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:
Answer & explanation
Correct answer: option 1
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.