Question

The perimeter of a triangle is 9m² $-$ 2n + 8 and its two sides are 4m² + 3n and 7m² + 5n $-$ 12. Find the third side of the triangle.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{be}4m^2+3n. \\ \mathrm{Second}\mathrm{side}\mathrm{of}\mathrm{triangle}=7m^2+5n-12 \\ \mathrm{Third}\mathrm{side}=p\left(x\right) \\ \mathrm{Perimeter}=9m^2-2n+8 \\ \mathrm{Now},4m^2+3n+7m^2+5n-12+p\left(x\right)=9m^2-2n+8 \\ \Rightarrow p\left(x\right)=9m^2-2n+8-\left(4m^2+3n+7m^2+5n-12\right) \\ \Rightarrow p\left(x\right)=9m^2-2n+8-4m^2-3n-7m^2-5n+12 \\ \Rightarrow p\left(x\right)=9m^2-4m^2-7m^2-2n-3n-5n+8+12 \\ \Rightarrow p\left(x\right)=-2m^2-10n+20 \\ \therefore \mathrm{Third}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{triangle}=-2m^2-10n+20 \end{gathered}$$