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07 Jan 2015, 07:38
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E
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Points A,B,C lie on the circle. Point O is the center of the circle and lies on the line segment AB. The area of triangle BOC is 8 square inches. If angle COB is 90 degrees then the diameter of the circle equals to
A. 1 inch
B. 2 inches
C. 4 inches
D. 6 inches
E. 8 inches
Attachment:
2015-01-07_1837.png [ 3.46 KiB | Viewed 4946 times ]
07 Jan 2015, 11:28
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Bunuel
Tough and Tricky questions: Geometry.
Points A,B,C lie on the circle. Point O is the center of the circle and lies on the line segment AB. The area of triangle BOC is 8 square inches. If angle COB is 90 degrees then the diameter of the circle equals to
A. 1 inch
B. 2 inches
C. 4 inches
D. 6 inches
E. 8 inches
Kudos for a correct solution.
Attachment:
2015-01-07_1837.png
1/2 b * h = 8 => b*h = 16
Since b=h as both are radius of the circle
b^2 = 16
b = 4 = r
Diameter d= 2r = 8
Ans E
08 Jan 2015, 00:41
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Answer = E. 8 inches
2015-01-07_1837.png [ 3.85 KiB | Viewed 4544 times ]
\(\triangle\) COB is an isosceles right angle triangle with CO = OB = radius of the circle
Let diameter = d
\(Area = \frac{1}{2} (\frac{d}{2})^2 = 8\)
\(\frac{d}{2} = 4\)
d = 8
Attachment:
2015-01-07_1837.png [ 3.85 KiB | Viewed 4544 times ]
\(\triangle\) COB is an isosceles right angle triangle with CO = OB = radius of the circle
Let diameter = d
\(Area = \frac{1}{2} (\frac{d}{2})^2 = 8\)
\(\frac{d}{2} = 4\)
d = 8
