Question

Question 1
A park, in the shape of a quadrilateral ABCD, has $\angle C={90}^{\circ },\text{AB}=9m,\text{BC}=12m,\text{CD}=5m\text{and AD}=8m$. How much area does it occupy?

Answer 1

Let us join BD.

In $\Delta BCD$, applying Pythagoras theorem,
${\text{BD}}^2={\text{BC}}^2+{\text{CD}}^2=(12{)}^2+(5{)}^2=144+25{\text{BD}}^2=169\text{BD}=\pm \sqrt{169}=\pm 13m=13m\left(sidecannotbenegative\right)\text{Area of}\Delta \text{BCD}=\frac{1}{2}\times \text{BC}\times \text{CD}=(\frac{1}{2}\times 12\times 5)m^2=30m^2\text{For}\Delta \text{ABD},\text{s =}\frac{\text{Perimeter}}{2}=\frac{(9+8+13)}{2}=15m\text{By Heron's formula}\text{Area of triangle =}\sqrt{s(s-a)(s-b)(s-c)}\text{Area of}\Delta \text{ABD}=\sqrt{s(s-a)(s-b)(s-c)}=\lfloor \sqrt{15(15-9)(15-8)(15-13)}\rfloor m^2=\left(\sqrt{15\times 6\times 7\times 2}\right)m^2=6\sqrt{35}{m}^2=(6\times 5.916)m^2=35.496m^2\text{Area of the park = Area of}\Delta \text{ABD}+\text{Area of}\Delta \text{BCD}=(35.496+30)m^2=65.496m^2=65.5m^2\text{(approximately)}$