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) \[6\frac{2}{5}\,{{\operatorname{cm}}^{2}}\]
Correct Answer: C
Solution :
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 \[\Delta BCD=\frac{1}{2}\times 3\times \frac{18}{5}=5\frac{2}{5}\,c{{m}^{2}}\]
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