# Q1.    A park, in the shape of a quadrilateral ABCD, has $\angle C = 90\degree$, $AB = 9 m$, $BC = 12 m$, $CD = 5 m$ and $AD = 8 m$. How much area does it occupy?

D Divya Prakash Singh

From the figure:

We have joined the BD to form two triangles so that the calculation of the area will be easy.

In triangle BCD, by Pythagoras theorem

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

$\Rightarrow BD^2 = 12^2+5^2 = 144+25 = 169$

$\Rightarrow BD = 13\ cm$

The area of triangle BCD can be calculated by,

$Area_{(BCD)} = \frac{1}{2}\times BC\times DC = \frac{1}{2}\times 12\times 5 = 30\ cm^2$

and the area of the triangle DAB can be calculated by Heron's Formula:

So, the semi-perimeter of the triangle DAB,

$s = \frac{a+b+c}{2} = \frac{9+8+13}{2} = \frac{30}{2} = 15\ cm$

Therefore, the area will be:

$A = \sqrt{s(s-a)(s-b)(s-c)}$

where, $a = 9 cm,\ b = 8cm\ and\ c = 13cm$.

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

$= \sqrt{12(6)(7)(2)} = \sqrt{1260} = 35.5\ cm^2$  (Approximately)

Then, the total park area will be:

$= Area\ of\ triangle\ BCD + Area\ of\ triangle\ DAB$

$\Rightarrow$ $Total\ area\ of\ Park = 30+35.35 = 65.5\ cm^2$

Hence, the total area of the park is $65.5\ cm^2.$

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