In the given figure, PQR is a triangle and quadrilateral ABCD is inscribed in it, QD = 2 cm, QC = 5 cm, CR = 3 cm, BR = 4 cm, PB = 6 cm, PA = 5 cm and AD = 3 cm. What is the area (in cm²) of the quadrilateral ABCD?

Solution(By Examveda Team)

$$\eqalign{ & {\text{Area}}\left( {\Delta \,1} \right) = \frac{{30}}{{10 \times 10}} = \frac{3}{{10}} = \frac{{24}}{{80}} \cr & {\text{Area}}\left( {\Delta \,2} \right) = \frac{{10}}{{10 \times 8}} = \frac{{10}}{{80}} \cr & {\text{Area}}\left( {\Delta \,3} \right) = \frac{{12}}{{8 \times 10}} = \frac{{12}}{{80}} \cr & {\text{Area}}\left( {ABCD} \right) = 80 - \left( {24 + 10 + 12} \right) = 34 \cr} $$

$$\eqalign{ & P{M^2} = {10^2} - {4^2} = 84 \cr & PM = \sqrt {84} \cr & {\text{Area}}\left( {PQR} \right) = \frac{1}{2} \times 8 \times \sqrt {84} = 4\sqrt {84} \cr & {\text{Area of quadrilateral }}ABCD \cr & = \frac{{34}}{{80}} \times {\text{ar}}\left( {\Delta PQR} \right) \cr & = \frac{{34}}{{80}} \times 4\sqrt {84} \cr & = \frac{{17\sqrt {21} }}{5}\,{\text{c}}{{\text{m}}^2} \cr} $$