Solids, Nets, and Spatial Relationships: Core practice
10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.
Difficulty: Core (core-course level)
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Problem 1 An eight-sided base
A pyramid has an octagonal base and an apex outside the base plane. Find its numbers of faces, edges, and vertices, and explain each count.
Count the complete solid, including its hidden parts. Text description of this figure
An octagonal pyramid has eight base vertices in one plane and a single apex above it. The front half of the octagonal base and its joining edges are solid; hidden back edges and apex connections are dashed. The sketch supplies no metric lengths.
- Hint 1
Separate the base from the parts that join it to the apex.
- Hint 2
An octagon has eight vertices and eight sides. Each base side belongs to one triangular face.
Answer
9 faces, 16 edges, and 9 vertices.
Full solution
There is one base and one triangular face for each of its eight edges.
Thus there are nine faces.
The eight base edges and eight edges leading to the apex are distinct.
Thus there are sixteen edges.
There are eight base vertices and one apex outside the base plane.
Thus there are nine vertices.
Answer
9 faces, 16 edges, and 9 vertices.
Key idea
Group faces, edges, and vertices by their roles rather than by which parts are visible.
- Hint 1
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Problem 2 Three descriptions
Classify each description, naming the base type where applicable. State whether each object is a polyhedron.
Case A. Two congruent pentagons in parallel planes are joined at corresponding vertices by parallel edges.
Case B. One hexagon has an apex outside its plane, with one triangular face along each hexagon edge.
Case C. Two congruent circular bases in parallel planes have corresponding circle points joined by parallel straight segments, forming a curved side.
- Hint 1
Decide whether the description uses two polygonal bases, one polygonal base and an apex, or circular bases.
- Hint 2
A polyhedron must have a surface made entirely of polygonal faces.
Answer
Case A: pentagonal prism, a polyhedron. Case B: hexagonal pyramid, a polyhedron. Case C: cylinder, not a polyhedron.
Full solution
Case A has the two congruent parallel polygonal bases and matching parallel joining edges of a prism.
Its pentagonal bases make it a pentagonal prism, and all its faces are polygons.
Case B has one base and a shared off-plane apex, so it is a hexagonal pyramid.
Its base and triangular sides are polygonal faces, making it a polyhedron.
Case C is a cylinder.
Its curved side prevents its complete surface from being made of polygonal faces.
Answer
Case A: pentagonal prism, a polyhedron. Case B: hexagonal pyramid, a polyhedron. Case C: cylinder, not a polyhedron.
Key idea
The base and joining structure determine the family; a curved side prevents classification as a polyhedron.
- Hint 1
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Problem 3 Complete lines and face planes
A rectangular box has bottom ABCD, top PQRS, and joining edges AP, BQ, CR, DS, with the upper vertex named directly above each corresponding lower one. Each face is a rectangle, and opposite face planes are parallel.
Case A. Classify complete lines AC and PQ, with justification. AC is a diagonal of the bottom face.
Case B. Classify complete lines AB and RS.
Case C. Identify the intersection of the top and right face planes.
Dashed marks show hidden edges and the bottom diagonal AC; classifications concern complete lines and planes. Text description of this figure
A rectangular box has lower vertices A, B, C, D in boundary order and upper vertices P, Q, R, S directly above them. Front face ABQP, right face BCRQ, and top face PQRS are visible. Edges AD, DC, and DS are hidden and dashed. No lengths are inferred from the projection. The bottom diagonal AC is added as a dashed blue guide because it is hidden inside the box.
- Hint 1
For Case A, ask what plane would have to contain A, P, and Q.
- Hint 2
For Case B, follow the pair of parallel top edges, then compare the front top edge with AB.
- Hint 3
For Case C, name the two vertices common to the top and right faces.
Answer
Case A: skew. Case B: parallel. Case C: line QR.
Full solution
Case A.
A plane containing AC and PQ would contain A, P, and Q, which determine the front plane.
C lies outside that plane, so the lines are noncoplanar and skew.
Case B.
AB is parallel to PQ in the front rectangle, and PQ is parallel to RS in the top rectangle.
Thus AB and RS have the same direction and are parallel distinct lines.
Case C.
The top plane PQRS and right plane BCRQ share Q and R.
Distinct intersecting planes meet in the complete line QR, not just the edge segment.
Answer
Case A: skew. Case B: parallel. Case C: line QR.
Key idea
Use named face planes and their geometric relationships to justify what a projection alone cannot decide.
- Hint 1
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Problem 4 A shifted upper base
A prism has congruent parallel triangular bases. Its upper base has been shifted so that the three joining edges are not perpendicular to the base planes.
Classify the prism as right or oblique. Find its face, edge, and vertex counts, and explain whether that shift changes these counts.
- Hint 1
Right and oblique describe the joining edges relative to the base planes.
- Hint 2
The shift preserves the two triangular boundaries and the three edges joining matching corners.
