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06 Nov 2019, 02:10
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
54% (02:55) correct
46%
(03:43)
wrong
based on 59
sessions
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[GMAT math practice question]
11.6ps.png [ 34.1 KiB | Viewed 2244 times ]
The figure shows the square with a side length of \(10 cm\), and a circle is inscribed by the square. What is the area of the shaded region?
A. \(40 cm^2\)
B. \(44 cm^2\)
C. \(46 cm^2\)
D. \(48 cm^2\)
E. \(50 cm^2\)
Attachment:
11.6ps.png [ 34.1 KiB | Viewed 2244 times ]
The figure shows the square with a side length of \(10 cm\), and a circle is inscribed by the square. What is the area of the shaded region?
A. \(40 cm^2\)
B. \(44 cm^2\)
C. \(46 cm^2\)
D. \(48 cm^2\)
E. \(50 cm^2\)
06 Nov 2019, 08:51
Kudos
Bookmarks
12.5pi+ (100-25pi)/2 = 50
Posted from my mobile device
Posted from my mobile device
06 Nov 2019, 10:04
Kudos
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Hey, i got 50. (E)
This is how:
Logic:
Area of [Big Square - Small Square - 2 edges(un-shaded area) - 1/4 of small circle + 1 edge]
Calculation:
100 - 25 - 2[ (100-25pi)/4 ] - 25pi/4 + [(100-25pi)/4]
Solve to get 50.
You can also approach this from the circle point of view:
Logic:
Area of [ Inscribed Circle - 1/4 of the circle - (un-shaded area in left bottom {which is further = area of corner circle - area of edge} ) + edge]
Calculation:
25pi - 25pi/4 - [ 25pi/4 -(25 - 25pi/4) ] + (25 - 25pi /4)
If anyone knows an easier way to solve this, please do.
Thanks,
Kaushal.
This is how:
Logic:
Area of [Big Square - Small Square - 2 edges(un-shaded area) - 1/4 of small circle + 1 edge]
Calculation:
100 - 25 - 2[ (100-25pi)/4 ] - 25pi/4 + [(100-25pi)/4]
Solve to get 50.
You can also approach this from the circle point of view:
Logic:
Area of [ Inscribed Circle - 1/4 of the circle - (un-shaded area in left bottom {which is further = area of corner circle - area of edge} ) + edge]
Calculation:
25pi - 25pi/4 - [ 25pi/4 -(25 - 25pi/4) ] + (25 - 25pi /4)
If anyone knows an easier way to solve this, please do.
Thanks,
Kaushal.
