Question

5.The diagonals ofa trapezium ABCD with AB||DC intersectat O.OA=2x+4,

OB=4x–2, OC=x+1andOD=4units,findx
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Answer 1

Answer:

3

Step-by-step explanation:

Given

trapezium ABCD with AB||DC intersectat O.

OA=2x+4

OB=4x–2

OC=x+1

OD=4

[The diagonals of a Trapezium divide each other proportionally]

OA/OC = OB/OD

(2x+4)/(x+1) = (4x-2)/4

(2x+4)× 4 = (4x-2) × (x+1) (cross multiply)

8x +16 = 4x^2 -2x +4x - 2

= 4x^2 +2x -2 -8x -16

= 4x^2 -6x -18

=2 ( 2x^2 -3x -9) ( common and eliminate it)

=2x^2-6x +3x -9 ( factorisation)

=2x(x-3) +3(x-3)

=(2x+3)(x-3)

2x+3=0 x-3 =0

2x= -3 x=3

x= -3/2

x=3

HOPE IT HELPS

Answer 2

Step-by-step explanation:

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