Chords AB and CD of a circle intersect externally at P. If AB = 7cm, CD = 1cm and PD = 5cm, then 50% of the length of PA(in cm) is:
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Correct answer: A
Explanation
The correct answer is option 1 i.e. 5
Chords AB and CD of a circle intersect externally at P.
AB = 7 cm, CD = 1 cm and PD = 5 cm
If two chords of a circle intersect externally, then the product of the lengths of the segments are equal.
Let, PB = x cm
PB × PA = PD × PC
x(x + 7) = 5(5 + 1)
x2 + 7x = 30
x2 + 7x - 30 = 0
x2 + 10x - 3x - 30 = 0
x(x + 10) - 3(x + 10) = 0
(x + 10)(x - 3) = 0
x + 10 = 0
x = - 10 [Not possible]
x - 3 = 0
x = 3
PA = x + 7 = 10
50% of PA = 10(50/100) = 5 cm