Let $A B C D$ be a trapezium with AD parallel to BC. Assume there is a point $M$ is interior of the segment $\mathrm{BC}$ such that $\mathrm{AB}=\mathrm{Am}$ and $\mathrm{DC}=\mathrm{DM}$. Then the ratio of the area of the trapezium to the area of triangle AMD is.

• A

  2
• Correct answer
  B

  3
• C

  4
• D

  not determinable from the data

Solution

Required $=\frac{\frac{1}{2}[(a+c)+2(a+c)] b}{\frac{1}{2}(a+c) b}=3$

Hence, the correct answer is option: 'B'.