A park, in the shape of a quadrilateral ABCD, has $\mathrm{?}C$ = ${90}^{?}$ , AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy?

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

= $\sqrt{15×6×2×7}$
= $\sqrt{3×5×3×2×2×7}$
= $3×2\sqrt{5×7}{m}^{2}$
= $6\sqrt{35}{m}^{2}$ = 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)

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

= Area of $\mathrm{?}ABD$ + Area of $\mathrm{?}BCD$
Area of quadrilateral ABCD = $35.4{m}^{2}+30{m}^{2}=65.4{m}^{2}$