Given, $$\Delta$$ be the area of $$\triangle \mathrm{PQR}$$ of side length

$$\begin{gathered} a=2, b=\frac{7}{2} \text { and } c=\frac{5}{2} \\ \Rightarrow s=\frac{a+b+c}{2}=\frac{2+\frac{7}{2}+\frac{5}{2}}{2}=4 \end{gathered}$$

$$\begin{aligned} \text { Now, } & \frac{2 \sin \mathrm{P}-\sin 2 \mathrm{P}}{2 \sin \mathrm{P}+\sin 2 \mathrm{P}} \\ &=\frac{2 \sin \mathrm{P}-2 \sin \mathrm{P} \cdot \cos \mathrm{P}}{2 \sin \mathrm{P}+2 \sin \mathrm{P} \cdot \cos \mathrm{P}} \\ & \quad(\sin 2 \theta=2 \sin \theta \cdot \cos \theta) \end{aligned}$$

$$\begin{aligned} & =\frac{1-\cos \mathrm{P}}{1+\cos \mathrm{P}} \\ & =\frac{1-\left(1-2 \sin ^2 \frac{\mathrm{P}}{2}\right)}{1+\left(2 \cos ^2 \frac{\mathrm{P}}{2}-1\right)} \\ & \quad\left(\cos 2 \theta=2 \cos ^2 \theta-1=1-2 \sin ^2 \theta\right) \\ & =\tan ^2 \frac{\mathrm{P}}{2} \\ & =\left(\sqrt{\left.\frac{(s-b)(s-c)}{s(s-a)}\right)^2}\right. \\ & =\left(\frac{(s-b)(s-c)}{\sqrt{s(s-a)(s-b)(s-c)}}\right)^2 \end{aligned}$$

$$\begin{aligned} & =\left(\frac{(s-b)(s-c)}{\Delta}\right)^2 \\ & \therefore \quad(\Delta=\sqrt{s(s-a)(s-b)(s-c)}) \\ & =\frac{\left(4-\frac{7}{2}\right)^2\left(4-\frac{5}{2}\right)^2}{\Delta^2} \\ & =\frac{9}{16 \Delta^2} \\ & =\left(\frac{3}{4 \Delta}\right)^2 \end{aligned}$$