InterviewSolution
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5. Three girls Reshma, Salma and Mandip areplaying a game by standing on a circle of radius5m drawn in a park. Reshma throws a ball toSalma, Salma to Mandip, Mandip to Reshma. Ifthe distance between Reshma and Salma andbetween Salma and Mandip is 6m each, what isthe distance between Reshma and Mandip? |
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Answer» Let O bethe centre of the circle. A, B and C represent the positions of Reshma, Salmaand Mandip. AB = 6cm and BC = 6cm. Radius OA = 5cm.(given) Draw BM ⊥ AC ABC is an isosceles triangle as AB = BC, M ismid-point of AC. BM isperpendicular bisector of AC and thus it passes through the centre of thecircle. Let AM = y and OM = x then BM = (5-x). In ΔOAM, OA²=OM²+AM² ( by Pythagorastheorem) 5²=x²+y²—(i) In ΔAMB, AB²=BM²+AM²(by Pythagoras theorem) 6²= (5-x)²+y²— (ii) Subtracting (i) from (ii), 36 – 25 = (5-x)² -x² 11 = (25+x²– 2×5×x) - x² 11= 25+x²-10x - x² 11= 25-10x 10x = 14 x= 7/5 Substituting the value of x in (i), we get y²+ 49/25 = 25y²=25 – 49/25y²=(625 – 49)/25y²=576/25 y = 24/5 Thus, AC = 2×AM AC = 2×y AC= 2×(24/5) m = 48/5 m = 9.6 m Hence, the Distance between Reshma and Mandipis 9.6 m. = thanxx but kuch zyada bada method hai |
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