Question

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

Answer 1

Since, the edges of a triangular board are a = 6cm, b = 8cm and c=10cm.
Now, semi – perimeter of a triangular board,
$s=\frac{a+b+c}{2}$
$=\frac{6+8+10}{2}=\frac{24}{2}=12cm$
Now, area of a triangular board $=\sqrt{s(s-a)(a-b)(s-c)}$ [by Heron’s formula]
$=\sqrt{12(12-6)(12-8)(12-10)}$
$=\sqrt{12\times 6\times 4\times 2}$
$=\sqrt{(12{)}^2\times (2{)}^2}$
$=12\times 2=24c{m}^2$
Given that, the cost of painting an area of 1 cm² = Rs.0.09
$\therefore$ The cost of painting an area of 24 $c{m}^2$ = 0.09 $\times$ 24 = Rs 2.16
Hence, the cost of painting the triangular board at the rate of 9 paise per $c{m}^2$ is Rs 2.16​​​​​​​.