Math, asked by Anonymous, 4 days ago

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 cost of polishing the floor at the rate of Rs.4 per m^2.

Answers

Answered by rittuahir91
3

Answer:

To find,

Cost of polishing=area\ of\ floor\times\cos t\ per\ m^2area of floor×cost per m2

Area of floor=3000\times area\ of\ tile3000×area of tile

Area of a tile=\frac{1}{2}\left(product\ of\ its\ diagonals\right)21(product of its diagonals)

=\frac{1}{2}\times45\times3021×45×30

=675\ cm^2675 cm2

Area of floor=3000\times675\ cm^23000×675 cm2

=2025000\ cm^2=2025000 cm2

=202.5\ m^2=202.5 m2

Cost of polishing=4\times202.54×202.5

=rs 810

Step-by-step explanation:

Hope's it helps you

Answered by Ehsanul885
2

Answer:

Answer is ₹810.

Step-by-step explanation:

SOLUTION

Given,

d1 = 45cm

d2 = 30cm

Area of one tile = Area of rhombus = 1/2 d1 × d2

=> (1/2 × 45 × 30) cm²

=> 675 cm².

Area of 3000 tiles = 3000 × area of one tile

=> (3000 × 675) cm²

=> 2025000 cm²

=> 2025000/100 × 100 m²

=> 202.5 m²

Hence,

The cost of polishing the floor

= (202.5 × 4) = 810.

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