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19 Feb 2025, 22:08
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
50% (02:39) correct
50% (03:12) wrong based on 4 sessions
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In the figure below, four semicircles are drawn, each one centered at the midpoint of one of the sides of square ABCD. Each of the four shaded “petals” is the intersection of two of the semicircles. If AB = 4, what is the total area of the shaded region?
A. \(8π\)
B. \(32 – 8π\)
C. \(16 – 8π\)
D. \(8π – 32\)
E. \(8π – 16\)
11.jpg [ 16 KiB | Viewed 419 times ]
A. \(8π\)
B. \(32 – 8π\)
C. \(16 – 8π\)
D. \(8π – 32\)
E. \(8π – 16\)
Attachment:
11.jpg [ 16 KiB | Viewed 419 times ]
22 Feb 2025, 10:31
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OE
Since AB = 4, the area of the square is 16, and so, obviously, the area of the shaded region must be much less.
Check each choice. Since π is slightly more than 3 \((π ≈ 3.14), 8π\) is somewhat greater than 24, approximately 25.
(A) \(8π ≈ 25.\) More than the area of the whole square: way too big.
(B) \(32 – 8π ≈ 32 – 25 = 7.\)
(C) \(16 – 8π\) is negative.
(D) \(8π – 32\) is also negative.
(E) \(8π – 16 ≈ 25 – 16 = 9.\)
Answer: E
Since AB = 4, the area of the square is 16, and so, obviously, the area of the shaded region must be much less.
Check each choice. Since π is slightly more than 3 \((π ≈ 3.14), 8π\) is somewhat greater than 24, approximately 25.
(A) \(8π ≈ 25.\) More than the area of the whole square: way too big.
(B) \(32 – 8π ≈ 32 – 25 = 7.\)
(C) \(16 – 8π\) is negative.
(D) \(8π – 32\) is also negative.
(E) \(8π – 16 ≈ 25 – 16 = 9.\)
Answer: E
