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 cost for metersquare is ₹4
​

Answer 1

Step-by-step explanation:

given d1 = 45cm ,d2 =30cm

no.of tiles =3000

rate per m^2=rs 4

area of rhombus =1/2 ×d1×d2

=1/2 ×45×30=675

area of 1 tile = 675 cm^2

=675/10000m^2

area of 3000 tiles

= 3000×675/10000

[ 1 cm^2= 1/1000m^2]

cost of polishing the floor

= total area ×rate

= (3000×675/4)×10000

=3×675/4×10

=3×67.5×4

= 12×67.5×4= rs 810

hence, the total cost of publishing the floor= rs 810

i hope its help you...

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