|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 16 Apr 2009
Posts: 252
Schools: Ross
Followers: 1
Kudos [?]:
10
[0], given: 10
|
Question Stats:
100% (02:08) correct
0% (00:00) wrong based on 0 sessions
. 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
Attachments
 park.doc [21.5 KiB]
Downloaded 76 times
_________________
Keep trying no matter how hard it seems, it will get easier.
|
|
|
|
|
|
|
Manager
Joined: 25 Jul 2009
Posts: 119
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
Followers: 2
Kudos [?]:
103
[0], given: 17
|
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 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
_________________
KUDOS me if I deserve it !! 
My GMAT Debrief - 740 (Q50, V39) | My Test-Taking Strategies for GMAT | Sameer's SC Notes
|
|
|
|
|
|
Current Student
Joined: 18 Jun 2009
Posts: 363
Location: San Francisco
Schools: Duke,Oxford,IMD,INSEAD
Followers: 7
Kudos [?]:
51
[0], given: 15
|
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
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
