Chords AB and CD of a circle intersect externally at P. If AB = 7 cm, CD = 1 cm and PD = 5 cm, then 50% of the length of PA (in cm) is:

5

Let PB = a cm

PB x PA = PD x PC

⇒ a (a + 7) = 5(5 + 1)

⇒ \( { a}^{2 } \) + 7a = 30

⇒ \( { a}^{2 } \) + 7a - 30 = 0

⇒ \( { a}^{2 } \) + 10a - 3a - 30 = 0

⇒ a(a + 10)(a - 3) = 0

⇒ a + 10 = 0

⇒ a = - 10 (not possible)

Therefore a - 3 = 0 , a = 3

PB = 3 cm

As, PA = PB + BA

PA = 3 + 7 = 10 cm

50% of PB = \(\frac{10}{2}\) = 5 cm.