A school wishes to use square tiles of artificial turf to cover an outdoor play area that measures 40 feet by 72 feet. Only whole tiles that are congruent squares of the largest possible size will be used. How many such square tiles are needed?

45 tiles of dimension 8 ft by 8ft

Step-by-step explanation:

Since the tile is going to be a square, then both sides must be equal.

This means the lenght of the side of the square must divide without remainder, the lenght and the breadth of the outdoor play area. The highest common factor of 40 and 72 is 8. This means that the side of each square should be 8 ft.

This means that 5 of the 8 ft sided tile will cover the 40 ft width of the play area, and 9 of the 8 ft sided tile will cover the lenght of the area.

The total amount of tiles needed will then be 9 x 5 = 45

unpaid balance is : $149.99-$50 = $99.99 new balance = 149.99+2.62-50 = $102.61 finance charge = 99.99- 0.21/12= $1.75 new balance = 99.99+1.75 = $101.74 average daily balance method average daily balance= 3399.70/30 = $113.32 finance charge = $ 0.21/12ă—113.32 = $1.98 new balance = $149.99-50+1.98 = $101.97

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