Question

ABCD is a trapezium such that AB || DC and the diagonals of ABCD intersect at point O. If OD = 5 cm, OB = 8 cm, AC = 26 cm, then the difference between the lengths of OA and OC is equal to

• A. 3 cm
• B. 6 cm
• C. 8 cm
• D. 12 cm

Answer 1

The correct option is B 6 cm

Given that ABCD is a trapezium such that AB || DC.



We know that the diagonals of trapezium divides proportionally.
$\Rightarrow \frac{AO}{OC}=\frac{OB}{OD}$
$\Rightarrow \frac{AO}{OC}+1=\frac{OB}{OD}+1$
$\Rightarrow \frac{AO}{OC}=\frac{OB+OD}{OD}$
$\Rightarrow \frac{26}{OC}=\frac{8+5}{5}$ $(\therefore OB=8cm,OD=5cm,AC=26cm)$
$\Rightarrow OC=\frac{26}{13}\times 5$
$\Rightarrow OC=10cm$
Now, $\frac{AO}{OC}=\frac{OB}{OD}$
$\Rightarrow \frac{AO}{10}=\frac{8}{5}$
$\Rightarrow AO=16cm$
$\therefore |AO-OC|=16-10=6cm$

Hence, the correct answer is option (2).