# The sides of the triangular board are 13 metres, 14 metres, 15metres. The cost of painting it at the rate of Rs. 8.75\[{{m}^{2}}\] per is?

a) Rs. 688.80

b) Rs. 735

c) Rs. 730.80

d) Rs. 722.50

Answer

Verified

360.3k+ views

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\[\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab},\cos {{C}^{2}}+\sin {{C}^{2}}=1\].

Last updated date: 22nd Sep 2023

•

Total views: 360.3k

•

Views today: 5.60k

Recently Updated Pages

What do you mean by public facilities

Please Write an Essay on Disaster Management

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE