Question

In the given fig., AB || DC and diagonals AC and BD intersects at O. If OA = 3x–1 and OB = 2x + 1, OC = 5x – 3 and OD = 6x – 5, find x.

Answer 1

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.

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

GIVEN:-

OA=3X-1            (as we know that in a trapezium diagonals are proportional)

OB=2X+1                OA/OC=OB/OD

OC=5X-3

OD=6X-5    

HENCE,

(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.

Hence, the value of  is x = 2.