Surface Area and Volume

Question

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Ralph is painting the barn, including the sides and roof. He wants to know how much paint to purchase. What is the total surface area that he is going to be painting? Round to the nearest hundredth.

(b) If one paint can covers 57 square feet, how many paint cans should he purchase?

(c) If each paint can costs $23.50, how much will the paint cost?

(d) Once the barn is finished being painted, there is going to be a party. Ralph wants to know how many people to invite to the party. What is the volume of the inside of the barn?

Answer 1

a) You have 6 rectangles, or 3 different pairs. The first pair has total surface area (of both combined) of $45 * 15 * 2=1350$. The second pair has a total surface area of $20 * 15 * 2 = 600$. Before I get to the last pair, the triangles have a total SA of $20 * 4 * 2 * \frac{1}{2} = 20 * 4 = 80$. By the Pythagorean theorem, the last rectangle pair have a side length $L$ which satisfies $4^2+10^2=L^2$, so $L = \sqrt{116} \approx 10.7703$. So the last pair has a total SA of $1350 + 600 + (L * 45 * 2) = 1350 + 600 + 969.327 = 2919.327$.

To the nearest hundredth, that's $2919.33 ft^2$.

b) Divide the total SA by 57, so $2919.33 / 57 \approx 51.22$ so you need a minimum of $52$ cans.

c) Multiply the number of cans by \$$23.50$, so $\$23.50 * 52 = \$1222$.

d) The volume is computed by the lower rectangular prism added to the triangular prism on top (the roof). So $20 * 45 * 15 = 13500$ and $20 * 4 * \frac{1}{2} * 45 = 1800$. So $13500 + 1800 = 15300$, so the total volume is $15300ft^3$.

Answer 2

The surface area of the front and back (as in the image) is two times the area of the rectangle and the triangle $$ 2 \left[ 20\cdot15 + \frac{1}{2} 20 \cdot 4 \right] $$ The area of a long sides is twice the large rectangle $$ 2 \left[ 45 \cdot 15 \right] $$ and the area of the roof is $$ 2 \left[ 45 \cdot \sqrt{10^2+4^2} \right] $$ where $\sqrt{10^2+4^2}$ is the length of the diagonal on the roof using the Pythagorean Theorem.
The total surface area is the sum of all of those.

As for the volume, you can calculate the volume of the rectangular prism with $$ 20 \cdot 45 \cdot 15 $$ and add that to the volume of the triangular prism which is $$ \frac{1}{2} 4 \cdot 15 \cdot 45 $$