Answer:
Required shingle bundles = [tex]\frac{1504.061}{2.25}[/tex] = 668.5 ≈ 669 bundles.
which costs 669 × 35.99 = 24077.31 dollars.
Step-by-step explanation:
Let us divide the roof into two parts 1) cuboid and 2) pyramid.
Now, lower part of the roof
1) lateral surface of cuboid.
each surface is a rectangle with length(l) 10.00 m and width(b) 2.32 m
⇒ area = l × b = 10 × 2.32 = 23.2 m²
now, total area of lower part of roof = 4 × 23.2 = 92.8 m²
(4 rectangles / faces)
Now, upper part of the roof.
2) lateral surface of pyramid.
given length of square base of pyramid (l) = 10.00 m
⇒ diagonal length (d) = 10√2 m.
now, given total height of roof = 4.36 m
⇒ height of only upper part (pyramid), h = 4.36 - 2.32 = 2.04 m.
Now, one of the pyramid edges (a) , height of the pyramid (h) and half of the diagonal of the square base (d/2) makes a right angled triangle.
From pythagorus theorem,
(height of pyramid)² + (half-diagonal)² = (length of pyramid edge)²
⇒ h² + (d/2)² = a²
⇒ a² = (2.04)² + (50√2)² = 5004.1616
⇒ a = 70.7400 m.
Length of all the edges of pyramid are same in this case,
length of edge, a = 70.7400 m.
Now, consider one triangular face of pyramid with two of the sides being
a = 70.7400 m and 10.00 m
⇒ area of one face of pyramid = [tex]\sqrt{p(p-a)(p-b)(p-c)}[/tex]
where, p = [tex]\frac{a+b+c}{2}[/tex]
(area of triangle with sides a,b,c is [tex]\sqrt{p(p-a)(p-b)(p-c)}[/tex])
Here p = [tex]\frac{70.7400+70.7400+10}{2}[/tex].
p = 75.7400 m.
area of one face = [tex]\sqrt{75.74(75.74-70.74)(75.74-70.74)(75.74-10)}[/tex]
= 352.815 m²
Total area of upper part of roof = 4 × 352.815 = 1411.261 m²
Now, total area of roof = 92.8 m² + 1411.261 m² = 1504.061 m².
Given that one bundle of shingles cover 2.25 m²
Required shingle bundles = [tex]\frac{1504.061}{2.25}[/tex] = 668.5 ≈ 669 bundles.
which costs 669 × 35.99 = 24077.31 dollars.
