Question

Find the cost of laying grass in a triangular field of sides \[50m,65m\]and \[65m\]at the rate of Rs 7 per \[{m^2}\].

Answer 1

Hint: We need to use the unitary method in this case. In the unitary method, if we know the price of a particular product we can find the price of the number of products by multiplication also we can find the price of a single product if we are given with price of the numbers of products with the division.

The area of the triangular field is found by using here’s formula i.e.
\[\sqrt {S(s - a)(s - b)(s - c)} \] where a, b, c are sides of the triangle
S is semi perimeter.
 Value of \[s = \dfrac{{a + b + c}}{2}\]
Therefore

Complete step by step answer:

Given sides of triangular grass field \[50n,65m\]and \[65m\] respectively
Let \[a = 50m\]
\[b = 65m\]
and \[c = 65m\]
Now \[S = \dfrac{{a + b + c}}{2} = \dfrac{{50 + 65 + 65}}{2} = 90m\]
Area of triangular field is found by heron’s formula
i.e. Area \[ = \sqrt {S(s - a)(s - b)(s - c)} \]
put \[s = 90m,a = 50m,b = 65m\] and \[c = 65m\]
Area \[ = \left[ {{{10}^2} \times {3^2} \times {2^2} \times {{25}^2}} \right]\]
\[ = \sqrt {90(40)(25)(25)} \]
\[ = \sqrt {10 \times 910 \times 4 \times 25 \times 25} \] \[\left[ {\because 90 = 9 \times 10\,\,40 = 4 \times 10} \right]\]
\[ = \left[ {{{10}^2} \times {3^2} \times {2^2} \times {{25}^2}} \right]\] \[\left[ {\because {3^2} = 9\,4 = {2^2}} \right]\]
\[ = 10 \times 3 \times 2 \times 25\]
Area\[ = 1500{m^2}\]
BY unitary method,
Cost of laying \[I{m^2}\] grass \[ = Rs.7\](Given)
Cost of laying \[1500{m^2}\] grass
\[ = Rs\,1500 \times 7 = Rs\,10500\]
Hence the cost of laying is Rs. \[1050\].

Note: The question can also be solved by using the concept of isosceles triangle of triangle In, isosceles triangle two sides, are equal.