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Standard VIII
Maths
Question

In the adjoining figure AB||DC and diagonal AC & BD intersect at point O. If AO=(3x−1)cm, OB=(2x+1)cm,OC=(5x−3)cm and OD=(6x−5)cm then find the value of x.
1526823_5fb7d90e4a1e453eb1a5155da3a09376.png

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Solution
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GIVEN:
OA= 3x-1 , OC= 5x-3, OD= 6x-5 , BO = 2x+1
AO/OC = BO/OD

[The diagonals of a Trapezium divide each other proportionally]

(3x-1)/(5x-3) =( 2x+1)/(6x-5)
(3x-1) (6x-5) = (5x-3) ( 2x+1)
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 -3
18x² -10x² -21x + x +5+3 =0
8x² - 20x +8= 0
4(2x² -5x +2) = 0
2x² -5x +2 = 0
2x² -4x -x +2= 0
[By factorization]
2x(x-2) -1(x-2)= 0
(2x-1) (x-2) = 0
(2x-1) = 0 or (x-2) = 0

x = ½ or x = 2

If we put x= ½ in OD , The value of OD is negative.

Hence, the value of is x = 2.

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1526823_5fb7d90e4a1e453eb1a5155da3a09376.png
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