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In the circle above, if the area of the rectangle set inside the circl

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In the circle above, if the area of the rectangle set inside the circl  [#permalink]

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New post 05 Jul 2018, 04:52
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In the circle above, if the area of the rectangle set inside the circle is 200 and b = 8a, what is the circumference of the circle?


A. \(25\pi \sqrt{65}\)

B. \(5 \sqrt{65}\)

C. \(5\pi \sqrt{13}\)

D. \(5\pi \sqrt{65}\)

E. \(5\pi\)


Attachment:
circle.png
circle.png [ 7.74 KiB | Viewed 384 times ]

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In the circle above, if the area of the rectangle set inside the circl  [#permalink]

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New post 05 Jul 2018, 05:02
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Given: b = \(8a\)(where \(a\) is the width of the rectangle) | Area(rectangle) = 200

The product of length and breadth is the area of the rectangle -> \(a*8a = 200\)

Solving for a, we will get \(a^2 = \frac{200}{8} = 25\) -> \(a = \sqrt{25} = 5\).

From the figure, we know that the length of the diagonal is the diameter of the circle.

Length of diagonal = \(\sqrt{a^2 + (8a)^2} = a\sqrt{65} = 5\sqrt{65}\)

Therefore, the circumference of the circle is \(\pi * d = 5\pi \sqrt{65}\) (Option D)
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In the circle above, if the area of the rectangle set inside the circl  [#permalink]

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New post 05 Jul 2018, 18:36
Bunuel wrote:
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In the circle above, if the area of the rectangle set inside the circle is 200 and b = 8a, what is the circumference of the circle?


A. \(25\pi \sqrt{65}\)

B. \(5 \sqrt{65}\)

C. \(5\pi \sqrt{13}\)

D. \(5\pi \sqrt{65}\)

E. \(5\pi\)


Attachment:
circle.png

(1) Use the area of the rectangle to find side lengths, in which \(L=b=8a\)
\(A=(W*L)=(a*b)=(a*8a)=200\)
\(8a^2=200\)
\(a^2=25\)
\(a=5\)
\(b=8a=40\)

(2) Use Pythagorean theorem to find hypotenuse = diameter of circle*
\(a^2+b^2=d^2\)
\(5^2+40^2=d^2\)
\(d^2=1,625\)
Look at the answer choices. \(\sqrt{65}\) shows up three times
\(1,625=65*25\), so
\(\sqrt{d^2}=\sqrt{25*65}\)
\(d=5\sqrt{65}\)

(3) Circumference= \(\pi* d=5\sqrt{65}\pi\)

Answer D

*Use variables instead of values to avoid 1,625
\(a^2+8a^2=d^2\)
\(a^2*(1+8^2)=d^2\)
\(\sqrt{a^2}*\sqrt{65}=\sqrt{d^2}\)
\(a*\sqrt{65}=d\)
\(d=5*\sqrt{65}\)

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In the circle above, if the area of the rectangle set inside the circl &nbs [#permalink] 05 Jul 2018, 18:36
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