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Tony will paint 16 right, cylindrical columns. The top face and the bottom face will be covered, so those parts will not be painted. Each column is 18 feet tall and has a diameter of 10 feet. One gallon of paint will cover 350 square feet. If paint is sold only in full gallons, how many gallons of paint must Tony buy to paint all 16 columns?

2007-09-17 02:16:07 · 7 answers · asked by Barbie 1 in Science & Mathematics Mathematics

7 answers

Surface area of 16 columns is given by:-
16 x π x 10 x 18 ft ²
= 9043.2 ft ²

No. of gallons
= 9043.2 / 350
= 26 gallons (rounded up to nearest gallon)

2007-09-17 06:18:17 · answer #1 · answered by Como 7 · 2 0

We need to know the surface area of each column.

Area = heigth * length

Height is given as 18 feet

Length is the circumference of a circle with radius 10 feet

Circumference of a circle is pi * d (or, as it's more often written 2 * pi * r).

So area of one cylinder is
18 * pi * 10

Area of 16 columns is 16 * 18 * 10 * pi which is just over 9047 sqaure feet

If one gallon does 350 sq feet, we would need
9047 / 350 gallons to do all columns

9047 / 350 is 25.85

But paint is only sold in FULL gallons, so we need to buy 26 gallons

2007-09-17 09:47:37 · answer #2 · answered by PeterT 5 · 0 1

Total outer surface area of 16 columns, 18 ft. tall and 10 ft. diameter
= 16 * 18 * (pi) *10 sq. ft.
Paint needed
= 1 gallon for each 350 sq. ft.

So, total paint needed
= 16 * 18 * (pi) * 10 / 350
= 26 gallon ( rounded of to the next higher integer )

2007-09-17 09:24:04 · answer #3 · answered by Madhukar 7 · 0 0

Tony needs ?? gallons of paint.
C = πD
ft sq = C * H
total ft sq = ft sq * total number of columns
total paint = total ft sq / paint probable coverage per unit

2007-09-17 09:21:35 · answer #4 · answered by credo quia est absurdum 7 · 0 0

26 gallons

2007-09-17 09:28:12 · answer #5 · answered by osho sai 2 · 0 0

alot

2007-09-17 09:21:20 · answer #6 · answered by Anonymous · 0 0

Kid, do your own homework.

2007-09-17 09:19:45 · answer #7 · answered by your_gurl_leah 5 · 0 0

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