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If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of

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If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of  [#permalink]

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New post 07 Feb 2019, 02:51
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

66% (01:29) correct 34% (01:42) wrong based on 35 sessions

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Re: If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of  [#permalink]

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New post 07 Feb 2019, 03:09
Bunuel wrote:
Image
If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of the circle, in terms of x, is

A. \(\sqrt{3} x\pi\)

B. \(\sqrt{5} x\pi\)

C. \(\sqrt{3x} \pi\)

D. \(\sqrt{5x} \pi\)

E. \(5x^2 \pi\)



If ABCD is a rectangle, lets calculate AC

AC =\(\sqrt{{x^2 + 4* x^2}}\)
AC = \(\sqrt{5}\)x

AC will be the diameter, radius will be \(\sqrt{5}\)x / 2

C = 2 \(\pi r\)
= 2 * \(\pi\)\(\sqrt{5}\)x / 2
= \(\sqrt{5} x\pi\)
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Re: If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of  [#permalink]

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New post 07 Feb 2019, 03:12
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image.jpg [ 28.22 KiB | Viewed 688 times ]


From the above fig, the diagonal of the rectangle serves as the diameter of the circle as the diagonal passes through the center of both the circle and the rectangle.

Since the diameter is of \(\sqrt{5}x\), radius of the circle \(\frac{\sqrt{5}x}{2}\)

The circumference of a circle is \(2\pi r\) = \(2*\pi*\frac{\sqrt{5}x}{2}\)

\(\sqrt{5}x \pi\)

OPTION: B
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Re: If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of  [#permalink]

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New post 07 Feb 2019, 03:52
diaganol of the rectangle = diameter of circle
d of rectangle = \sqrt{5}x

radius = \sqrt{5}x/2
2 * pi * \sqrt{5}x/2

=>\(\sqrt{5} x\pi\)
IMO B

Bunuel wrote:
Image
If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of the circle, in terms of x, is

A. \(\sqrt{3} x\pi\)

B. \(\sqrt{5} x\pi\)

C. \(\sqrt{3x} \pi\)

D. \(\sqrt{5x} \pi\)

E. \(5x^2 \pi\)


Attachment:
Geometry_015.jpg
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Re: If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of   [#permalink] 07 Feb 2019, 03:52
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