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# In the figure above, RSTV is a square inscribed in a circle with radiu

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In the figure above, RSTV is a square inscribed in a circle with radiu  [#permalink]

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14 Nov 2017, 22:41
00:00

Difficulty:

25% (medium)

Question Stats:

91% (01:52) correct 9% (02:19) wrong based on 42 sessions

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In the figure above, RSTV is a square inscribed in a circle with radius r. In terms of r, what is the total area of the shaded regions?

(A) r^2(π – 2)
(B) 2r(2 – π)
(C) π(r^2 – 2)
(D) πr^2– 8r
(E) πr^2 – 4r

Attachment:

2017-11-15_1035_001.png [ 7.25 KiB | Viewed 1711 times ]

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Re: In the figure above, RSTV is a square inscribed in a circle with radiu  [#permalink]

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14 Nov 2017, 23:44
Since RSTV is a circle the relationship between r and itrs sides would be 2^1/2 to 1
Area of squre=( 2^1/2*r)^2=2r^2
Area of circle=P*r^2
Area of shaded region=P*r^2 - 2r^2=r^2*(P-2)
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Re: In the figure above, RSTV is a square inscribed in a circle with radiu  [#permalink]

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15 Nov 2017, 08:09
Bunuel wrote:

In the figure above, RSTV is a square inscribed in a circle with radius r. In terms of r, what is the total area of the shaded regions?

(A) r^2(π – 2)
(B) 2r(2 – π)
(C) π(r^2 – 2)
(D) πr^2– 8r
(E) πr^2 – 4r

Attachment:
2017-11-15_1035_001.png

Diagonal of the square = $$2r$$ , so area of the square = $$4r^2/2$$ ir, $$2r^2$$

Area of the circle = $$πr^2$$

So, Area of the shaded region = $$πr^2 - 2r^2$$

Or, Area of the shaded region = $$r^2 ( π- 2 )$$ , answer will be (A)
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In the figure above, RSTV is a square inscribed in a circle with radiu  [#permalink]

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16 Nov 2017, 17:36
Bunuel wrote:

In the figure above, RSTV is a square inscribed in a circle with radius r. In terms of r, what is the total area of the shaded regions?

(A) r^2(π – 2)
(B) 2r(2 – π)
(C) π(r^2 – 2)
(D) πr^2– 8r
(E) πr^2 – 4r

Attachment:
2017-11-15_1035_001.png

The total area of the shaded regions equals
(Area of Circle) - (Area of Square)

Area of circle= $$πr^2$$

Area of square
The square's diagonal = 2r

The relationship between a square's diagonal, d, and its side, s, is
$$s^2 = d$$, so
$$s = \frac{d}{\sqrt{2}}$$
$$s = \frac{2r}{\sqrt{2}}$$

Area of square =
$$s^2 =\frac{2r}{\sqrt{2}}* \frac{2r}{\sqrt{2}} =\frac{(4)r^2}{2}= 2r^2$$

(Circle area) - (square area)
$$(πr^2 - 2r^2) =$$
$$r^2(π - 2)$$

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Re: In the figure above, RSTV is a square inscribed in a circle with radiu  [#permalink]

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13 Sep 2018, 13:37
Bunuel wrote:

In the figure above, RSTV is a square inscribed in a circle with radius r. In terms of r, what is the total area of the shaded regions?

(A) r^2(π – 2)
(B) 2r(2 – π)
(C) π(r^2 – 2)
(D) πr^2– 8r
(E) πr^2 – 4r

Attachment:
2017-11-15_1035_001.png

A faster approach to get the area of the square, is to treat it as a rombus.
A = (d1 x d2)/2 = 2r x 2r / 2 = 2r^2
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Re: In the figure above, RSTV is a square inscribed in a circle with radiu &nbs [#permalink] 13 Sep 2018, 13:37
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