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21 Nov 2017, 23:19
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If a square is inscribed in a circle of radius r as shown above, then the area of the square region is
(A) r^2/2π
(B) πr^2/2
(C) πr^2
(D) r^2
(E) 2r^2
Attachment:
2017-11-21_1033.png [ 4.39 KiB | Viewed 26611 times ]
Luckisnoexcuse
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21 Nov 2017, 23:32
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Bunuel
If a square is inscribed in a circle of radius r as shown above, then the area of the square region is
(A) r^2/2π
(B) πr^2/2
(C) πr^2
(D) r^2
(E) 2r^2
Attachment:
2017-11-21_1033.png
Diagonal of Square = 2r = \(x\sqrt{2}\)
Area of square = \((\sqrt{2}r)^2\)
=2r^2
E
26 Nov 2017, 22:43
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Bunuel
If a square is inscribed in a circle of radius r as shown above, then the area of the square region is
(A) r^2/2π
(B) πr^2/2
(C) πr^2
(D) r^2
(E) 2r^2
Attachment:
2017-11-21_1033.png
Diagonal, d, of square = \(2r\)
Side, s, of square*:
\(s\sqrt{2} = d\)
\(s = \frac{d}{\sqrt{2}}\)
\(s=\frac{2r}{\sqrt{2}}\) *\(\frac{\sqrt{2}}{\sqrt{2}}\)
\(s= \frac{2r\sqrt{2}}{2}=r\sqrt{2}\)
Area of square = \(s^2\)
A= \((r\sqrt{2})^2 = 2r^2\)
Answer E
* derived from
\(s^2 + s^2 = d^2\)
\(2s^2 = d^2\)
\(\sqrt{2}*\sqrt{s^2} = \sqrt{d^2}\)
\(s\sqrt{2} = d\)
\(s = \frac{d}{\sqrt{2}}\)
