question_answer

In the given figure, ABCD is a field in the form of a quadrilateral whose sides are indicated. If \[\angle DAB=90{}^\circ .\]The area of field is:

A)  \[306\,{{m}^{2}}\]             

B)  \[307\,{{m}^{2}}\]

C)  \[308\,{{m}^{2}}\]                         

D)  \[309\,{{m}^{2}}\]

Correct Answer: A

Solution :

    Draw the perpendicular CM on AB from C. From the figure, \[BM=AB-AM=40-28=12m\] Area of the field = area of rectangle AMCD + area of right angled triangle MBC \[=28\times 9+\frac{1}{2}\times 12\times 9=306\,{{m}^{2}}.\]