Math, asked by ashmeetbrar13dec, 3 months ago

in a trapezium ABCD, AB||DC . its diagonals AC and BD are joined to intersect at O. if AO=(x+5) cm , BO=(x-3) cm CO=(x+3) cm , then find the values of x

Answers

Answered by pekka9
2

Answer:

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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