The edges of a triangular board are 6cm, 8cm and 10cm. The cost of painting it at the rate of 9 paise per ${\text{cm}}^2$ is

$b=8$ cm

$c=10$ cm

$s=\frac{a+b+c}{2}=\frac{6+8+10}{2}=\frac{24}{2}$

Area $=\sqrt{s\left(s-a\right)\times \left(s-b\right)\times \left(s\times -c\right)}$

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

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

$=24{\text{ cm}}^2$

Cost of painting it at the rate of 9 paise $=24\times 9=216$ paise$=\frac{216}{100}$ rupees

$=\text{Rs}\mathrm{.2.16}$