Question

ABCD is a given field whose sides are indicated. If $\angle$DAB = 90${}^o$, then area of the field is ............

• A. 180 m${}^2$
• B. 126 m${}^2$
• C. 306 m${}^2$
• D. 300 m${}^2$

Answer 1

Answer: (C)

306 m${}^2$

IN $△DBC$ $\angle DAB={90}^0$
Then $D{B}^2=(40{)}^2+(9{)}^2=1600+81=1681$
OR DB=41 m
Area of $△DBC=\sqrt{42(42-41)(42-28)(42-15)}=\sqrt{15876}=126$ $m^2$
Area of$△ADB=\frac{1}{2}(40\times 9)=\frac{360}{2}=180$$m^2$
Area of ABCD=$△DBC+△ADB$
Area of ABCD=$126+180=306m^2$