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In the figure above, if the area of the inscribed rectangle is 32, the

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Math Expert
Joined: 02 Sep 2009
Posts: 42649

Kudos [?]: 135955 [0], given: 12717

In the figure above, if the area of the inscribed rectangle is 32, the [#permalink]

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13 Nov 2017, 01:26
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In the figure above, if the area of the inscribed rectangle is 32, then the circumference of the circle is

(A) 20π
(B) 4π√5
(C) 4π√3
(D) 2π√5
(E) 2π√3

[Reveal] Spoiler:
Attachment:

2017-11-13_1325.png [ 8.18 KiB | Viewed 264 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 135955 [0], given: 12717

Intern
Joined: 28 Dec 2010
Posts: 49

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Re: In the figure above, if the area of the inscribed rectangle is 32, the [#permalink]

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13 Nov 2017, 02:41
Area = 32
$$2x^2 = 32$$
x = 4

Sides, 4 & 8
Diameter is hypotenuse of the triangle.
Diameter = $$\sqrt{4^2 + 8^2}$$
Diameter = $$4\sqrt{5}$$

Circumference = $$4π√5$$

B.
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Intern
Joined: 01 Nov 2017
Posts: 4

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Re: In the figure above, if the area of the inscribed rectangle is 32, the [#permalink]

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13 Nov 2017, 02:47
2x*x=32
=> x=4
therefore, length= 8 & width= 4
we know, diagonal of rectangle= root(8^2+4^2)
=root(80)
=4root5

Now, circum= 2πr=2π*2root5= 4π root 5
so ans is B

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Re: In the figure above, if the area of the inscribed rectangle is 32, the   [#permalink] 13 Nov 2017, 02:47
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