Question

A park, in shape of a quadrilateral ABCD has $\angle C={90}^0$, AB= 9m, BC= 12m, CD= 5m and AD = 8m. How much area does it occupy?

Answer 1

In quadrilateral ABCD,
AB= 9m, BC = 12m, CD = 5m
and DA = 8m, $\angle C={90}^0$
Join BD,
Now in right $\Delta BCD$,
$B{D}^2=B{C}^2+C{D}^2=(12{)}^2+(5{)}^2$
= $144+25=169=(13{)}^2$
$\therefore BD=13m$
Now Area of $\Delta BCD=\frac{1}{2}\times 12\times 5=30m^2$
and in $\Delta ABD$,
$s=\frac{a+b+c}{2}=\frac{9+13+8}{2}=\frac{30}{2}=15$
$\therefore$ Area of $\Delta ABD=\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{15(15-9)(15-13)(15-8)}$
= $\sqrt{15\times 6\times 2\times 7}=\sqrt{180\times 7}{m}^2$
= $\sqrt{36\times 5\times 7}=6\sqrt{35}{m}^2=6\left(5.92\right)$
= $35.52m^2$
$\therefore$ Total area of quad. ABCD = 30+35.52 $m^2$
= $65.52m^2$