The edges of a triangular board are 6 cm,

(b) Since,the edges of a triangular board are $a=6 \mathrm{~cm} b=8 \mathrm{~cm}$ and $c=10 \mathrm{~cm}$.

Now, semi-perimeter of a triangular board,

$s=\frac{a+b+c}{2}$

$=\frac{6+8+10}{2}=\frac{24}{2}=12 \mathrm{~cm}$

Now, area of a triangular board $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]

$=\sqrt{12(12-6)(12-8)(12-10)}$

$=\sqrt{12 \times 6 \times 4 \times 2}$

$=\sqrt{(12)^{2} \times(2)^{2}}$

$=12 \times 2=24 \mathrm{~cm}^{2}$

Since, the cost of painting for area $1 \mathrm{~cm}^{2}=₹ 0.09$

$\therefore$ Cost of paint for area $24 \mathrm{~cm}^{2}=0.09 \times 24=₹ 2.16$

Hence, the cost of a triangular board is $₹ 2.16$.