Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
The dimensions of the field shown in the given figure are $A C=150 \mathrm{m}, A H=120 \mathrm{m}$, $A G=80 \mathrm{m}, A F=50 \mathrm{m}$, $E F=30 \mathrm{m}, G B=50 \mathrm{m}$, $\mathrm{HD}=20 \mathrm{m}$ The area of this field is $6500 \mathrm{m}^{2}$
(b)$6550 \mathrm{m}^{2}$
(c)$6600 \mathrm{m}^{2}$
(d)$6650 \mathrm{m}^{2}$
Answer
Answer (as printed): B
Explanation
Area of the field $=\operatorname{ar}(\triangle A F E)+\operatorname{ar}(\triangle A G B)+\operatorname{ar}(\triangle B G C)$ + ar $(\triangle D H C)$ + or (trap DEFH) $$\begin{aligned} =\left(\frac{1}{2} \times A F \times E F\right)+ & \left(\frac{1}{2} \times A G \times B G\right)+\left(\frac{1}{2} \times C G \times B G\right) \\ & +\left(\frac{1}{2} \times D H \times C H\right)+\left\{\frac{1}{2} \times(D H+E F) \times H F\right\} \end{aligned}$$ $$\begin{aligned} {[\because C G} & =(A C-A G)=(150-80) \mathrm{m} \\ & =70 \mathrm{m}, C H=(A C-A H) \\ & =(150-120) \mathrm{m}=30 \mathrm{m}, \end{aligned}$$ $H F=(A H-A F)=(120-50) \mathrm{m}=70 \mathrm{m}]$ $$=(750+2000+1750+300+1750) \mathrm{m}^{2}=6550 \mathrm{m}^{2} .$$