Question

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:

• A. $Rs.2.00$
• B. $Rs.2.16$
• C. $Rs.2.48$
• D. $Rs.3.00$

Answer 1

The correct option is B

$Rs.2.16$

Explanation for the correct option:

Find the area of the triangular board

To find the area of the triangle, we will use Heron's formula.

Heron's formula can be given that: $A=\sqrt{s(s-a)(s-b)(s-c)}$

Where $s=\frac{a+b+c}{2}$ and $a,b,c$ are the sides of the triangle.

So the semi-perimeter of the triangle will be,

$$\begin{gathered} s=\frac{6+8+10}{2} \\ \Rightarrow s=\frac{24}{2} \\ \Rightarrow s=12cm \end{gathered}$$

Now, the area of the triangular board will be,

$$\begin{gathered} A=\sqrt{12(12-6)(12-8)(12-10)} \\ \Rightarrow A=\sqrt{12\left(6\right)\left(4\right)\left(2\right)} \\ \Rightarrow A=24c{m}^2 \end{gathered}$$

Find the cost of painting

Since it is given that the rate of cost of painting is $9$ paise per $c{m}^2$.

So the cost of painting will be,

$$\begin{gathered} =24\times 9paise \\ =216paise \\ =Rs.\frac{216}{100} \\ =Rs.2.16 \end{gathered}$$

Hence, the correct option is (b) $Rs.2.16$.