Question

(i) In Fig. (a), if AB $\parallel$ CD, find the value of x.

(ii) In Fig. (b), if AB $\parallel$ CD, find the value of x.

(iii) If Fig. (c), AB $\parallel$ CD. if OA = 3x - 19, OB = x - 4, OC = x - 3 and OD = 4,find x.

Answer 1

$\left(1\right)$ For $Fig.a$,

$AB\parallel CD$ [ Given ]

$\therefore$ Quadrilateral $ABCD$ is a trapezium.

$\therefore$ $\frac{AO}{CO}=\frac{BO}{DO}$ [ Diagonals of trapezium divides each other proportionally ]

$\Rightarrow$ $\frac{4}{4x-2}=\frac{x+1}{2x+4}$

$\Rightarrow$ $4(2x+4)=(x+1)(4x-2)$

$\Rightarrow$ $8x+16=4x^2-2x+4x-2$

$\Rightarrow$ $4x^2-6x-18=0$

$\Rightarrow$ $2x^2-3x-9=0$

$\Rightarrow$ $(x-3)(2x+3)=0$

$\therefore$ $x=3$ or $x=\frac{-3}{2}$

$\left(2\right)$ For $Fig.\left(b\right)$,

$AB\parallel CD$ [ Given ]

$\therefore$ Quadrilateral $ABCD$ is a trapezium.

$\therefore$ $\frac{AO}{CO}=\frac{BO}{DO}$ [ Diagonals of trapezium divides each other proportionally ]

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

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

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

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

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

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

$\therefore$ $x=2$ or $x=\frac{1}{2}$

$\left(3\right)$ For $Fig.\left(c\right)$,

$AB\parallel CD$ [ Given ]

$\therefore$ Quadrilateral $ABCD$ is a trapezium.

$\therefore$ $\frac{AO}{CO}=\frac{BO}{DO}$ [ Diagonals of trapezium divides each other proportionally ]

$\Rightarrow$ $\frac{3x-19}{x-3}=\frac{x-4}{4}$

$\Rightarrow$ $(3x-19)\left(4\right)=(x-4)(x-3)$

$\Rightarrow$ $12x-76=x^2-3x-4x+12$

$\Rightarrow$ $x^2-19x+88=0$

$\Rightarrow$ $(x-8)(x-11)=0$

$\therefore$ $x=8$ or $x=11$