How does the ratio of corresponding side lengths of the two quadrilaterals compare with the ratio of the areas of the quadrilate
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PART A
The barn is constructed by the following 2D shapes
Two triangles of height = 4' and base = 20'
Area = 2 × (0.5×4×20) = 80
Two rectangles of length = 20' and width = 15'
Area = 2 × 20 × 15 = 600
Two rectangles of length = 45' and width = 15'
Area = 2 × 45' × 15' = 1350
One base of length = 45' and width = 15'
Area = 45 × 15 = 675
Two rectangles on each on one side of the roof. We have the length = 45' but not the width. We can work out the width by using Pythagoras theorem
w² = 4² + 10²
w² = 16 + 100
w² = 116
w = √116 = 10.77
Area of the two rectangles on the roof part is = 2 ×10.77 × 45 = 969.33
Total area to paint = 969.33+675+1350+600+80 = 3674.33 ≈ 3700 (to the nearest hundreth)
PART B
Numbers of paint cans needed = 3700 ÷ 57 = 64.9 ≈ 65 cans
PART C
Total cost of paint = 65 × 23.50 = $1527.50
PART D
The barn is constructed by a cuboid and a rectangular prism
V of cuboid = length × width × height
V of cuboid = 20 × 45 × 15
V of cuboid = 13500
V of triangular prism = Area of cross section × depth
V = [0.5×4×20] × 45
V = 1800
Total volume = 1800 + 13500 = 15300
The barn is constructed by the following 2D shapes
Two triangles of height = 4' and base = 20'
Area = 2 × (0.5×4×20) = 80
Two rectangles of length = 20' and width = 15'
Area = 2 × 20 × 15 = 600
Two rectangles of length = 45' and width = 15'
Area = 2 × 45' × 15' = 1350
One base of length = 45' and width = 15'
Area = 45 × 15 = 675
Two rectangles on each on one side of the roof. We have the length = 45' but not the width. We can work out the width by using Pythagoras theorem
w² = 4² + 10²
w² = 16 + 100
w² = 116
w = √116 = 10.77
Area of the two rectangles on the roof part is = 2 ×10.77 × 45 = 969.33
Total area to paint = 969.33+675+1350+600+80 = 3674.33 ≈ 3700 (to the nearest hundreth)
PART B
Numbers of paint cans needed = 3700 ÷ 57 = 64.9 ≈ 65 cans
PART C
Total cost of paint = 65 × 23.50 = $1527.50
PART D
The barn is constructed by a cuboid and a rectangular prism
V of cuboid = length × width × height
V of cuboid = 20 × 45 × 15
V of cuboid = 13500
V of triangular prism = Area of cross section × depth
V = [0.5×4×20] × 45
V = 1800
Total volume = 1800 + 13500 = 15300