Question

(ii) In the given figure, if AB || CD, find the value of x.

Answer 1

SOLUTION :

(ii)  GIVEN : AB || CD, OA = 3x-1 , OC = 5x-3, OB = 2x+1, OD = 6x-5

Since, the diagonals of a trapezium divide each other proportionally.

Now, DO/OB = CO/OA

(6x–5)/(2x+1) = (5x–3)/(3x–1)

(6x – 5)(3x – 1) = (2x + 1)(5x – 3)

3x(6x – 5) – 1(6x – 5) = 2x(5x – 3) + 1(5x – 3)

18x² -15x - 6x + 5 = 10x² - 6x +5x -3

18x² -21x + 5 = 10x² - x +5x -3

18x² - 10x² – 21x +x + 5 +3 = 0

18x² - 10x² – 20x + 8 = 0

8x² – 20x + 8 = 0

4(2x² - 5x +2)= 0

2x² - 5x +2 = 0

2x² - 4x -1x +2 = 0

[By middle term splitting]

2x(x - 2) - 1(x - 2) = 0

(2x – 1)(x – 2) = 0

(2x – 1)= 0  or (x – 2) = 0

x = 1/2   or  x = 2

If we put x = ½  ,then  the diagonals of a trapezium does not  divide each other proportionally.

Hence, the value of  x = 2

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Answer 2

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