Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circles is
A. √r
B. √2 r AB
C. √3 r
D. √3/2 r
Answers
Given :Two equal circles of radius r intersect such that each passes through the centre of the other.
To find : The length of the common chord of the circle.
Solution :
Let O and O' be the centre of two circles and OA and O'A be radius of the two circles.
AB be the common chord of both the circles
Draw OM ⊥ AB and, O'M ⊥ AB.
∆AOO' is an equilateral triangle.
AM is the Altitude of ∆AOO'.
Since, the Height of equilateral ∆ is √3/2 × side.
∴ Height of equilateral ∆AOO' , AM = √3/2 × r
AB = 2 AM
AB = 2 × √3/2 × r
AB = √3r
Hence, the length of the common chord of the circle is √3r.
Among the given options option (C) √3r is correct.
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Answer:
Step-by-step explanation:
Let O and O' be the centre of two circles and OA and O'A be radius of the two circles.
AB be the common chord of both the circles
Draw OM ⊥ AB and, O'M ⊥ AB.
∆AOO' is an equilateral triangle.
AM is the Altitude of ∆AOO'.
Since, the Height of equilateral ∆ is √3/2 × side.
∴ Height of equilateral ∆AOO' , AM = √3/2 × r
AB = 2 AM
AB = 2 × √3/2 × r
AB = √3r