Question

The floor of a building consists of around 3000 tiles which are rhombus shaped and each of its diagonals are 45cm and 30cm in length.Find the total cost of flooring if each tile costs rupees 20 per sq.m

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Diagonal 1 = 45cm

Diagonal 2 = 30cm

Number of tiles

= 3000

Rate per m² = ₹20

As we know that :-

$\tt{\rightarrow Area\;of\;Rhombus=\dfrac{1}{2}\times D1\times D2}$

$\tt{\rightarrow\dfrac{1}{2}\times 45\times 30}$

= 675

Area of 1 tile :-

= 675 cm²

Change it in m²

$\tt{\rightarrow\dfrac{675}{10000}}$

Hence,

Area of 3000 tiles

$\tt{\rightarrow 3000\times\dfrac{675}{10000}}$

${\boxed{\sf\:{1cm^2=\dfrac{1}{10000}m^2}}}$

Cost of flooring :-

= Total Area × rate

$\tt{\rightarrow\dfrac{3000\times 675\times 20}{10000}}$

$\tt{\rightarrow\dfrac{3\times 675\times 20}{10}}$

= 4050

${\boxed{\sf\:{Total\;cost\;of\; flooring}}}$

= ₹4050

Answer 2

Solution:

Given:

=> No of tiles = 3000

=> D1 = 45 cm.

=> D2 = 30 cm

=> Cost of flooring = 20 Rs per m²

To Find:

=> Total cost of flooring.

Formula used:

$\sf{\implies Area\;of\;Rhombus=\dfrac{1}{2} \times D1 \times D2}$

So,

$\sf{\implies Area\;of\;Rhombus=\dfrac{1}{2} \times D1 \times D2}$

$\sf{\implies \dfrac{1}{2} \times 45 \times 30}$

$\sf{\implies 675\;cm^{2}}$

There are 3000 tiles. So, area of 3000 tiles = 675 × 3000

                                                                         = 2025000 cm²

Then we will convert it into m². We know that 1 cm² = 1/10000 m²

=> 2025000/10000 m²

=> 202.5 m²

If the Cost of flooring of 1 m² = 20 Rs.

So, cost of flooring of 202.5 m² = 202.5 × 20

                                                     = 4050 Rs.

So, the total cost of flooring = 4050 Rs.