Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

7 a rectangular box is to be covered in material. how much is required …

Question

7 a rectangular box is to be covered in material. how much is required to cover the entire box if it has the dimensions \\(1.3\text{ m}\\), \\(1.5\text{ m}\\) and \\(1.9\text{ m}\\)?

8 two wooden boxes, both with dimensions \\(80\text{ cm}\\), \\(1\text{ m}\\) and \\(25\text{ cm}\\), are placed on the ground, one on top of the other as shown. the entire outside surface is then painted. find the area of the painted surface.

9 the four walls and roof of a barn (shown) are to be painted.
a find the surface area of the barn, not including the floor.
b if \\(1\text{ litre}\\) of paint covers \\(10\text{ m}^2\\), find how many litres are required to complete the job.

10 an open top rectangular box \\(20\text{ cm}\\) wide, \\(25\text{ cm}\\) long and \\(10\text{ cm}\\) high is made from wood \\(1\text{ cm}\\) thick. find the surface area:
a outside the box (do not include the top edge)
b inside the box (do not include the top edge)

Explanation:

Calculate surface area of the rectangular box

Using the Rectangular Prism Surface Area knowledge point

$$ LATEXBLOCK0 $$

Calculate painted surface area of stacked boxes

Using the Surface Area of Composite Solids knowledge point

$$ LATEXBLOCK1 $$

Calculate barn surface area and paint required

The barn consists of a rectangular base and a triangular roof prism.

  • Base walls: Two sides of \(4 \times 2 = 8\text{ m}^2\) each, front and back walls up to the roof line of \(2.4 \times 2 = 4.8\text{ m}^2\) each.
  • Roof gables (front/back triangles): Base \(2.4\text{ m}\), height \(3.5 - 2 = 1.5\text{ m}\). Area of two triangles: \(2 \times \frac{1}{2} \times 2.4 \times 1.5 = 3.6\text{ m}^2\).
  • Roof panels: Two rectangular panels of \(4 \times 1.9 = 7.6\text{ m}^2\) each.
  • Total surface area (excluding floor):
$$ A = 2(8) + 2(4.8) + 3.6 + 2(7.6) = 16 + 9.6 + 3.6 + 15.2 = 44.4\text{ m}^2 $$
  • Paint required:
$$ V = \frac{44.4}{10} = 4.44\text{ litres} $$

Calculate outside surface area of open box

The outside dimensions are \(l = 25\text{ cm}\), \(w = 20\text{ cm}\), \(h = 10\text{ cm}\).

  • Outside area (excluding top edge and top opening):
$$ A_{\text{out}} = lw + 2lh + 2wh = 25 \times 20 + 2(25 \times 10) + 2(20 \times 10) = 500 + 500 + 400 = 1400\text{ cm}^2 $$

Calculate inside surface area of open box

The wood is \(1\text{ cm}\) thick.

  • Inside length: \(25 - 2(1) = 23\text{ cm}\)
  • Inside width: \(20 - 2(1) = 18\text{ cm}\)
  • Inside height (open top, so only subtract bottom thickness): \(10 - 1 = 9\text{ cm}\)
  • Inside area (excluding top edge):
$$ A_{\text{in}} = l_{\text{in}}w_{\text{in}} + 2l_{\text{in}}h_{\text{in}} + 2w_{\text{in}}h_{\text{in}} = 23 \times 18 + 2(23 \times 9) + 2(18 \times 9) = 414 + 414 + 324 = 1152\text{ cm}^2 $$

Answer:

Question 7

\(14.54\text{ m}^2\)

Question 8

\(2.6\text{ m}^2\)

Question 9

a) \(44.4\text{ m}^2\)
b) \(4.44\text{ litres}\)

Question 10

a) \(1400\text{ cm}^2\)
b) \(1152\text{ cm}^2\)