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?

Answers (1)

From the figure:

quadrilateral 2

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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