Practicing Success
In a circle with centre O, AB is a chord of length 10 cm. Tangents at points A and B intersect outside the circle at P. If OP = 2 OA, then find the length (in cm) of AP. |
12.5 10 12 15 |
10 |
Let the radius of the circle be R cm. So, OA = OB = R then, OP = 2 x OA = 2R Tangent segments AP, BP and radii OA, OB from a kite So, OP bisects the chord AB. Thn, AQ = BQ = 5 cm Now, \(\Delta \)OAP & \(\Delta \)OBP are two right angled triangles. From \(\Delta \)OAP, cos \(\angle\)AOP = \(\frac{OA}{OP}\) = cos \(\angle\)AOP = \(\frac{R}{2R}\) = cos \(\angle\)AOP = \(\frac{1}{2}\) = \(\angle\)AOP = cos 60 So, \(\angle\)OPA = 180 - (90 + 60) = 30 Also, \(\Delta \)AQO & \(\Delta \)BQO are two right-angled triangles. So, \(\angle\)OAQ = 180 - 90 - 60 = 30 From \(\Delta \)AQO, cos \(\angle\)QAO = \(\frac{AQ}{OA}\) = cos 30 = \(\frac{5}{OA}\) = \(\sqrt {3 }\)/2 = \(\frac{5}{OA}\) = OA = 10\(\sqrt {3 }\) So, AP = (10\(\sqrt {3 }\)) x \(\sqrt {3 }\) = 10 cm Therefore, AP is 10 cm. |