Question

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

Answer 1

SOLUTION :  

(i) GIVEN  AB || CD, OA = 4 , OC = 4x-2, OB = x+1, OD = 2x+4

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

OC/OA = OD/OB

(4x–2)/4 = (2x+4)/(x+1)

4(2x + 4) = (4x – 2)(x +1)

8x + 16 = x(4x – 2) + 1(4x – 2)

8x + 16 = 4x² – 2x + 4x – 2

- 4x² + 8x + 16 + 2 – 2x = 0

- 4x² + 6x + 8 = 0

4x² – 6x – 18 = 0

4x² – 12x + 6x – 18 = 0

[By middle term splitting ]

4x(x – 3) + 6(x – 3) = 0

(4x + 6) (x – 3) = 0

x = – 6/4 or x = 3

[sides can't be negative]

Hence,the value of x = 3 .

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

tri(OAB) & tri(OCD) are similar triangles
thus sides are proportional
OA/OB = OC/OD
4/(x+1 )=(4x-2)/(2x+4)
4/(x+1) = (2x-1)/(x+2)
4(x+2)= (2x-1)(x+1)
4x+8 = 2x^2+x-1
2x^2-3x-9=0
x=3 or -3/2
-ve not possible