Using identical cubes, make a solid that gives the following projections:

Question

Homi · Accepted Answer

Understand the Question

To construct a solid from orthogonal projections (Top, Front, and Side views), represent the position of each unit cube using 3D coordinates $(x, y, z)$ :

$x$ : Width (left to right)
$y$ : Depth (front to back)
$z$ : Height (bottom to top)

A unit cube exists at $(x, y, z)$ if and only if its corresponding projected cell is present in all three views:

Top View: $(x, y)$
Front View: $(x, z)$
Side View: $(y, z)$

(i) Using identical cubes, make a solid that gives the projections in Set 1.

Step 1 · Determine Cube Positions for Set 1

Diagram 1

From the given projections for Set 1:

Top View (i) $(x, y)$ : $(0, 0), (1, 0), (2, 0), (0, 1), (1, 1)$
Front View (ii) $(x, z)$ : $(0, 0), (1, 0), (2, 0), (0, 1)$
Side View (iii) $(y, z)$ : $(0, 0), (1, 0), (1, 1)$

Testing each coordinate $(x, y, z)$ :

For $(x, y) = (0, 0)$ $(x, y) = (0, 0)$ :
- $(0, 0, 0)$ : Present in all views $\implies$ $(0, 0, 0)$ exists
- $(0, 0, 1)$ : Side $(0, 1)$ missing $\implies$ does not exist
For $(x, y) = (1, 0)$ $(x, y) = (1, 0)$ :
- $(1, 0, 0)$ : Present in all views $\implies$ $(1, 0, 0)$ exists
- $(1, 0, 1)$ : Front $(1, 1)$ missing $\implies$ does not exist
For $(x, y) = (2, 0)$ $(x, y) = (2, 0)$ :
- $(2, 0, 0)$ : Present in all views $\implies$ $(2, 0, 0)$ exists
- $(2, 0, 1)$ : Front $(2, 1)$ missing $\implies$ does not exist
For $(x, y) = (0, 1)$ $(x, y) = (0, 1)$ :
- $(0, 1, 0)$ : Present in all views $\implies$ $(0, 1, 0)$ exists
- $(0, 1, 1)$ : Present in all views $\implies$ $(0, 1, 1)$ exists
For $(x, y) = (1, 1)$ $(x, y) = (1, 1)$ :
- $(1, 1, 0)$ : Present in all views $\implies$ $(1, 1, 0)$ exists
- $(1, 1, 1)$ : Front $(1, 1)$ missing $\implies$ does not exist
For $(x, y) = (2, 1)$ : Not in Top View $\implies$ no cubes

Answer

(i) The solid consists of $6$ cubes at positions: $(0, 0, 0), (1, 0, 0), (2, 0, 0), (0, 1, 0), (1, 1, 0), (0, 1, 1)$

(ii) Using identical cubes, make a solid that gives the projections in Set 2.

Step 1 · Determine Cube Positions for Set 2

Diagram 2

From the given projections for Set 2:

Top View (iv) $(x, y)$ : $(0, 0), (1, 0), (0, 1)$
Front View (v) $(x, z)$ : $(0, 0), (1, 0), (0, 1)$
Side View (vi) $(y, z)$ : $(0, 0), (1, 0), (1, 1)$

Testing each coordinate $(x, y, z)$ :

For $(x, y) = (0, 0)$ $(x, y) = (0, 0)$ :
- $(0, 0, 0)$ : Present in all views $\implies$ $(0, 0, 0)$ exists
- $(0, 0, 1)$ : Side $(0, 1)$ missing $\implies$ does not exist
For $(x, y) = (1, 0)$ $(x, y) = (1, 0)$ :
- $(1, 0, 0)$ : Present in all views $\implies$ $(1, 0, 0)$ exists
- $(1, 0, 1)$ : Front $(1, 1)$ missing $\implies$ does not exist
For $(x, y) = (0, 1)$ $(x, y) = (0, 1)$ :
- $(0, 1, 0)$ : Present in all views $\implies$ $(0, 1, 0)$ exists
- $(0, 1, 1)$ : Present in all views $\implies$ $(0, 1, 1)$ exists
For $(x, y) = (1, 1)$ : Not in Top View $\implies$ no cubes

Answer

(ii) The solid consists of $4$ cubes at positions: $(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 1, 1)$

(iii) Using identical cubes, make a solid that gives the projections in Set 3.

Step 1 · Check Consistency of Set 3 Projections

Analyze the maximum dimensions along the $y$ -axis (depth) for Set 3:

Top View (vii) shows positions along the $y$ -axis at $y = 0, 1, 2$ , which requires a total depth of $2\text{ units}$ .
Side View (ix) shows positions along the $y$ -axis at $y = 0, 1$ , which corresponds to a total depth of only $1\text{ unit}$ .

Since the depth of the Top View conflicts with the depth of the Side View, the projections are contradictory.

Answer

(iii) No solid can be formed because the Top View has a maximum depth of $2\text{ units}$ while the Side View has a maximum depth of $1\text{ unit}$ .

Common Mistakes

Overlooking 3D Intersection: Placing a cube based on only one or two views without confirming it satisfies the third view.
Ignoring Dimension Consistency: Forgetting that corresponding dimensions (such as depth $y$ across Top and Side views) must match exactly across all projections.

Question 16

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