Answer
An oblique triangular prism; 5 faces, 9 edges, and 6 vertices. The shift does not change these counts.
Full solution
The joining edges are explicitly not perpendicular to the bases, so the triangular prism is oblique.
This decision uses the stated spatial relationship, not its page orientation.
Two triangular bases and three joining parallelograms give five faces.
The two triangles have six boundary edges altogether, and three edges join corresponding vertices, giving nine edges.
Each triangle contributes three vertices, giving six vertices.
Shifting the base changes angles but neither adds nor removes a face, edge, or vertex.
Answer
An oblique triangular prism; 5 faces, 9 edges, and 6 vertices. The shift does not change these counts.
Key idea
A change in angle can change right to oblique without changing the solid's incidence structure.
- Hint 1
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Problem 5 Opposite faces after closing
Six equal squares form this cube net: A, B, C, D run left to right; E is above B and F is below B. Each shared edge is a fold.
List all three pairs of opposite faces. Explain why A, C, and E cannot all meet at a single cube vertex.
The positions in the statement fix the net orientation. Text description of this figure
A cube net has four squares A, B, C, D in a horizontal row, with E above B and F below B. Solid outer edges are cuts; dashed internal edges are folds.
- Hint 1
Hold B underneath and raise its four neighbors.
- Hint 2
The square beyond C closes the remaining opening. Opposite faces cannot share a cube vertex.
Answer
Opposite pairs: A and C; B and D; E and F. A, C, and E cannot meet at one vertex.
Full solution
With B underneath, A and C rise to the left and right, so they are opposite.
E and F rise at the back and front, so they are opposite.
D folds over from C to close the top, opposite B.
These are all three opposite pairs.
A and C are opposite faces.
They have no common vertex, so adding E cannot make the three meet at one vertex.
Answer
Opposite pairs: A and C; B and D; E and F. A, C, and E cannot meet at one vertex.
Key idea
Track folded positions before deciding which face pairs meet or remain opposite.
- Hint 1
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Problem 6 A straight strip
Six equal squares J, K, L, M, N, P form one straight row, in that order. The five shared edges remain joined while folding.
Can the strip be the net of a cube? Justify your conclusion by following the row around the cube.
All five internal lines are fold edges. Text description of this figure
Six equal squares labeled J, K, L, M, N, P run in one straight horizontal row. The five shared edges are dashed fold lines, and every square is joined only to its immediate neighbors.
- Hint 1
Consecutive square faces around this straight strip must turn about parallel fold edges.
- Hint 2
Follow the first four squares as a belt around the four sides of a cube. Ask where the next square would land.
Answer
No. J, K, L, M would form a four-face belt, and N would overlap J instead of closing a new face.
Full solution
The shared fold edges are parallel.
The next hinge is opposite the previous hinge on each square, so consecutive faces can follow only a four-face belt on a cube.
Reversing a turn would fold back over an earlier face.
To avoid overlap, J, K, L, and M must therefore occupy the four faces of that belt.
The next quarter-turn puts N on the same face position as J.
The interiors would overlap, leaving the two ends of the belt unclosed.
Thus the straight six-square strip is not a cube net.
Answer
No. J, K, L, M would form a four-face belt, and N would overlap J instead of closing a new face.
Key idea
The number of surface pieces is insufficient unless their fold directions let them reach distinct face positions.
- Hint 1
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Problem 7 The display case drawings
A display case is modeled as a rectangular box. Its front drawing shows width 9 cm and height 4 cm. Its top drawing shows width 9 cm and depth 6 cm. Its right drawing shows depth 6 cm and height 4 cm.
State the three box dimensions, and identify the paired views that check each one. The panels are schematic and not drawn to one common scale.
Use stated dimensions rather than measuring the schematic drawing. Text description of this figure
Three schematic orthographic views of a rectangular display case. Front is 9 cm wide and 4 cm high; top is 9 cm wide and 6 cm deep; right is 6 cm deep and 4 cm high. Vertical labels are in centimeters.
- Hint 1
Each view contains two dimensions, and each dimension occurs in two views.
- Hint 2
Compare front with top for width, top with right for depth, and front with right for height.
Answer
Width 9 cm, checked by front and top; depth 6 cm, checked by top and right; height 4 cm, checked by front and right.
Full solution
The front and top agree on width 9 cm.
The top and right agree on depth 6 cm.
The front and right agree on height 4 cm.
Every shared dimension is consistent, so the stated views fit a 9 cm wide, 6 cm deep, 4 cm high rectangular box.
Answer
Width 9 cm, checked by front and top; depth 6 cm, checked by top and right; height 4 cm, checked by front and right.
Key idea
Check repeated dimensions across the named views before combining them into one solid.
- Hint 1
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Problem 8 Read a height plan
A solid has filled vertical columns of unit cubes on a three-by-two footprint, with no gaps. Viewed from above, its back row of heights is 1, 3, 2 and its front row is 2, 1, 1, each read left to right.
Give the front silhouette heights from left to right.
