Question

Ralph is painting the barn below, including the sides and roof. He wants to know how much paint to purchase.

Answer 1

I believe the correct answer is B

Answer 2

Answer:

a. In the given figure,

The triangular prism is placed above a cuboid,

$\because \sqrt{4^2+10^2}=\sqrt{16+100}=\sqrt{116}$

∴ Surface area of the triangular prism,

$A_1= 2\times \frac{1}{2}\times 20\times 4+2\times \sqrt{116}\times 45$

$=(80 + 90\sqrt{116})\text{ square feet}$

Now, the surface of the cuboid having dimensions 20' × 45' × 15'

So, the surface area of the cuboid,

$A_2=2\times 20\times 15 + 2\times 45\times 15$

$=1950\text{ square feet}$

Hence, the total surface area of the house,

$A=A_1+A_2$

$=80+90\sqrt{116}+1950$

$=2999.32966528$

$\approx 2999.33\text{ square feet}$

b. ∵ 1 paint = 57 square feet,

⇒ 1 square feet = $\frac{1}{57}$ paints,

⇒ The number of cans required for 2999.33 square feet = $\frac{2999.33}{57}$ ≈ 53

c. Since, the cost of one can = $ 23.50,

So, the cost of 53 cans = $ 1245.5,

d. The volume inside the barn = Volume of triangular prism + volume of cuboid

= 1/2 × 20 × 4 × 45 + 20 × 45 × 15

= 15,300 cube feet.