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08 Dec 2018, 10:23
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C
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Difficulty:
65%
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Question Stats:
62% (02:26) correct
38%
(03:03)
wrong
based on 78
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Two concentric circles have their centers at point O such that a line segment AB having its end points on the outer circle touches the inner circle at point C. The length of the line segment AB is 2*\(\sqrt{2}\) times the radius of the inner circle. If an equilateral triangle is drawn such that the area of the triangle is equal to the ratio of the radii of the outer circle and the inner circle respectively, what is the length of the side of the triangle?
A)\(\sqrt{2}\)
B)\(\sqrt{3}\)
C)2
D)3
E)2*\(\sqrt{3}\)
08 Dec 2018, 10:50
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r = radius of inner circle, R = radius of outer circle
From Pythagoras theorem, R^2 = 2r^2 + r^2, giving R = r *sqrt(3)
Thus R/r = sqrt(3).
Area of equilateral triangle = sqrt(3) x side^2/4.
Therefore side length is 2
From Pythagoras theorem, R^2 = 2r^2 + r^2, giving R = r *sqrt(3)
Thus R/r = sqrt(3).
Area of equilateral triangle = sqrt(3) x side^2/4.
Therefore side length is 2
09 Dec 2018, 10:02
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Let the radius of bigger circle be R
Let the radius of smaller circle be r.
R^2 = r^2 + (√2 r)^2 = 3r^2
R= √3r
Also let side of equilateral triangle be a
So, area = √3/4 * a^2 = R/r = √3
a^2 = 4
a = 2
Answer C
Let the radius of smaller circle be r.
R^2 = r^2 + (√2 r)^2 = 3r^2
R= √3r
Also let side of equilateral triangle be a
So, area = √3/4 * a^2 = R/r = √3
a^2 = 4
a = 2
Answer C
