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EXERCISE 12.4Two circles of radi 5 cm and 3 cm intersect at two points and the distance betweentheir centres is 4 em. Find the length of the common chord. |
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Answer» Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O' Hence OA = OB = 5 cm O'A = O'B = 3 cm OO' is the perpendicular bisector of chord AB. Therefore, AC = BC Given OO' = 4 cm Let OC = x Hence O'C = 4 − x In right angled ΔOAC, by Pythagoras theorem OA^2 = OC^2 + AC^2 ⇒ 5^2 = x^2 + AC^2 ⇒ AC^2 = 25 − x^2.......(1) In right angled ΔO'AC, by Pythagoras theorem O'A^2 = AC^2 + O'C^2 ⇒ 3^2 = AC^2 + (4 – x)^2 ⇒ 9 = AC^2 + 16 + x^2 − 8x ⇒ AC^2 = 8x − x^2 − 7.......(2) From (1) and (2), we get 25 − x^2 = 8x − x^2 − 7 8x = 32 Therefore, x = 4 Hence the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle. AC^2 = 25 − x^2 = 25 − 42 = 25 − 16 = 9 Therefore, AC = 3 m Length of the common chord, AB = 2AC = 6 m |
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