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Bunuel

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

Total area of shaded regions
(Circle area) - (square area)
\((πr^2 - 2r^2) =\)
\(r^2(π - 2)\)

Answer A
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Bunuel

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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Bunuel

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

Total Area - Area of square = Shaded area
Area of square: Found using this https://gmatclub.com/forum/if-a-square- ... l#p2285889

Answer= PI (r^2) - 2 (r^2) = (A) r^2(π – 2)
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