Answer :

In the quadrilateral ABCD,
we have, right angled triangle $\mathrm{?}BCD$,



We have,
${BD}^2={BC}^2+{CD}^2$
(By Pythagoras theorem)

= ${12}^2+5^2=144+25=169$
$?$ ${BD}^2={13}^2$
$?$ BD = 13m

Also, In $\mathrm{?}ABD$, we have,

AB = 9m, BD = 13m, DA = 8m

Now, we know that,
$s=\frac{a+b+c}{2}$
$?$ $s=\frac{9+13+8}{2}=\frac{30}{2}=15m$

Now, Area of triangle
= $\sqrt{15(15?9)(15?13)(15?8)}$
(Since, Heron's formula [area = $\sqrt{s(s?a)(s?b)(s?c)}$])

= $\sqrt{15\times 6\times 2\times 7}$
= $\sqrt{3\times 5\times 3\times 2\times 2\times 7}$
= $3\times 2\sqrt{5\times 7}{m}^2$
= $6\sqrt{35}{m}^2$ = 6 × 5.9$m^2$ = 35.4$m^2$(approx) ...(i)

Since, $\mathrm{?}BCD$ is an right angled triangle,

Area of $\mathrm{?}BCD$ = $\frac{1}{2}$ × BC × CD
(Since, Area of triangle = $\frac{1}{2}$ × Base × Height)

= $\frac{1}{2}\times 12\times 5=30m^2$ ...(ii)

Area of quadrilateral ABCD
= Area of $\mathrm{?}ABD$ + Area of $\mathrm{?}BCD$

Hence from (i) and (ii),

Area of quadrilateral ABCD = $35.4m^2+30m^2=65.4m^2$