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A and C lie on a circle center O with radius `5sqrt2`. The point B inside the circle is such that angle `/_`ABC = 90, AB = 6, BC = 2. Find `sqrt(OB^2 -1)` |
| Answer» A and C lie on a circle with radius `5*sqrt2` as radius.There is a point B inside the circle making `/_ABC = 90^circ`We need to find, `sqrt(OB^2 - 1)`Let center of circle be O.`:. OA = 5*sqrt2`Given,AB = 6BC = 2`AC^2 = AB^2 + BC^2``= 36 + 4 = 40 rArr AC = 2*sqrt10`Let us draw a perpendicular from O on line AB cutting it at D and equal to length y (say).BD = x (say)Now we will draw lines parallel to OD and BD in order to form a rectangle OEBD. OB will be the diagonal of this rectangle.In `/_ OEC``OC^2 = OE^2 + EC^2``rArr 50 = x^2 + (2+y)^2``rArr 50 = x^2 + 4 + 4y + y^2``rArr 46 = x^2 + y^2 + 4y` .....(1)In `/_ OAD``OA^2 = AD^2 + OD^2``rArr 50 = y^2 + (6-x)^2``rArr 50 = x^2 + y^2 - 12y + 36``rArr 14 = x^2 + y^2 - 12x` ......(2)Solving (1) & (2)`x = 1 , y = 5``OB^2 = x^2 + y^2 = 26`Hence, `sqrt( OB^2 - 1) = sqrt(25) = 5` | |