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huongguyen
Joined: 23 May 2020
Last visit: 26 Nov 2022
Posts: 158
Given Kudos: 143
Concentration: Statistics, Finance
Schools: Foster '22 Marshall '25 Tepper '24
GPA: 3.41
WE:Advertising (Advertising and PR)
14 Sep 2021, 23:51
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D
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:
55% (03:03) correct
45%
(03:46)
wrong
based on 33
sessions
History
Date
Time
Result
Not Attempted Yet
Given two squares ABCD and MNPQ as shown in the figure.
BD = 12. EO = OK = AK = AE = r
pi = 3.14
What is the shaded area?
A. 56.52
B. 36
C. 15.48
D. 20.52
E. 47.52
BD = 12. EO = OK = AK = AE = r
pi = 3.14
What is the shaded area?
A. 56.52
B. 36
C. 15.48
D. 20.52
E. 47.52
Attachments
Q22.png [ 56.86 KiB | Viewed 1355 times ]
16 Sep 2021, 04:28
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huongguyen
Area of shaded region = area of circle - area of smaller square.
I. Side of the outer square = 2r, and diagonal = \(2\sqrt{2}r=12.............r=3\sqrt{2}\)
Thus, area of circle = \(\pi * (3\sqrt{2})^2=3.14*18= ~ 56\)
II. Diagonal of the inner square = 2r, so side of the inner square = \(r\sqrt{2}=3\sqrt{2}*\sqrt{2}=6\).
Thus, area of the square = 6*6 = 36
Area of shaded region = area of circle - area of smaller square = ~(56-36) = ~ 20.
D
Attachments
Q22.png [ 71.02 KiB | Viewed 1194 times ]
