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?

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?

Answer: Option C

Solution

Answer: Option C

Solution:

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

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