Question

2. The inner dimensions of a swimming pool are
50 ft x 20 ft x 10 ft. The inner sides and the floor
of the swimming pool have to be fixed with tiles
of size 2 ft x 2 ft. Find the cost of fixing the tiles at
the rate of 15 each.​

Answer 1

Answer:

area of four walls of pool=2(L+B)H

=2(50+20)10

=1400ft

area of floor=LB

=50by20

= 1000ft

total area=1400+1000

=2400ft

area of tiles=2by2

=4ft

no of tiles =2400÷4

=600tiles

cost of fixing the tiles=15by600

=Rs9000

Answer 2

Answer:

₹9000

Step-by-step explanation:

Lateral surface area of a cuboid =  $2(l+b)h$

                                                     = $2(50+20)10$

                                                      = $1400$

Area of floor = $l * b$

                      = $50 * 20 = 1000 ft$²

Total area = $2400$

Area of tiles = 2 * 2 = 4 ft^2

No. of tiles = 2400/4=600

Total cost = Rs 9000