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Calculate the Total Surface Area (T.S.A.) of an open cuboid with length 6 dm, width 5 dm, and height 3 dm. Also, find the number of edges and vertices.

Mathematics

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3 parts
a

96 dm296 \text{ dm}^2

b(i)
12
b(ii)
8

Step-by-step solution

3 steps

Here's the solution to the problem:

Part a· 3 steps

Calculate its T.S.A.

  1. Identify the dimensions of the cuboid.

    The length (ll) is 6 dm6 \text{ dm}.
    The width (ww) is 5 dm5 \text{ dm}.
    The height (hh) is 3 dm3 \text{ dm}.

  2. Determine the formula for the Total Surface Area (T.S.A.) of an open cuboid.

    An open cuboid means one face is missing. From the diagram, the top face is missing.
    The T.S.A. will be the sum of the areas of the bottom face, two side faces (front and back), and two end faces (left and right).

    T.S.A.=(l×w)+2(l×h)+2(w×h)\text{T.S.A.} = (l \times w) + 2(l \times h) + 2(w \times h)

  3. Substitute the dimensions into the formula and calculate the T.S.A.

    T.S.A.=(6 dm×5 dm)+2(6 dm×3 dm)+2(5 dm×3 dm)\text{T.S.A.} = (6 \text{ dm} \times 5 \text{ dm}) + 2(6 \text{ dm} \times 3 \text{ dm}) + 2(5 \text{ dm} \times 3 \text{ dm}) T.S.A.=30 dm2+2(18 dm2)+2(15 dm2)\text{T.S.A.} = 30 \text{ dm}^2 + 2(18 \text{ dm}^2) + 2(15 \text{ dm}^2) T.S.A.=30 dm2+36 dm2+30 dm2\text{T.S.A.} = 30 \text{ dm}^2 + 36 \text{ dm}^2 + 30 \text{ dm}^2 T.S.A.=96 dm2\text{T.S.A.} = 96 \text{ dm}^2

    The T.S.A. of the open cuboid is:

    96 dm2\boxed{96 \text{ dm}^2}

Part b(i)12

edges

Find the number of;

An open cuboid still has all the lines forming its structure. A standard cuboid has 12 edges. Removing a face does not remove the edges that define its boundaries.
The number of edges is:

12\boxed{12}

Part b(ii)8

vertices

A standard cuboid has 8 vertices (corners). An open cuboid retains all its corners.
The number of vertices is:

8\boxed{8}
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