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In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.

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#### Solution

Given: PA and PB are the tangents to the circle.

PA = 12 cm

QC = QD = 3 cm

To find: PC + PD

PA = PB = 12 cm (The lengths of tangents drawn from an external point to a circle are equal)

Similarly, QC = AC = 3 cm

and QD = BD = 3 cm.

Now, PC = PA − AC = 12 − 3 = 9 cm

Similarly, PD = PB − BD = 12 − 3 = 9 cm

Hence, PC + PD = 9 + 9 = 18 cm.

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