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Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
10 Sep 2014, 22:22
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:44) correct
40%
(02:57)
wrong
based on 120
sessions
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Not Attempted Yet
As shown in figure, Square ABCD has arc AC centred at B and arc BD centred at C. If AB = 4, area of shaded region is
A: \(\frac{16\pi}{3} - 4\sqrt{3}\)
B: \(4\sqrt{3} - \frac{4\pi}{3}\)
C: \(4 + 4\sqrt{3} + \frac{4\pi}{3}\)
D: \(16 + 2\sqrt{3} - \frac{8\pi}{3}\)
E: \(16 - 4\sqrt{3} - \frac{8\pi}{3}\)
square.png [ 9.08 KiB | Viewed 10845 times ]
A: \(\frac{16\pi}{3} - 4\sqrt{3}\)
B: \(4\sqrt{3} - \frac{4\pi}{3}\)
C: \(4 + 4\sqrt{3} + \frac{4\pi}{3}\)
D: \(16 + 2\sqrt{3} - \frac{8\pi}{3}\)
E: \(16 - 4\sqrt{3} - \frac{8\pi}{3}\)
Attachment:
square.png [ 9.08 KiB | Viewed 10845 times ]
11 Sep 2014, 03:50
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Took sometime, but learned some new concepts. There is always alternate ways of solution one problem.
Step 1: lets name the point of intersection of arc AC and arc BD as O. Connecting BO, we will get sector AOB with AB,OB as radius and arc AO. Similiarly connecting OC, we will get sector sector COD with CD ,OC as radius and arc DO.
Step 2. OB = OC = BC = radius of the both the circle = side of square ABCD. BOC will form a equilateral triangle.
Step 3: Area of the shaded area = Area of Square - (Area of sector AOB + Area of triangle BOC + Area of sector COD)
Step 4: Calculate area of sector AOB. angle AOB = 30° (Since BOC is equilateral triangle, all angles will be 60°).
Area of sector will be (30/360) of area of circle. = (30/360)Πr² = (1/12)Π×4² = 4Π/3
Step 5: Area of equilateral triangle BOC with side r = (1/2)*(r√3/2)*r = 4√3
Step 6. Area of sector COD will be equal to area of Sector AOB (same radius, same angle).
Step 7. Area of shaded area = 16 - (4Π/3 + 4√3 + 4Π/3 ) = 16 - 8Π/3 - 4√3
Step 1: lets name the point of intersection of arc AC and arc BD as O. Connecting BO, we will get sector AOB with AB,OB as radius and arc AO. Similiarly connecting OC, we will get sector sector COD with CD ,OC as radius and arc DO.
Step 2. OB = OC = BC = radius of the both the circle = side of square ABCD. BOC will form a equilateral triangle.
Step 3: Area of the shaded area = Area of Square - (Area of sector AOB + Area of triangle BOC + Area of sector COD)
Step 4: Calculate area of sector AOB. angle AOB = 30° (Since BOC is equilateral triangle, all angles will be 60°).
Area of sector will be (30/360) of area of circle. = (30/360)Πr² = (1/12)Π×4² = 4Π/3
Step 5: Area of equilateral triangle BOC with side r = (1/2)*(r√3/2)*r = 4√3
Step 6. Area of sector COD will be equal to area of Sector AOB (same radius, same angle).
Step 7. Area of shaded area = 16 - (4Π/3 + 4√3 + 4Π/3 ) = 16 - 8Π/3 - 4√3
General Discussion
10 Sep 2014, 23:03
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PareshGmat
Make the complete diagram to understand the relevance of the point where the two arcs intersect.
Attachment:
Ques4.jpg [ 20.02 KiB | Viewed 9752 times ]
The area you are looking for is enclosed between arc AC, arc BC and line BA. ABC is an equilateral triangle. So angle ABC is 60
Shaded region \(= (1/6)*\pi*r^2 + ((1/6)*\pi*r^2 - \sqrt{3}r^2/4)\)
