Two circles of radii10 cm and 8 cm intersect and the length of the common chord is 12 cm. The distance between their centres is

6 cm

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8 cm

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3 cm

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5 cm

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Solution

The correct option is A
6 cm

OP BC
Therefore, P is the mid-point of BC. (since, perpendicular form the centre bisects the chord)
‘O" is the mid-point of AB. (since, ‘O" is the centre)
Therefore, in DABC, from mid-point theorem, OP = CA
Þ AC = 2OP
= 2 × 3 cm
= 6 cm

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Two circles of radii10 cm and 8 cm intersect and the length of the common chord is 12 cm. The distance between their centres is

The correct option is A 6 cmOP BC Therefore, P is the mid-point of BC. (since, perpendicular form the centre bisects the chord) ‘O" is the mid-point of AB. (since, ‘O" is the centre) Therefore, in DABC, from mid-point theorem, OP = CA Þ AC = 2OP = 2 × 3 cm = 6 cm

The correct option is A
6 cm

OP BC
Therefore, P is the mid-point of BC. (since, perpendicular form the centre bisects the chord)
‘O" is the mid-point of AB. (since, ‘O" is the centre)
Therefore, in DABC, from mid-point theorem, OP = CA
Þ AC = 2OP
= 2 × 3 cm
= 6 cm