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The diagram shows a cuboid. On the 1 cm² grid, complete the net of the cuboid. One face has been drawn for you. A cube has a surface area of 384 cm². Find the length of one of its sides.

Mathematics

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20. (a) The diagram shows a cuboid. On the 1 cm² grid, complete the net of the cuboid. One face has been drawn for you.

The cuboid has dimensions: length = 6 cm, width = 2 cm, height = 3 cm.
The face already drawn is a 6 cm by 3 cm rectangle. This can be considered the front face.
A complete net of a cuboid needs 6 faces:

  • Two faces of 6 cm×3 cm6 \text{ cm} \times 3 \text{ cm} (front and back)
  • Two faces of 6 cm×2 cm6 \text{ cm} \times 2 \text{ cm} (top and bottom)
  • Two faces of 2 cm×3 cm2 \text{ cm} \times 3 \text{ cm} (left and right sides)

The student has already drawn the central 6 cm×3 cm6 \text{ cm} \times 3 \text{ cm} face, the two 2 cm×3 cm2 \text{ cm} \times 3 \text{ cm} side faces (to the left and right), and the two 6 cm×2 cm6 \text{ cm} \times 2 \text{ cm} faces (above and below). This accounts for 5 faces.

  1. Identify the missing face.

    The missing face is the back face, which is a 6 cm×3 cm6 \text{ cm} \times 3 \text{ cm} rectangle.

  2. Attach the missing face to the existing net.

    To complete the net, draw the final 6 cm×3 cm6 \text{ cm} \times 3 \text{ cm} rectangle. A common way to attach it is to the 6 cm6 \text{ cm} edge of the 6 cm×2 cm6 \text{ cm} \times 2 \text{ cm} face that is positioned above the central 6 cm×3 cm6 \text{ cm} \times 3 \text{ cm} face. This means extending the net upwards from the top face.

    (Since I cannot draw on the image, I am describing the completion of the net based on the visible drawing.)

    20. (b) A cube has a surface area of 384 cm². Find the length of one of its sides.

  3. Recall the formula for the surface area of a cube.

    The surface area AA of a cube with side length ss is given by A=6s2A = 6s^2, because a cube has 6 identical square faces.

  4. Substitute the given surface area into the formula.

    We are given A=384 cm2A = 384 \text{ cm}^2.

    384=6s2384 = 6s^2
  5. Solve for s2s^2.

    Divide both sides by 6:

    s2=3846s^2 = \frac{384}{6} s2=64s^2 = 64
  6. Solve for ss.

    Take the square root of both sides:

    s=64s = \sqrt{64} s=8 cms = 8 \text{ cm}

    The length of one of its sides is 8 cm\boxed{8 \text{ cm}}.

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