Question

The sides of a quadrilateral 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.

Answer 1

Let $AB=26m,BC=27m,CD=7m,DA=24m$

Now, in $\Delta ADC,$

$A{C}^2=A{D}^2+C{D}^2$

$A{C}^2=(24{)}^2+(7{)}^2=576+49=625$

$AC=25m$

Perimeter of $\Delta ABC=2s=AB+BC+CA$

$2s=26m+27m+25m$

$s=39m$

$\therefore$ Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{39(39-26)(39-27)(39-25)}$

$=\sqrt{39\times 13\times 12\times 14}$

$=\sqrt{13\times 3\times 12\times 2\times 2\times 3\times 2\times 7}$

$=13\times 2\times 3\times \sqrt{14}$

$=291.84m^2$

Now, area of $\Delta ADC=\frac{1}{2}\times \text{Base}\times \text{Height}$

$=\frac{1}{2}\times 7\times 24=84m^2$

$\therefore$ Area of field $ABCD$

$=\text{Area of}\Delta ABC+\text{Area of}\Delta ADC$

$=291.84m^2+84m^2$

$=375.8m^2$

Hence, area of rectangular field $ABCD=375.8m^2.$