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26 Aug 2015, 00:46
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (02:13) correct
32%
(02:09)
wrong
based on 149
sessions
History
Date
Time
Result
Not Attempted Yet
The above picture depicts a circle perfectly inscribed in a square. Is the shaded region > 4?
(1) The area of the large rectangle equals 64.
(2) The perimeter of the shaded region equals 8 + 2π.
Kudos for a correct solution.
Attachment:
circle-in-square-red2.GIF [ 2.24 KiB | Viewed 4012 times ]
26 Aug 2015, 09:59
Kudos
Bookmarks
IMO: D
Consider the following fig:
Capture.JPG [ 15.19 KiB | Viewed 3731 times ]
Question Stem: Area of the shaded region = \(\frac{[Area of the square - Area of the cirlce]}{4}\)
Area of square = 2r*2r = \(4r^2\)
Area of Circle = \(πr^2\)
Area of the shaded region = \(\frac{[4r^2 - πr^2]}{4}\)
IS \(\frac{r^2[4 - π]}{4}\) > 4 ?
IS \(r^2[4 - π]\) > 16 ? -- (i)
St 1: The area of the large rectangle equals 64.
I think the question is flawed. PSTU is a rectangle and PQRS is also a rectangle(Every square is a rectangle). The use of word large is ambiguous. It can either represent PSTU or PQRS. If largest is used instead then PQRS can be considered which would be sufficient alone as well.
Area of PSTU = 64
Where PS = 2r & SU = r
Thus, 2r*r = 64
\(r^2\)= 32 ---(ii)
Sub in Eq - (i)
\(32[4 - π]\) > 16 ? YES
Hence Suff
St 2: The perimeter of the shaded region equals 8 + 2π
Perimeter of the shaded region = Length of the arc TV + Side TQ + Side QV
Length of the arc = \(\frac{θ}{360} * 2πr\)
= \(\frac{90}{360} * 2πr\)
= \(\frac{πr}{2}\)
Thus,
\(8 + 2π = \frac{πr}{2} + r + r\)
r = 4
Sub in eq--(i)
\(r^2[4 - π]\) > 16 ?
\(16[4 - π]\) > 16 ? YES
Hence Suff
Consider the following fig:
Attachment:
Capture.JPG [ 15.19 KiB | Viewed 3731 times ]
Question Stem: Area of the shaded region = \(\frac{[Area of the square - Area of the cirlce]}{4}\)
Area of square = 2r*2r = \(4r^2\)
Area of Circle = \(πr^2\)
Area of the shaded region = \(\frac{[4r^2 - πr^2]}{4}\)
IS \(\frac{r^2[4 - π]}{4}\) > 4 ?
IS \(r^2[4 - π]\) > 16 ? -- (i)
St 1: The area of the large rectangle equals 64.
I think the question is flawed. PSTU is a rectangle and PQRS is also a rectangle(Every square is a rectangle). The use of word large is ambiguous. It can either represent PSTU or PQRS. If largest is used instead then PQRS can be considered which would be sufficient alone as well.
Area of PSTU = 64
Where PS = 2r & SU = r
Thus, 2r*r = 64
\(r^2\)= 32 ---(ii)
Sub in Eq - (i)
\(32[4 - π]\) > 16 ? YES
Hence Suff
St 2: The perimeter of the shaded region equals 8 + 2π
Perimeter of the shaded region = Length of the arc TV + Side TQ + Side QV
Length of the arc = \(\frac{θ}{360} * 2πr\)
= \(\frac{90}{360} * 2πr\)
= \(\frac{πr}{2}\)
Thus,
\(8 + 2π = \frac{πr}{2} + r + r\)
r = 4
Sub in eq--(i)
\(r^2[4 - π]\) > 16 ?
\(16[4 - π]\) > 16 ? YES
Hence Suff
26 Aug 2015, 19:11
Kudos
Bookmarks
shaded region > 4?
If we can determine the radius, we can determine the side of the square and the rectangle, allowing us to find the area of the square, rectangle and circle. What is the radius?
(1) The area of the large rectangle equals 64.
r*2r=64
Sufficient
(2) The perimeter of the shaded region equals 8 + 2π.
8 + 2π = (2πr)/4 + 2r
Sufficient
Answer: D
If we can determine the radius, we can determine the side of the square and the rectangle, allowing us to find the area of the square, rectangle and circle. What is the radius?
(1) The area of the large rectangle equals 64.
r*2r=64
Sufficient
(2) The perimeter of the shaded region equals 8 + 2π.
8 + 2π = (2πr)/4 + 2r
Sufficient
Answer: D
