# Q2.    Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

From the figure:

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

The area of the triangle ABC can be calculated by Heron's formula:

So, the semi-perimeter, where $a = 3 cm,\ b =4cm\ and\ c = 5cm.$

$s = \frac{a+b+c}{2} = \frac{3+4+5}{2} = 6cm$

Heron's Formula for calculating the area:

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

$= \sqrt{6(6-3)(6-4)(6-5)}= \sqrt{6(3)(2)(1)} = \sqrt{36} = 6\ cm^2$

And the sides of the triangle ACD are $a' =4cm,\ b' = 5cm\ and\ c' = 5cm.$

So, the semi-perimeter of the triangle:

$s' = \frac{a'+b'+c'}{2} = \frac{4+5+5}{2} = \frac{14}{2} = 7cm$

Therefore, the area will be given by, Heron's formula

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

$= \sqrt{7(7-4)(7-5)(7-5)}$

$= \sqrt{7(3)(2)(2)} = 2\sqrt{21} = 9.2\ cm^2\ \ \ \ (Approx.)$

Then, the total area of the quadrilateral will be:

$= Area\ of\ triangle\ ABC + Area\ of\ triangle\ ACD$

$\Rightarrow$ $Total\ area\ of\ quadrilateral\ ABCD = 6+9.2 = 15.2\ cm^2$

Hence, the area of the quadrilateral ABCD is $15.2 \ cm^2.$

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