Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the sides of the triangle.
The perimeter of triangle $(2 \mathrm{~s})=\mathrm{a}+\mathrm{b}+\mathrm{c}$
Let $\mathrm{a}=4$ and $2 \mathrm{~s}=10$ i.e. $\mathrm{s}=5$
$\therefore 10=4+b+c \Rightarrow b=6-c$
Now, area of triangle $=\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{5(5-4)(5-6+c)(5-c)}$
$\Delta=\sqrt{5(1)(\mathrm{c}-1)(5-\mathrm{c})} \Rightarrow \Delta^{2}=5(\mathrm{c}-1)(5-\mathrm{c})$
$=5\left(5 c-c^{2}-5+c\right)=5\left(-c^{2}+6 c-5\right)$
Let $f(c)=5\left(-c^{2}+6 c-5\right) \Rightarrow f^{\prime}(c)=5(-2 c+6) \Rightarrow f^{\prime \prime}(c)=5(-2)=-10$
For extreme value, $f^{\prime}(c)=0$ i.e. $5(-2 c+6)=0 \Rightarrow c=3$
Thus $f^{\prime \prime}(3)=-10 < 0 \Rightarrow f(c)$ has maximum value at $c=3$
$\therefore \mathrm{b}=6-3=3$ i.e. the lengths of the remaining sides are $3 \mathrm{~cm}$ and $3 \mathrm{~cm}$.