In Fig. if AP = 10 cm, then BP =A. \(\sqrt{109}\,cm\)B. \(\sqrt{127

1.	In Fig. if AP = 10 cm, then BP =A. \(\sqrt{109}\,cm\)B. \(\sqrt{127}\,cm\)C . \(\sqrt{119}\,cm\)D. \(\sqrt{109}\,cm\)
Answer» Answer is B. \(\sqrt{127}\,cm\) Given: AP = 10 cm OA = 6 cm OB = 3 cm Property : The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. By above property, ∆PAO is right-angled at ∠PAO (i.e., ∠PAO = 90°) and ∆PBO is right-angled at ∠PBO (i.e., ∠PBO = 90°). Therefore by Pythagoras theorem in ∆PAO, OP² = OA² + AP² ⇒ OP² = 6² + 10² ⇒ OP² = 36 + 100 ⇒ OP = √136 Now by Pythagoras theorem in ∆PBO, OP² = OB² + BP² BP² = OP² – OB² ⇒ BP² = (√136) ² – 3² ⇒ BP² = 136 – 9 ⇒ BP= √127 Hence, BP= √127 cm

In Fig. if AP = 10 cm, then BP =A. \(\sqrt{109}\,cm\)B. \(\sqrt{127}\,cm\)C . \(\sqrt{119}\,cm\)D. \(\sqrt{109}\,cm\)

Answer»

Answer is B. \(\sqrt{127}\,cm\)

Given:

AP = 10 cm

OA = 6 cm

OB = 3 cm

Property : The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, ∆PAO is right-angled at ∠PAO (i.e., ∠PAO = 90°) and ∆PBO is right-angled at ∠PBO (i.e., ∠PBO = 90°).

Therefore by Pythagoras theorem in ∆PAO,

OP² = OA² + AP²

⇒ OP² = 6² + 10²

⇒ OP² = 36 + 100

⇒ OP = √136

Now by Pythagoras theorem in ∆PBO,

OP² = OB² + BP²

BP² = OP² – OB²

⇒ BP² = (√136) ² – 3²

⇒ BP² = 136 – 9

⇒ BP= √127

Hence, BP= √127 cm

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