question_answer

In the given figure, ABCD is a field in the form of a quadrilateral whose sides are indicated. If\[304c{{m}^{2}}\]. The area of field is:

A)  \[550c{{m}^{2}}\]           

B)  \[154c{{m}^{2}}\]

C)  \[4746{{m}^{2}}\]            

D)  \[4748{{m}^{2}}\]            

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 \[=\left( \frac{1}{2}\times BD\times AL \right)+\left( \frac{1}{2}\times BD\times CM \right)\]