Calculate the length of PQ

1.	Calculate the length of PQ
Answer» We have AP = 9 cm, CQ = 6 cm and AB = 12 cm. For triangles △BAP & △BCQ, ∠BAP = ∠BCQ = 90°(\(\because\) Angle between radius and tangent at point of contact in 90°) ∠ABP = ∠CBQ (Vertically opposite angles) ∠APB = ∠BQC (sum of angles in a triangle in 180°) \(\therefore\) △ BAP \(\sim\) △BCQ (By AAA similarity criteria) \(\therefore\) AP/CQ = AB/BC ⇒ BC = \(\frac{AB\times CQ}{AP}\) = \(\frac{12\times6}9\) = 4 x 2 = 8 cm Now, in △ ABP, BP² = AP² + AB² = 9² + 12² = 81 + 144 = 225 = 15² \(\therefore\) BP = 15 cm Now, in △BCQ, BQ² = BC² + QC² = 8² + 6² = 64 + 36 = 100 = 10² \(\therefore\) BP = 10 cm Now, PQ = BP + BQ = 15 + 10 = 25 cm

Answer»

We have AP = 9 cm, CQ = 6 cm and AB = 12 cm.

For triangles △BAP & △BCQ,

∠BAP = ∠BCQ = 90°(\(\because\) Angle between radius and tangent at point of contact in 90°)

∠ABP = ∠CBQ (Vertically opposite angles)

∠APB = ∠BQC (sum of angles in a triangle in 180°)

\(\therefore\) △ BAP \(\sim\) △BCQ (By AAA similarity criteria)

\(\therefore\) AP/CQ = AB/BC

⇒ BC = \(\frac{AB\times CQ}{AP}\) = \(\frac{12\times6}9\)

= 4 x 2 = 8 cm

Now, in △ ABP,

BP² = AP² + AB²

= 9² + 12²

= 81 + 144

= 225

= 15²

\(\therefore\) BP = 15 cm

Now, in △BCQ,

BQ² = BC² + QC²

= 8² + 6²

= 64 + 36 = 100 = 10²

\(\therefore\) BP = 10 cm

Now, PQ = BP + BQ = 15 + 10

= 25 cm

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