Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
The readings in a field book are: It is subsequently realised that the distances to $C$ and $D$ had been interchanged by mistake. The are of the actual field 1300 sq. m
(b)1500 sq. m
(c)1800 sq. m
(d)2000 sq. m
Answer
Answer (as printed): A
Explanation
Interchanging the distances to C and D, the field may be drawn as shown in the adjoining figure. We have : $A B=40 \mathrm{m}, A F=30 \mathrm{m}, A G=20 \mathrm{m}, A H=10$ $\mathrm{m}, C H=30 \mathrm{m}, D G=20 \mathrm{m}, E F=30 \mathrm{m}$. Area of the field $=\operatorname{ar}(\triangle A H C)+\operatorname{ar}($ rect $C E F H)+\operatorname{ar}(\triangle B F E)$ $+\operatorname{ar}(\Delta B G D)+\operatorname{ar}(\Delta A G D)$ $$\begin{aligned} & \begin{array}{r} =\left(\frac{1}{2} \times A H \times C H\right)+(C H \times F H)+\left(\frac{1}{2} \times B F \times E F\right) \\ +\left(\frac{1}{2} \times B G \times D G\right)+\left(\frac{1}{2} \times A G \times D G\right) \end{array} \\ & =\left(\frac{1}{2} \times 10 \times 30\right)+(30 \times 20)+\left(\frac{1}{2} \times 10 \times 30\right)+\left(\frac{1}{2} \times 20 \times 20\right) \\ & +\left(\frac{1}{2} \times 20 \times 20\right) \\ & =(150+600+150+200+200) \mathrm{m}^{2}=1300 \mathrm{m}^{2} . \end{aligned}$$