• # question_answer ABCD is a trapezium in which $AB||CD,$ if $AB=7\,cm,\,\,DC=3\,cm$ cm and the area of the trapezium is 18 cm2, then the area of ABCD is A)  $8\frac{2}{5}\text{c}{{\text{m}}^{\text{2}}}$ B)  $7\frac{2}{5}\,{{\operatorname{cm}}^{2}}$ C)  $5\frac{2}{5}\,{{\operatorname{cm}}^{2}}$ D)  $6\frac{2}{5}\,{{\operatorname{cm}}^{2}}$

Area of trapezium $=\frac{1}{2}\,(7+3)\times h$ $\Rightarrow$ $18=\frac{1}{2}\times 10\times h$ $h=\frac{18}{5}\,cm$ $\therefore$ Area of $\Delta BCD=\frac{1}{2}\times 3\times \frac{18}{5}=5\frac{2}{5}\,c{{m}^{2}}$