The sides of a quadrilateral ABCD are tangents to the circle with centre ‘O’. If AB = 8 cm and CD = 5 cm find AD + BC.

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Solution

We know that the lengths of tangents to any circle drawn from an external point are equal.

Here, AB = 8 cm, CD = 5 cm, AS = AP, BP = BQ, CR = CQ and DR = DS

Using this information, we get:

AD + BC = (AS + SD) + (BQ + CQ)

⇒ AD + BC = AP + RD + BP + CR (AS = AP, SD = RD, BP = BQ and CQ = CR)

⇒ AD + BC = (AP + BP) + (RD + CR)

⇒ AD + BC = AB + CD

⇒ AD + BC = 8 cm + 5 cm = 13 cm

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