Question

The floor of a building consists of 3000 tiles which are rhombus shaped. The diagonals of
each of the tiles are 45 cm and 30 cm. Find the total cost of polishing the floor, if cost per
m’is2.250​

Answer 1

Answer:

Area of 1 tile = 1/2 digonal1*diagonal 2

=1/2*45*30

=675cm sq.

Therefore area of 3000 tile is

3000* 675

=2,025,000

Cost of polishing floor is

2025000*225. (1m=100cml

=4,556,250,00

Answer 2

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

Given:

Diameter (d1) = 45 cm

Also,

Diameter (d2) = 30 cm

Number of tiles = 3000

Rate per m²= ₹2.250

Area of Rhombus

$\tt{\rightarrow\dfrac{1}{2}\times d1\times d2}$    

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

= 675 cm²

Area of 1 tile

= 675 cm²

Convert it into m²

As we know that :-

[1cm²= 1/10000m²]

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

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

Cost of Polishing the floor

= Total Area × rate

$\tt{\rightarrow\dfrac{3000\times 675\times 2.250}{10000}}$  

= ₹ 455.625

Therefore,

Total cost of polishing the floor

= ₹ 455.625