Answer

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Hint: To solve the question, we have to calculate the area of the given triangular board to calculate the portion of area to be painted. The obtained when is multiplied with the rate of cost of painting, will give you the total cost of painting the triangular board.

Complete step-by-step answer:

We know that the portion of area to be painted is equal to the area of the triangular board.

To calculate the area of the triangular, we use the Heron's formula of area of a triangle which is equal to \[\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]

Where a, b, c are three sides of the triangle and s is the semi-perimeter of the triangle.

We know that the formula for semi-perimeter of triangle with sides a, b, c is given by \[s=\dfrac{a+b+c}{2}\]

The given sides of the triangular board are 13 metres, 14 metres, 15metres. By substituting the values of sides of triangle in the above mentioned formula of semi-perimeter, we get

\[s=\dfrac{13+14+15}{2}=\dfrac{42}{2}\]

\[\Rightarrow s=21\]

By substituting the values of semi-perimeter and the sides of triangle in the above mentioned formula of area of a triangular board, we get

\[A=\sqrt{21\left( 21-13 \right)\left( 21-14 \right)\left( 21-15 \right)}\]

Where A represents the area of a triangular board.

\[A=\sqrt{21\left( 8 \right)\left( 7 \right)\left( 6 \right)}\]

By writing the values in terms of prime numbers we get

\[A=\sqrt{\left( 7\times 3 \right)\left( 2\times 2\times 2 \right)\left( 7 \right)\left( 3\times 2 \right)}\]

By rearranging the values we get

\[A=\sqrt{\left( 7\times 7 \right)\left( 2\times 2\times 2\times 2 \right)\left( 3\times 3 \right)}\]

\[A=\sqrt{{{7}^{2}}\times {{2}^{4}}\times {{3}^{2}}}\]

\[\begin{align}

& A=\sqrt{{{\left( 7\times {{2}^{2}}\times 3 \right)}^{2}}} \\

& =7\times {{2}^{2}}\times 3 \\

& =21\times 4 \\

& =48{{m}^{2}} \\

\end{align}\]

Thus, the area of the triangular board of sides 13 metres, 14 metres, 15metres is equal to 48\[{{m}^{2}}\]

The cost of painting the triangular board at the rate of Rs. 8.75 per\[{{m}^{2}}\] = 8.75 \[\times \] (area of the triangular board of given sides).

The cost of painting the triangular board at the rate of Rs. 8.75 per\[{{m}^{2}}\] = \[8.75\times 48\] = Rs. 420.

Note: The possibility of mistake can be not able to analyse the need to calculate the area of the triangled board to calculate the total cost of painting of the triangular board of given sides. The alternative way to calculate the area of triangle of sides a, b, c is to use the formula \[\dfrac{1}{2}ab\sin C\] where \[\sin C\] can be calculated by using\[\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab},\cos {{C}^{2}}+\sin {{C}^{2}}=1\].

Complete step-by-step answer:

We know that the portion of area to be painted is equal to the area of the triangular board.

To calculate the area of the triangular, we use the Heron's formula of area of a triangle which is equal to \[\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]

Where a, b, c are three sides of the triangle and s is the semi-perimeter of the triangle.

We know that the formula for semi-perimeter of triangle with sides a, b, c is given by \[s=\dfrac{a+b+c}{2}\]

The given sides of the triangular board are 13 metres, 14 metres, 15metres. By substituting the values of sides of triangle in the above mentioned formula of semi-perimeter, we get

\[s=\dfrac{13+14+15}{2}=\dfrac{42}{2}\]

\[\Rightarrow s=21\]

By substituting the values of semi-perimeter and the sides of triangle in the above mentioned formula of area of a triangular board, we get

\[A=\sqrt{21\left( 21-13 \right)\left( 21-14 \right)\left( 21-15 \right)}\]

Where A represents the area of a triangular board.

\[A=\sqrt{21\left( 8 \right)\left( 7 \right)\left( 6 \right)}\]

By writing the values in terms of prime numbers we get

\[A=\sqrt{\left( 7\times 3 \right)\left( 2\times 2\times 2 \right)\left( 7 \right)\left( 3\times 2 \right)}\]

By rearranging the values we get

\[A=\sqrt{\left( 7\times 7 \right)\left( 2\times 2\times 2\times 2 \right)\left( 3\times 3 \right)}\]

\[A=\sqrt{{{7}^{2}}\times {{2}^{4}}\times {{3}^{2}}}\]

\[\begin{align}

& A=\sqrt{{{\left( 7\times {{2}^{2}}\times 3 \right)}^{2}}} \\

& =7\times {{2}^{2}}\times 3 \\

& =21\times 4 \\

& =48{{m}^{2}} \\

\end{align}\]

Thus, the area of the triangular board of sides 13 metres, 14 metres, 15metres is equal to 48\[{{m}^{2}}\]

The cost of painting the triangular board at the rate of Rs. 8.75 per\[{{m}^{2}}\] = 8.75 \[\times \] (area of the triangular board of given sides).

The cost of painting the triangular board at the rate of Rs. 8.75 per\[{{m}^{2}}\] = \[8.75\times 48\] = Rs. 420.

Note: The possibility of mistake can be not able to analyse the need to calculate the area of the triangled board to calculate the total cost of painting of the triangular board of given sides. The alternative way to calculate the area of triangle of sides a, b, c is to use the formula \[\dfrac{1}{2}ab\sin C\] where \[\sin C\] can be calculated by using\[\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab},\cos {{C}^{2}}+\sin {{C}^{2}}=1\].

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