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

Given:

A park, in 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 quadrilateral $ABCD$,

$AB = 9\ m, BC = 12\ m, CD = 5\ m$ and $DA = 8\ m, \angle C = 90^o$

Join $BD$.

In right angled triangle $BCD$,

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

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

$= 144 + 25$

$= 169$

$= (13)^2$

$\Rightarrow BD = 13\ m$

Area of triangle $\mathrm{BCD}=\frac{1}{2} base \times altitude$

$=\frac{1}{2} \times \mathrm{BC} \times \mathrm{CD}$

$=\frac{1}{2} \times 12 \times 5$

$=30 \mathrm{~m}^{2}$

In triangle $\mathrm{ABD}$,

$s=\frac{a+b+c}{2}$

$=\frac{9+13+8}{2}$

$=\frac{30}{2}$

$=15$

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

$=\sqrt{15(15-9)(15-13)(15-8)}$

$=\sqrt{15 \times 6 \times 2 \times 7}$

$=\sqrt{180 \times 7}$

$=\sqrt{36 \times 5 \times 7}$

$=6 \sqrt{35}$

$=6(5.92)$

$=35.52 \mathrm{~m}^{2}$

Total area of quadrilateral $\mathrm{ABCD}=30+35.52$

$=65.52 \mathrm{~m}^{2}$.

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Updated on: 10-Oct-2022

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