Question

3) PQRS is a Trapezium with PQ || SR and diagonals intersect at point O. Also OP = 6x – 5, OR = 2x + 1, OQ = 3x – 1, OS = 5x – 3. Determine the value of ‘x’. Diagram is essential.

Answer 1

In ΔPOQ and ΔROS,

∠POQ = ∠ROS..................[Vertically Opposite Angles]

∵PQ║SR and PR is the transversal,

∠QPO = ∠SRO.................[Alternate Angles]

∵PQ║SR and SQ is the transversal,

∠PQO = ∠RSO.................[Alternate Angles]

∴ ΔPOQ ≈ ΔROS.............[AAA Test]

∴   $\frac{OP}{OR} = \frac{OQ}{OS}$......................[CPST]

Put the values of OP, OR, OQ and OS

∴ We get,

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

      $(6x-5)(5x-3) = (3x-1)(2x+1)$

      $6x^{2} - 11x +4=0$

   ∴$(2x-1)(3x-4)=0$

Thus,

$x=1/2 OR x= 4/3$

∵ Ratio of lengths of any two sides can never be negative,

   $x = 4/3$