Question

11. In the adjoining figure, ABCD is a trapezium in which
AB || DC. The diagonals AC and BD intersect at O. Prove

that
AO/OC = OD/BO

Using the above result, find the value(s) of x if
OA = 3x - 19, OB = x - 4, OC = x - 3 and OD = 4.
​

Answer 1

Answer:

=2

Step-by-step explanation:

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.