Question

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

• A. Rs.200
• B. Rs.2.16
• C. Rs.2.48
• D. Rs.3.00

Answer 1

The correct option is B

Rs.2.16

Since, the edges of a triangular board are a = 6 cm, b = 8 cm and c = 10 cm.
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 $=\sqrt{s(s-a)(s-b)(s-c)}$
[by Heron’s formula]
$=\sqrt{12(12-6)(12-8)(12-10)}$
$=\sqrt{12\times 6\times 4\times 2}$
$=\sqrt{(12{)}^2\times (2{)}^2}$
$=12\times 2=24c{m}^2$
Since, the cost of painting for area 1 cm² = ₹ 0.09
$\therefore$ Cost of paint for area 24 $c{m}^2$ = 0.09 $\times$ 24 = ₹ 2.16
Hence, the cost of painting the triangular board is ₹ 2.16.