Question

In trapezium $\square PQRS$, seg $PQ\parallel$ seg $SR$, diagonal $PR$ and diagonal $QS$ intersects each other at point $O.$ If $PO=x+4;QO=x+2;RO=x+10$ and $SO=x+7$, then $PQ:RS=$?

• A. $3:2$
• B. $2:3$
• C. $4:9$
• D. $9:4$

Answer 1

The correct option is A $3:2$

In trapezium $\square PQRS$, seg $PQ\parallel$ seg $SR$

$PO=x+4;QO=x+2;RO=x+10$ and $SO=x+7$

The given trapezium is isosceles trapezium

$⟹\frac{PO}{OR}=\frac{QO}{OS}$

$⟹\frac{x+4}{x+10}=\frac{x+2}{x+7}$

$⟹x^2+11x+28=x^2+12x+20$

$⟹x=8$

In $△OPQ$ and $△OSR$,

$\angle SOR\cong \angle POQ$,

$\angle OSR\cong \angle OQP$ ........... opposite angles are congruent

$\angle ORS\cong \angle OPQ$ ........... opposite angles are congruent

$⟹△POQ\cong △SOR$

$⟹\frac{PQ}{RS}=\frac{OP}{OR}=\frac{x+4}{x+10}$

$=\frac{8+4}{8+10}=\frac{12}{18}=\frac{2}{3}$

$\therefore \frac{PQ}{RS}=\frac{2}{3}$

Hence, option B is correct.