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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, 00:23
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  45% (medium)

Question Stats:

63% (01:35) correct 38% (02:24) wrong based on 24 sessions

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Kudos [?]: 135644 [0], given: 12705

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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, 01:22
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(#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, 02:29
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.

Kudos [?]: 8 [0], given: 37

BSchool Forum Moderator
User avatar
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Joined: 26 Feb 2016
Posts: 1702

Kudos [?]: 750 [0], given: 20

Location: India
Concentration: General Management, Leadership
WE: Sales (Retail)
Premium Member CAT Tests
If (#x) represents the area of a semicircle with diameter x, then (#2) [#permalink]

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New post 14 Nov 2017, 04:11
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!
_________________

Stay hungry, Stay foolish

2017-2018 MBA Deadlines

Class of 2020: Rotman Thread | Schulich Thread
Class of 2019: Sauder Thread

Kudos [?]: 750 [0], given: 20

Intern
Intern
User avatar
B
Joined: 16 Aug 2016
Posts: 15

Kudos [?]: 8 [0], given: 37

Location: India
GMAT 1: 460 Q35 V19
GPA: 3.6
WE: Brand Management (Retail)
GMAT ToolKit User CAT Tests
Re: If (#x) represents the area of a semicircle with diameter x, then (#2) [#permalink]

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New post 14 Nov 2017, 04:30
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 :)

Kudos [?]: 8 [0], given: 37

Re: If (#x) represents the area of a semicircle with diameter x, then (#2)   [#permalink] 14 Nov 2017, 04:30
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If (#x) represents the area of a semicircle with diameter x, then (#2)

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