Question 59

Chords AB and CD of a circle, when produced, meet at a point P out side the circle. If AB = 6 cm, CD = 3 cm and PD = 5 cm, then PB is equal to:

Solution

Given that,

AB = 6 cm, CD = 3 cm and PD = 5 cm $$PB=?$$

We k now that intersecting secant theorem says that, when two secant lines AB and CD intersect outside the circle at a point P,

then $$PA\times PB =PC\times PD$$ -- (1)

PA = PB+AB and PC = PD+CD
(1) ==> (PB+AB)(PB) = (PD+CD)(PD)
$$PB\times(PB+6) = (5+3)\times3$$
$$PB^2+6PB = 8\times3$$
$$PB^2+6PB-24=0$$
$$PB = \dfrac{-6\pm\sqrt{36+96}}{2}$$

$$PB = \dfrac{11.48-6}{2}$$ or $$PB = \dfrac{-6-11.48}{2}$$
$$PB = 2.744 cm$$ or $$PB = -8.74$$
Length cannot be negative.
Therefore, Length of PB = 2.744 cm.


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