we know that diagonal of a square = √2 * Side OR
b = √2 * a.. so side of smaller square 'a' = b/√2.

So now we have a relation between 'a' and 'b'. We need to find b*b (area of larger square).

(1) a=4. Which means b=4*√2. So area can be found. Sufficient.

(2) radius of circle =2√2, so diameter=4√2. This is same as side 'b' of larger square. So area can be found. Sufficient.

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Re: In the figure above, a circle is inscribed in a square with side b and  [#permalink]

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16 Dec 2017, 17:52
Bunuel wrote:

In the figure above, a circle is inscribed in a square with side b and a square with side a is inscribed in the circle. What is the area oF the square with side b ?

(1) a = 4
(2) The radius of the circle is $$2\sqrt{2}$$

Attachment:
square.gif

Notes:
Diagonal of a square = $$Side*\sqrt{2}$$

Radius of a circle = $$\frac{Diameter}{2}$$

(1) a = 4. So, Diagonal of small square = Diameter of the circle = Side of large square = $$b = 4\sqrt{2}$$. Area of large square is: $$b^2=(4\sqrt{2})^2=(16)(2)=32$$, sufficient.

(2) The radius of the circle is $$2\sqrt{2}$$. Then, Diameter of circle = $$2*Radius$$ = $$2*(2\sqrt{2})$$ = Side of large square = $$b = 4\sqrt{2}$$. Area of large square is: $$b^2=(4\sqrt{2})^2=(16)(2)=32$$, sufficient.

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Re: In the figure above, a circle is inscribed in a square with side b and  [#permalink]

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28 Jan 2019, 02:12
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Re: In the figure above, a circle is inscribed in a square with side b and   [#permalink] 28 Jan 2019, 02:12
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