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If (#x) represents the area of a semicircle with diameter x, then (#2)

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If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

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New post 14 Nov 2017, 01:23
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (02:03) correct 35% (02:19) wrong based on 30 sessions

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If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

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New post 14 Nov 2017, 02:22
1
1
(#x) represents the area of a semicircle with diameter x.

Formula:
Area of a semicircle = \(\frac{1}{2}\)*pi*r^2 where r-radius or diameter/2

(#2) = \(\frac{1}{2}*pi*1^2\)
(#4) = \(\frac{1}{2}*pi*2^2\)

(#2) + (#4) = \(\frac{1}{2}*pi*(1+4)\) = \(\frac{1}{2}*pi*{5}\)

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*{2\sqrt{5}}^2*\frac{1}{2^2}\) = \(\frac{1}{8}*pi*{4*5}\) = \(\frac{1}{2}*pi*{5}\)

Therefore, (#2) + (#4) = \((#2 \sqrt{5})\)(Option D)
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Re: If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

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New post 14 Nov 2017, 03:29
pushpitkc wrote:
(#x) represents the area of a semicircle with diameter x.

Formula:
Area of a semicircle = \(\frac{1}{2}\)*pi*r^2 where r-radius or diameter/2

(#2) = \(\frac{1}{2}*pi*1^2\)
(#4) = \(\frac{1}{2}*pi*2^2\)

(#2) + (#4) = \(\frac{1}{2}*pi*(1+4)\) = \(\frac{1}{2}*pi*{5}\)

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*{2\sqrt{5}}^2*\frac{1}{2^2}\) = \(\frac{1}{8}*pi*{4*5}\) = \(\frac{1}{2}*pi*{5}\)

Therefore, (#2) + (#4) = \((#2 \sqrt{5})\)(Option D)



Hi, kindly explain why (2\sqrt{5})^2 expression been multiplied by 1/2^2 in the 2nd last line.
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If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

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New post 14 Nov 2017, 05:11
pulkitarya wrote:
pushpitkc wrote:
(#x) represents the area of a semicircle with diameter x.

Formula:
Area of a semicircle = \(\frac{1}{2}\)*pi*r^2 where r-radius or diameter/2

(#2) = \(\frac{1}{2}*pi*1^2\)
(#4) = \(\frac{1}{2}*pi*2^2\)

(#2) + (#4) = \(\frac{1}{2}*pi*(1+4)\) = \(\frac{1}{2}*pi*{5}\)

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*{2\sqrt{5}}^2*\frac{1}{2^2}\) = \(\frac{1}{8}*pi*{4*5}\) = \(\frac{1}{2}*pi*{5}\)

Therefore, (#2) + (#4) = \((#2 \sqrt{5})\)(Option D)



Hi, kindly explain why (2\sqrt{5})^2 expression been multiplied by 1/2^2 in the 2nd last line.



Hi pulkitarya

The reason I have written it like that is for better readability. It should have been

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*(2\sqrt{5}/2)^2\) = \(\frac{1}{2}*pi*{5}\)

Hope that helps!
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Re: If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

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New post 14 Nov 2017, 05:30
pushpitkc wrote:
pulkitarya wrote:
pushpitkc wrote:
(#x) represents the area of a semicircle with diameter x.

Formula:
Area of a semicircle = \(\frac{1}{2}\)*pi*r^2 where r-radius or diameter/2

(#2) = \(\frac{1}{2}*pi*1^2\)
(#4) = \(\frac{1}{2}*pi*2^2\)

(#2) + (#4) = \(\frac{1}{2}*pi*(1+4)\) = \(\frac{1}{2}*pi*{5}\)

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*{2\sqrt{5}}^2*\frac{1}{2^2}\) = \(\frac{1}{8}*pi*{4*5}\) = \(\frac{1}{2}*pi*{5}\)

Therefore, (#2) + (#4) = \((#2 \sqrt{5})\)(Option D)



Hi, kindly explain why (2\sqrt{5})^2 expression been multiplied by 1/2^2 in the 2nd last line.



Hi pulkitarya

The reason I have written it like that is for better readability. It should have been

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*(2\sqrt{5}/2)^2\) = \(\frac{1}{2}*pi*{5}\)

Hope that helps!


ok, my bad..missed that totally..
thanks pushpitkc :)
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Re: If (#x) represents the area of a semicircle with diameter x, then (#2) &nbs [#permalink] 14 Nov 2017, 05:30
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If (#x) represents the area of a semicircle with diameter x, then (#2)

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