Give the right silhouette heights from front to back. Only the outer silhouettes are requested.
Numbers give column heights, not labels on a flat surface. Text description of this figure
A height plan has back row 1, 3, 2 and front row 2, 1, 1. Every site is occupied by a filled column of unit cubes.
- Hint 1
Columns behind one another project to the same front-view width position.
- Hint 2
For the right view, compare all three columns in the front row, then all three in the back row.
Answer
Front heights: 2, 3, 2 units. Right heights from front to back: 2, 3 units.
Full solution
For the front view, compare each front height with the back height behind it.
The three pairs are 2 with 1, 1 with 3, and 1 with 2.
Their maxima give front heights 2, 3, 2.
For the right view, the tallest front-row column has height 2.
The tallest back-row column has height 3.
With front at the left of this view, its heights are 2, 3.
The tallest column along each sight line sets the silhouette. Answer
Front heights: 2, 3, 2 units. Right heights from front to back: 2, 3 units.
Key idea
Orthographic silhouettes use the greatest height along each sight line, rather than a sum of heights.
- Hint 1
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Problem 9 Two hidden column heights
A solid has filled vertical columns of unit cubes on a full two-by-two footprint. All column heights are positive integers.
The front silhouette has heights 3, 2 from left to right. The right silhouette has heights 3, 2 from front to back.
The back row of the height plan is b, 2 and the front row is 3, a, with both rows read left to right. Determine every possible value of a and of b. Can these choices be made independently?
The silhouettes specified in the statement constrain the two unknown columns. Text description of this figure
A full two-by-two height plan has back row b, 2 and front row 3, a. Both unknowns are positive integers.
- Hint 1
A hidden column cannot exceed the specified silhouette in either view.
- Hint 2
The right width position already contains a height-two column. The back depth position already contains a height-two column.
- Hint 3
Test the positive integer choices below or equal to those bounds against both view directions.
Answer
Each of a and b may independently be 1 or 2; all four pairs are possible.
Full solution
The front silhouette at the right position has height 2.
Since a lies at that width position, it cannot exceed 2.
The known back-right height 2 already supplies the required maximum.
The right silhouette at the back position has height 2.
Thus b cannot exceed 2, and the known back-right height 2 already supplies that maximum.
All heights are positive integers, so a and b may each be 1 or 2.
The fixed front-left height 3 supplies the left front-view height and the front right-view height.
No remaining condition couples the two choices.
Every one of the four pairs keeps both maxima and the full top footprint correct.
Answer
Each of a and b may independently be 1 or 2; all four pairs are possible.
Key idea
A silhouette can bound a hidden column without determining its exact height.
- Hint 1
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Problem 10 Construct from all three views
A solid is made of filled vertical columns of unit cubes, with no gaps inside any column. Its full top footprint has three positions across and two positions from front to back, all occupied.
Its front silhouette heights are 2, 3, 2 from left to right. Its right silhouette heights are 3, 2 from front to back. No hidden edges are specified.
Construct two different compatible solids by giving two height plans. List each back row and front row from left to right. Verify both view directions and name a site at which your plans differ.
Construct columns before opening the solution; several height plans may fit. Text description of this figure
Three unit-grid silhouette views. Front heights are 2, 3, 2 from left to right; right heights are 3, 2 from front to back. The top footprint is a full three-by-two grid. Grid lines are guides; no hidden edges are specified.
- Hint 1
Every width position needs at least one column reaching its front-view height.
- Hint 2
The front row must reach height 3 somewhere, and the back row must reach height 2 but no higher.
- Hint 3
After making one plan, try shortening a column whose top is hidden behind a taller column in both views.
Answer
For example, Plan A has back 1, 2, 2 and front 2, 3, 1. Plan B has back 2, 1, 2 and front 2, 3, 2. They differ at back-left, back-middle, and front-right. Other verified plans are accepted.
Full solution
One plan uses back row 1, 2, 2 and front row 2, 3, 1.
Each site has a positive integer height, so its top footprint is full.
For this plan, the maxima at the three width positions are 2, 3, 2.
The front row reaches 3, and the back row reaches 2, giving the required right silhouette.
A second plan uses back row 2, 1, 2 and front row 2, 3, 2.
Its width-position maxima are again 2, 3, 2, and its front and back row maxima are again 3 and 2.
All six sites remain occupied.
The back-left height changes from 1 to 2, the back-middle height from 2 to 1, and the front-right height from 1 to 2.
These changes give different filled-column solids while preserving all three specified views.
Each plan has width-position maxima 2, 3, 2 and depth-position maxima 3, 2 from front to back. Answer
For example, Plan A has back 1, 2, 2 and front 2, 3, 1. Plan B has back 2, 1, 2 and front 2, 3, 2. They differ at back-left, back-middle, and front-right. Other verified plans are accepted.
Key idea
Construct a compatible solid by satisfying every sight-line constraint, then check which hidden choices remain free.
- Hint 1