ABCD is a trapezium in which AB $$\parallel$$ DC and its diagonals intersect at P. If AP = (3x-1) cm, PC = (5x-3) cm, BP = (2x+1) cm and PD = (6x-5) cm, then the length of DB is:

Solution

Given that,

ABCD is a trapezium, in which AB $$\parallel$$ DC and its diagonals intersect at P.

AP = (3x-1) cm, PC = (5x-3) cm, BP = (2x+1) cm and PD= (6x-5) cm

We know that the diagonals of the trapezium is always intersect in the same ratio.

So, $$\dfrac{AP}{PC}=\dfrac{PB}{PD}$$

Now, substituting the values,

$$\dfrac{3x-1}{5x-3}={2x+1}{6x-5}$$

$$(3x-1)(6x-5)=(2x+1)(5x-3)$$

$$18x^2-15x-6x+5=10x^2-6x+5x-3$$

$$18x^2-21x+5=10x^2-x-3$$

$$18x^2-10x^2-21x+x+5+3$$

$$8x^2-20x+8=0$$

$$2x^2-5x+2=0$$

$$2x^2-4x-x+2=0$$

$$2x(x-2)-1(x-2)=0$$

$$(x-2)(x-1)=0$$

x=1 or x=2

So, length of DB= BP+PD=2x+1+6x-5=8x-4

Hence DB=4 cm or 12 cm