Question

The sides of quadrangular field taken in order are $26,27,7,24m$ respectively. The angle contained by the last $2$ sides is a right angle. Find its area.

Answer 1

The sides of a quadrilateral field taken order as

$AB=26m$

$BC=27m$

$CD=7m$

and $DA=24m$

Diagonal AC is joined

Now $△ADC$

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

$AC=\sqrt{{24}^2+7^2}$

$AC=\sqrt{625}$

$AC=25m$

Now area of $△ABC$

$S=\frac{1}{2}(AB+BC+CA)$

$=\frac{1}{2}(26+27+25)$

$=\frac{78}{2}$

$=39m$

By using heron's formula

Area of $△ABC=\sqrt{S(S-AB)(S-BC)(S-CA)}$

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

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

$=\sqrt{85179}$

$=291.85m^2$

Now area of $△ADC$

$S=\frac{1}{2}(AD+CD+AC)$

$=\frac{1}{2}(25+24+7)$

$=\frac{56}{2}$

$=28m$

By using heron's formula

Area of $△ADC=\sqrt{S(S-AD)(S-DC)(S-CA)}$

$=\sqrt{28(28-24)(28-7)(28-25)}$

$=\sqrt{28\times 4\times 21\times 3}$

$=\sqrt{7056}$

$=84m^2$

hence, total area is$375.85$ $m^2$