Question

The sides of a quadrangular field, taken in order are $26m,27m,7m$ and $24m$ respectively. The angle contained by the last two sides is a right angle. Find its area.
[Take $\sqrt{14}=3.751$]

Answer 1

Area of $\Delta ADC=\frac{1}{2}(AD\times DC)$
$=\frac{1}{2}\times 24\times 7m^2=84m^2$
In $\Delta ADC,$ we have
$A{C}^2=A{D}^2+C{D}^2$
$\Rightarrow A{C}^2={24}^2+7^2$
$\Rightarrow A{C}^2={25}^2$
$\Rightarrow AC=25m$
Thus, in $\Delta ABC,$ we have
$a=BC=27m,b=CA=25m$ and $c=AB=26m$
Let $2s$ be the perimeter of $\Delta ABC$. Then,
$2s=a+b+c$
$2s=27+25+26=78$
$\Rightarrow s=39m$
$\therefore$ Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{39\times 12\times 14\times 13}$
$\Rightarrow$ Area of $\Delta ABC$
$=\sqrt{13\times 3\times 3\times 4\times 2\times 7\times 13}$
$=78\sqrt{14}c{m}^2=292.57m^2$
Area of quadrialteral $ABCD=84+292.57$
$=376.57m^2$.