Question

Kevin is planning to paint a building whose dimensions are shown below. The front of the building has a door measuring 63 in by 80 in. The left and right sides of the building each have 2 windows measuring 6 ft by 3 ft each. The back of the building has no window or door. Use the given information to answer the questions. Each tab shows a different view of the building. (a) Assuming the roof, windows, and door require no paint, what is the area in square feet that needs paint? (Do not round any intermediate computations and give your answer as a whole number.) Conversion facts for length 1 foot (ft) = 12 inches (in) 1 yard (yd) = 3 feet (ft) 1 yard (yd) = 36 inches (in) Front-right view Back-left view (b) Kevin will use paint that is sold in cans. Each can contains enough paint to cover 370 ft. Assume Kevin doesn't have any paint yet and he can't buy partial cans. How many cans will he need to buy in order to paint the building? cans 9 ft
Just final answer please, no need to explain. Thank you :)

Answer 1

0 of the four walls (inclucling door and winelons), Perimeter height of the building = (30.ft +285t + 30++28ft) x 9 off the base x 2 (30ft + 28ft) x 9ft = ax58x 9 ft ² = 116x9 ft2 A, - 1044.ft? Area De door, Az = 63 in x 80 in +(63x++) x (80*80 63x20 12x3 ft² fift=12 in >> linata ft) 7x5 ft2 21x5 ft 2 = A2 = 35 ft² Area of a window, Az = 6ft x 3ft - 18-f42 Next and that needs to be painted, A = Area of the four walls me door - 4 (Area of a window) Area 9

Same-sized = A = A, -A 4 (A₂) (Remember there are 4 h Windows, 2 on Left and aon right) = 1044ft - 35 st ² - 4 (18 ft²) = (1009 - 72) ft² = (A = [937 f + 7 Carea that need to be painted) #q cans of paint needed a (area to be painted) (area (area covered by one can g party ban) А 937 2.5324 370ft? 370 cans Since, Kevin connot buy postical cons, therefore to pano the entire builclary, he will need: 3 cans, I Total Cost = one can # g con neecked X COST of 3 x $36.50 $ 109.50 " Kevin need to spend : $109.50