Perpendicular From Centre of Circle Bisects the Chord and Vice Versa
Perpendicular From Centre of Circle Bisects the Chord and Vice Versa
Theorem 1 The perpendicular from the centre of a circle to a chord bisects the chord.
Given A chord AB of a circle C (O,r) and perpendicular OD to the chord AB. TO PROVE AD = DB CONSTRUCTION Join OA and OB. PROOF In triangles AOD and BOD, we have and, So, by RHS - criterion of congruence, we have
Hence Proved |  |
Theorem 2: A line drawn from the centre of a circle to the mid point of the chord is perpendicular to the chord
Given A chord AB of a circle C (O,r) and a line OD which bisects the chord AB. TO PROVE OD is perpendicular to AB CONSTRUCTION Join OA and OB. PROOF In triangles AOD and BOD, we have and, So, by SSS - criterion of congruence, we have
But they are linear pairs so Hence Proved |  |
Illustration: If the length of a chord of a circle is 24 cm and is at a distance of 9 cm from the centre of the circle, Find the radius of the circle . The length of a chord of a circle AB = 24 cm Perpendicular distance from the centre OD = 9 cm As the perpendicular bisects the chord AD = 12 cm OA = 15 cm
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A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is | |||
| Right Option : D | |||
| View Explanation | |||
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| Right Option : D | |||||
| View Explanation | |||||
If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is | |||
| Right Option : C | |||
| View Explanation | |||
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