Let t be the area of a regular pentagon with each side equal to 1 ....

\text { Area of regular pentagon }=\frac{a^2 n}{4 \tan \left(\frac{180}{n}\right)} $n \Rightarrow$ No. of sides $\mathrm{a} \Rightarrow$ Length of side $\therefore$ For regular pentagon by side length 1 , $$\begin{aligned} \text { Area }(t) & =\frac{5}{4 \tan 36^{\circ}} \\ & =\frac{5}{4(0.73)} \\ & =1.71 \end{aligned}$$ Now, $\mathrm{P}(1.71)=0$ to be found with least degree and integer coefficient soon that gcd of all coefficient is 1 . Let $\quad \mathrm{x}=1.71$ $100 x=171$ $\therefore \mathrm{P}(\mathrm{x})=100 \mathrm{x}-171=0$ is the polynomial which satisfied all the conditions. $$\begin{aligned} & \therefore \mathrm{m}=100+171=271 \\ & \therefore \sqrt{\mathrm{~m}}=16.46 \\ & \therefore \text { Nearest integer }=16 \end{aligned}$$ But this question can have multiple solutions as student can take \tan 36 as $0.72,0.726$ or even 0.7. Every time we will get different answers. So this question should be Bonus.