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# park

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Manager
Joined: 16 Apr 2009
Posts: 233
Schools: Ross

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14 Aug 2009, 08:00
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100% (02:08) correct 0% (00:00) wrong based on 3 sessions

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. The figure above represents a rectangular parking lot that is 30 meters by 40 meters and an attached semicircular driveway that has an outer radius of 20 meters and an inner radius of 10 meters. If the shaded region is not included, what is the area, in square meters, of the lot and driveway?
(A) 1,350pie
(B) 1,200 + 400pie
(C) 1,200 + 300pie
(D) 1,200 + 200pie
(E) 1,200 + 150pie
[Reveal] Spoiler:
E

Attachments

park.doc [21.5 KiB]

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Manager
Joined: 25 Jul 2009
Posts: 116
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN

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14 Aug 2009, 08:58
vannu wrote:
. The figure above represents a rectangular parking lot that is 30 meters by 40 meters and an attached semicircular driveway that has an outer radius of 20 meters and an inner radius of 10 meters. If the shaded region is not included, what is the area, in square meters, of the lot and driveway?
(A) 1,350pie
(B) 1,200 + 400pie
(C) 1,200 + 300pie
(D) 1,200 + 200pie
(E) 1,200 + 150pie
[Reveal] Spoiler:
E

The required area is: A(Rectangle) + A(Larger Semicircle) - A(Smaller Semicircle)
= 30*40 + (pi*20^2)/2 - (pi*10^2)/2
= 1200 + 150*pi

ANS: E
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Joined: 18 Jun 2009
Posts: 356
Location: San Francisco

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14 Aug 2009, 09:01
Area of the rectangular region = 30*40 = 1200 ... 1

Area of the semi circular region = pie*r^2/2 = pie*400/2 = pie*200

Area of shaded semi circular region = pie*r^2/2 = pie*100/2 = pie*50

Area of unshaded region semi circular region = Area of the semi circular region - Area of shaded semi circular region = pie*200/2 - pie*50 = pie 150

Area of unshaded region = Area of the rectangular region + Area of unshaded region semi circular region = 1200 + 150 pie ... E
Re: park   [#permalink] 14 Aug 2009, 09:01
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