Question

For given figures find the value of x , where the diagonals are in terms of x.
(i). In Fig. (a), $$AB\parallel CD,$$ find the value of x.
(ii). In Fig. (b), $$AB\parallel CD,$$ find the value of x.
(iii). In Fig. (c), $$AB\parallel CD,$$ if $OA=3x-19,OB=x-4,OC=x-3$ and $OD=4,$ find x.

Answer 1

Hint- A quadrilateral with one pair of parallel sides is a trapezium. In each part of this question, the given 4-sided figure is a trapezium. Use the property of trapezium, diagonals of trapezium divide each other proportionally. Simply find the value of x by simplifying the equations so obtained.

Complete step-by-step solution -
(i) In fig (a), Given $$AB\parallel CD,$$
So, ABCD is a trapezium.
Since, diagonals of trapezium divide each other proportionally that’s why AC and BD divide each other proportionally.
$\Rightarrow {\displaystyle \frac{AO}{CO}}={\displaystyle \frac{BO}{DO}}$
Now, substitute the values of AO, CO, BO and DO

$\Rightarrow {\displaystyle \frac{2x+4}{x+1}}={\displaystyle \frac{4x-2}{4}}$
On cross multiplying we get,
$\Rightarrow 4(2x+4)=(4x-2)(x+1)$
$\Rightarrow 8x+16=4x^2+2x-2$
$\Rightarrow 0=4x^2+2x-8x-2-16$
$\Rightarrow 0=4x^2-6x-18$
or, $4x^2-6x-18=0$ …………………………. (a)
$\Rightarrow (x-3)(2x+3)=0$
$\therefore$ $x=3$ or $x={\displaystyle \frac{-3}{2}}$
Hence, the value of x = 3 or $x={\displaystyle \frac{-3}{2}}$

(ii) In fig (b), Given $$AB\parallel CD,$$

So, ABCD is a trapezium.
Since, diagonals of trapezium divide each other proportionally that’s why AC and BD divide each other proportionally.
$\Rightarrow {\displaystyle \frac{AO}{CO}}={\displaystyle \frac{BO}{DO}}$
Now, substitute the values of AO, CO, BO and DO
$\Rightarrow {\displaystyle \frac{3x-1}{5x-3}}={\displaystyle \frac{2x+1}{6x-5}}$
On cross multiplying we get,
$\Rightarrow (6x-5)(3x-1)=(2x+1)(5x-3)$
$\Rightarrow 18x^2-21x+5=10x^2-x-3$
$\Rightarrow 18x^2-10x^2-21x+x+5+3=0$
$\Rightarrow 8x^2-20x+8=0$
or, $2x^2-5x+2=0$ ………………………….. (b)
$\Rightarrow (x-2)(2x-1)=0$
$\therefore$ $x=2$ or $x={\displaystyle \frac{1}{2}}$
Hence, the value of x = 2 or $x={\displaystyle \frac{1}{2}}$

(iii) In fig (c), Given $$AB\parallel CD,$$

So, ABCD is a trapezium.
Since, diagonals of trapezium divide each other proportionally that’s why AC and BD divide each other proportionally.
$\Rightarrow {\displaystyle \frac{AO}{CO}}={\displaystyle \frac{BO}{DO}}$
Now, substitute the values of AO, CO, BO and DO
$\Rightarrow {\displaystyle \frac{3x-19}{x-3}}={\displaystyle \frac{x-4}{4}}$
On cross multiplying we get,
$\Rightarrow 4\times (3x-19)=(x-4)(x-3)$
$\Rightarrow 12x-76=x^2-7x+12$
$\Rightarrow x^2-19x+88=0$
$\Rightarrow (x-8)(x-11)=0$ ……………………………….. (c)
$\therefore$ $x=8$ or $x=11$
Hence, the value of x = 8 or $x=11$

Note- In such types of questions, just keep in mind the basic proportionality of diagonal components in a trapezium and also solving the quadratic equation so obtained by using the splitting the middle term method, completing the square method or discriminant method.