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

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Given:

A park, in the shape of a quadrilateral $ABCD$ has $\angle C=90^o$, $AB=9\ m$, $BC=12\ m$, $CD=5\ m$ and $AD=8\ m$.

To do:

We have to find the area it occupies.

Solution:

In $\triangle BCD$, using Pythagoras theorem,

$BD^2=BC^2+CD^2$

$BD^2=(12)^2+(5)^2$

$BD^2=144+25=169$

​$BD^2=(13)^2$

$BD=13\ m$

Therefore,

Area of $\triangle BCD=\frac{1}{2}\times12\times5=6\times5=30\ m^2$

In $\triangle ABD$,

$s=\frac{1}{2}(8+9+13)=15\ m$

Area of $\triangle ABD=\sqrt{s(s-a)(s-b)(s-c)}$

​Area of $\triangle ABD=\sqrt{15(15-8)(15-9)(15-13)}$

​Area of $\triangle ABD= \sqrt{15\times7\times6\times2}=6\sqrt{35}\ m^2$

Therefore,

Area of quadrilateral $ABCD=$ Area of $\triangle ABD+$ Area of $\triangle BCD$

Therefore,

Area of quadrilateral $ABCD=30+6\sqrt35=6(5+\sqrt{35})\ m^2$.

The area occupied by the park is $6(5+\sqrt{35})\ m^2$.

Updated on 10-Oct-2022 13:42:05