Question

The floor of a building consists of $3000$ tiles which are rhombus shaped and each of its diagonals are $45$ cm and $30$ cm in length. Find the total cost of polishing the floor. If the cost per ${m^2}$ is $4$ Rupees.

Answer 1

Hint: Rhombus looks like a diamond with four equal straight sides and opposite sides are parallel and opposite angles are equal.
Area of the Rhombus$= \dfrac{1}{2}{d_1}{d_2}$ where, ${d_1},{d_2}$ are the diagonals of the Rhombus.

Complete step by step solution:
Given that : The Tile is in the shape of the rhombus.
Let, one Diagonal be ${d_1} = 45cm$
Other diagonal be ${d_2} = 30cm$
Area of the Rhombus shaped tile $= \dfrac{1}{2}{d_1}{d_2}$
Put values,
$\therefore Area = \dfrac{1}{2}\times 45 \times 30$
$\therefore Area = 45 \times 15$
$\therefore Area = 675c{m^2}$
Therefore, Area of $1$ tile $= 675c{m^2}$
Therefore Area of $3000$ tiles$= 3000 \times 675$
$= 2025000$ $c{m^2}$
Converting $c{m^2}$ in ${m^2}$
As we know that,
$10000c{m^2} = 1{m^2}$
$2025000c{m^2} = ?$
$\begin{array}{l} = \dfrac{{2025000 \times 1}}{{10000}}\\\ = 202.5{m^2} \end{array}$
Therefore, flooring of the building is $202.5{m^2}$ covered by $3000$ tiles.
We are given that the floor is polished at the cost of per ${m^2}$ is $4$ Rupees.
$1{m^2} = 4$ Rupees
$\begin{array}{l} \therefore 202.5{m^2} = ?\\\ \end{array}$
Take cross-multiplication-
Total cost $= 202.5 \times 4$
Total cost $= 810$ Rupees
Thus, the required solution , the total cost for polishing the floor is $810$ Rupees.

Note: Check the units given, observe all the units of the given parameters and the answer asked. Calculations are only possible if the given units are of the same format. As and are far different. As we know that $1m = 100cm$. Study in detail MKS (Meter, Kilogram, second ) and CGS (Centimetre, gram , second) units.