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Example 8 A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected? Let P be the pole Gates A & B are diametrically opposite So, AB = Diameter of circle = 13 m Also given that, Difference of the distance of the pole from the two gates is 7 metres BP – AP or AP – BP is 7 m Let us take AP – BP = 7 i.e. AP = BP + 7 Let BP = x So, AP = BP + 7 = x + 7 Now, Since AB is a diameter ∠ APB = 90° Now, Δ APB is a right angle triangle Using Pythagoras theorem Hypotenuse2 = Height2 + Base2 BP2 + AP2 = AB2 x2 + (x + 7)2 = 132 x2 + x2 + 49 + 14x = 169 2x2 + 14x + 49 – 169 = 0 2x2 + 14x – 120 = 0 Divide the equation by 2 2𝑥2/2+14𝑥/2−120/2 = 0 x2 + 7x – 60 = 0 Comparing equation with ax2 + bx + c = 0 Here, a = 1, b = 7, c = – 60 We know that D = b2 – 4ac D = (7)2 – 4×(1)×(−60) = 49 – 4×(−60) = 49 + 240 = 289 Hence , roots to equation are x = (−𝑏 ± √𝐷)/2𝑎 Putting values x = (− 7 ± √289)/(2 × 1) x = (− 𝟕 ± 𝟏𝟕 )/𝟐 Solving So, x = 5, & x = –12 x = (−7 + 17)/2 x = 10/2 x = 5 x = (−7 −17)/2 x = (−24)/2 x = –12 Since x is distance, it must be positive Hence, x = 5. Hence, BP = x = 5 m & AP = x + 7 = 5 + 7 = 12 m

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo