Question

In adjoining figure,ABCD Is a trapezium in which CD parallel to AB and its diagonals intersect at 0.if AO=5X-7cm,OC=2X+1cm,DO=7X-5cm and OB=7X+1cm,find the value of x.

Answer 1

Here we have been given,

ABCD Is a trapezium in which CD parallel to AB and its diagonals intersect at 0. AO=5x-7cm,OC=2x+1cm,DO= 7x-5cm and OB=7x+1cm.  

We know that the diagonals of a Trapezium Divide each other proportionally.

So, $\frac{AO}{OC} = \frac{DO}{OB}$

⇒ $\frac{5x - 7}{2x + 1} = \frac{7x - 5}{7x + 1}$

⇒ $35x^{2} - 49x + 5x - 7 = 14x^{2} + 7x - 10x -5$

⇒ $21x^{2} - 41x - 2 = 0$

⇒ $21x^{2} - 42x + x -2 = 0$

⇒ $21x(x-2) + 1(x-2) = 0$

⇒ $(21x + 1) (x-2) = 0$

⇒ $x - 2 = 0  or 21x + 1 = 0$

⇒ $x = 2  or x = \frac{-1}{21}$

$x = \frac{-1}{21}  is Rejected$

Therefore $x = 2$.

Answer 2

Here we have been given,

ABCD Is a trapezium in which CD parallel to AB and its diagonals intersect at 0. AO=5x-7cm,OC=2x+1cm,DO= 7x-5cm and OB=7x+1cm.

We know that the diagonals of a Trapezium Divide each other proportionally.

So, \frac{AO}{OC} = \frac{DO}{OB}

OC

AO

=

OB

DO

⇒ \frac{5x - 7}{2x + 1} = \frac{7x - 5}{7x + 1}

2x+1

5x−7

=

7x+1

7x−5

⇒ 35x^{2} - 49x + 5x - 7 = 14x^{2} + 7x - 10x -535x

2

−49x+5x−7=14x

2

+7x−10x−5

⇒ 21x^{2} - 41x - 2 = 021x

2

−41x−2=0

⇒ 21x^{2} - 42x + x -2 = 021x

2

−42x+x−2=0

⇒ 21x(x-2) + 1(x-2) = 021x(x−2)+1(x−2)=0

⇒ (21x + 1) (x-2) = 0(21x+1)(x−2)=0

⇒

⇒

Therefore x = 2x=2 .