In the given figure, PA and PB are tangents to the circle at A and B respectively and the chord BC is parallel to tangent PA. If AC = 6 cm, and length of the tangent AP is 9 cm, then what is the length of the chord BC?
Answer / Ranjan Kumar Singh
In this problem, we have angles ACP, BCP, and ACB as right angles because they are inscribed in the circle and the tangents PA and PB intersect at the center. Since AC = 6 cm, angle APB is a straight angle (180°). Using these facts, we can construct a triangle ABC with side lengths AB = AP + PB, BC = AC + AP, and angle BAC = 90°. Now, we can apply the law of cosines to triangle ABC: AB² = AP² + PB² - 2 * AP * PB * cos(90°), which simplifies to AB² = AP² + PB². Given AP = 9 cm and AC = 6 cm, we can find PB using the Pythagorean theorem on triangle APC: PB² = AP² - AC², so PB² = 81 - 36 = 45. Thus, AB² = 9² + 45 = 121, and AB = sqrt(121) ≈ 10.96 cm (approximately). Since BC is parallel to AP, angle ABC is a right angle, so triangle ABC is a right-angled triangle with hypotenuse AB and legs AC and BC. Using the Pythagorean theorem again: BC² = AB² - AC² = 121 - 36 = 85, and BC ≈ 9.21 cm (approximately).
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