Question

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

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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

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.