Question

3. Chords AB and CD of the circle intersect outside the circle in point P. PA = 9, PB = 4,
PC= 10 then PD = ?
(a) 3.6
(b) 4.0
(c) 5.0 (d) 6.5​

Answer 1

Answer:

PA.PB=PC.PD

9x4=10xPD

3.6=PD

Answer 2

Step-by-step explanation:

Answer is 36 because

In △PAC and △PDB,

∠BAC=180

∘

−∠PAC (linear pairs)

∠PDB=∠CDB=180

∘

−∠BAC

=180

∘

−(180

∘

−∠PAC)=∠PAC)

∠PAC=∠PDB

∠APC=∠BPD ...[Common]

∴ By AA-criterion of similarity,

△PAC∼△DPB

(ii) △PAC∼△DPB

So, sides are proportional

PD

PA

=

PB

PC

⇒PA.PB=PC.PD

and then 9×4=10×PD

36=10×3.6

36=36