Question

ABCD is a trapezium such that BC || AD and AD = 4 cm. If the diagonals AC and BD intersect at O such that $\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{DO}}{\mathrm{OB}}=\frac{1}{2}$, then BC =

(a) 7 cm
(b) 8 cm
(c) 9 cm
(d) 6 cm

Answer 1

Given: ABCD is a trapezium in which BC||AD and AD = 4 cm

The diagonals AC and BD intersect at O such that

To find: DC

In ΔAOD and ΔCOB

$$\begin{gathered} \angle \mathrm{OAD}=\angle \mathrm{OCB}\left(\mathrm{Alternate}\mathrm{angles}\right) \\ \angle \mathrm{ODA}=\angle \mathrm{OBC}\left(\mathrm{Alternate}\mathrm{angles}\right) \\ \angle \mathrm{AOD}=\angle \mathrm{BOC}\left(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right) \\ \mathrm{So},∆\mathrm{AOD}~∆\mathrm{COB}\left(\mathrm{AAA}\mathrm{similarity}\right) \\ \mathrm{Now},\mathrm{correponding}\mathrm{sides}\mathrm{of}\mathrm{similar}∆\text{'}\mathrm{s}\mathrm{are}\mathrm{proportional}. \\ \frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{AD}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{AD}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{2}=\frac{4}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=8\mathrm{cm} \end{gathered}$$

Hence the correct answer is