In the adjoining figure, chords AB and CD intersect at P.
If AB = 16 cm, PD = 8 cm, PC = 6 and AP >PB, then AP =

Given: AB = 16 cm, PD = 8 cm, and PC = 6 cm
We know that two chords AB and CD intersect at P inside the circle with centre at O.
Then PA× PB = PC× PD.
Let PA = x then PB = AB – AP = 16 – x
⇒ x × (16–x) = 6× 8
⇒ 16x – x2 =48
⇒ x2 – (12+4)x + 48 = 0
⇒ (x–12)(x –4) =0
⇒ x = 4 and 12
x = 4 is not possible because AP >PB
⇒ AP = 12 cm
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