India's #1 AI Tutorimportant questions · Mathematics · Chapter 12हिंदी में पढ़ें → Class 9 Mathematics Chapter 12: Visualising Solid Shapes – Important Questions & Answers
Chapter 12 (Visualising Solid Shapes) tests your spatial reasoning—a core skill for the 2024–25 CBSE board exam. This chapter covers 3D geometry fundamentals: identifying and counting faces, edges, and vertices; drawing and interpreting nets; and visualising top, front, and side views of solid objects. These concepts appear across 1-mark MCQs, 2-mark short answers, and 5-mark problem-solving questions. We've curated 18 questions spanning all difficulty levels and question types likely to appear in your board assessment. Work through each category systematically—MCQs build confidence, 2-marks develop application skill, and 5-marks sharpen reasoning. Use this guide to self-assess, then drill weak areas with daily AI-powered practice at cbsetutor.ai.
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Start 3-day free trial →Why These Questions Matter in the 2024–25 CBSE Board Pattern
Visualising Solid Shapes (Chapter 12) is a mandatory part of the Class 9 Mathematics syllabus and carries approximately 6–8 marks in the board exam. The revised CBSE assessment pattern emphasizes not just formula recall but spatial comprehension—your ability to mentally rotate objects, count hidden edges, and convert 2D nets into 3D mental models. Examiners test this through: (1) 1-mark MCQs asking you to identify a shape from a net or count vertices; (2) 2-mark questions requiring you to match views to objects or calculate edge/face/vertex counts; (3) 3-mark questions about nets, cross-sections, or reasoning about unfolded shapes; and (4) 5-mark integrated problems combining multiple solid shapes and their properties. Success in this chapter also strengthens your foundation for coordinate geometry and trigonometry in later classes. The questions below reflect the exact cognitive demand and language style of recent CBSE board papers.
1-Mark MCQs: Quick Recall & Visual Recognition
**Question 1:** A cube has how many faces, edges, and vertices?
(a) 6 faces, 12 edges, 10 vertices
(b) 6 faces, 12 edges, 8 vertices
(c) 8 faces, 12 edges, 6 vertices
(d) 6 faces, 8 edges, 12 vertices
**Answer:** (b) 6 faces, 12 edges, 8 vertices. A cube is a regular polyhedron with 6 square faces, 12 equal edges, and 8 corners (vertices).
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**Question 2:** Which of the following nets can fold into a cube?
(a) A cross shape with 6 connected squares arranged as a '+' sign
(b) A straight line of 6 squares
(c) A 2×3 rectangle of squares
(d) Both (a) and (c)
**Answer:** (d) Both (a) and (c). The cross pattern (net 1) and the 2×3 grid are valid cube nets. A straight line of 6 squares cannot fold into a cube—opposite faces would overlap.
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**Question 3:** A triangular pyramid (tetrahedron) has how many vertices?
(a) 3 (b) 4 (c) 5 (d) 6
**Answer:** (b) 4. A triangular pyramid has a triangular base (3 vertices) plus one apex vertex, totalling 4 vertices.
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**Question 4:** The front view of a cylinder is a ___.
(a) Circle (b) Rectangle (c) Triangle (d) Ellipse
**Answer:** (b) Rectangle. When viewed from the front, a cylinder appears as a rectangle (the height and diameter form the outline).
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**Question 5:** A rectangular prism (cuboid) with length 5 cm, width 4 cm, and height 3 cm has how many edges?
(a) 12 (b) 8 (c) 6 (d) 10
**Answer:** (a) 12. Every cuboid has 12 edges: 4 along length, 4 along width, and 4 along height.
2-Mark Short-Answer Questions: Application & Reasoning
**Question 1:** Draw a net of a rectangular prism (cuboid) and label its dimensions as length = 3 cm, width = 2 cm, and height = 1.5 cm. How many rectangles of each type would appear in the unfolded net?
**Answer:** When unfolded, a cuboid has 6 rectangular faces:
- Two faces of 3 × 2 cm (top and bottom)
- Two faces of 3 × 1.5 cm (front and back)
- Two faces of 2 × 1.5 cm (left and right)
One valid net arranges these 6 rectangles in a cross-like pattern. Students should show the unfolding with correct dimensions labelled.
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**Question 2:** A triangular prism has a triangular base with 3 sides. How many vertices, edges, and faces does it have? Verify using Euler's formula (V – E + F = 2).
**Answer:**
- Vertices (V) = 6 (3 on top triangle, 3 on bottom triangle)
- Edges (E) = 9 (3 on top, 3 on bottom, 3 connecting top and bottom)
- Faces (F) = 5 (2 triangular, 3 rectangular)
Verification: V – E + F = 6 – 9 + 5 = 2 ✓ (Euler's formula holds)
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**Question 3:** A square pyramid has a square base with 4 sides. Draw the shape and write down the number of faces, vertices, and edges.
**Answer:**
- Faces (F) = 5 (1 square base + 4 triangular sides)
- Vertices (V) = 5 (4 at the base corners + 1 apex)
- Edges (E) = 8 (4 forming the base + 4 from base corners to apex)
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**Question 4:** If the top view of a solid is a circle and the front view is also a circle, what solid could it be? Give two examples.
**Answer:** The solid is a **sphere**. Any 2D projection of a sphere (from any angle) appears as a circle. Alternatively: a **cone** viewed from the top shows a circle, but its front view would be a triangle, so a cone alone doesn't fit. The answer is primarily a **sphere**, and any object with circular symmetry in all axial directions (e.g., a globe, a ball bearing).
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**Question 5:** A polyhedron has 7 faces and 10 vertices. Using Euler's formula (V – E + F = 2), calculate the number of edges.
**Answer:**
Using V – E + F = 2:
10 – E + 7 = 2
17 – E = 2
E = 15
The polyhedron has **15 edges**.
3-Mark Questions: Deeper Understanding & Multi-Step Logic
**Question 1:** A net of a cube is shown below (imagine a cross with 6 squares). If the central square is numbered 1 and the four adjacent squares are numbered 2, 3, 4, 5 (in order around the cross), what number appears opposite to square 1?
**Answer:** When the cross net folds into a cube, the four squares (2, 3, 4, 5) form the four sides, and the remaining square (the tail of the cross) becomes the opposite face to square 1. If the net shows square 6 at the end of the cross, then **6 is opposite to 1**. (Note: For any standard cube cross-net, the square at the far end of any arm of the cross is opposite to the central square.)
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**Question 2:** A pentagonal prism is a prism with a pentagonal (5-sided) base. Without drawing, determine the number of vertices, edges, and faces, then verify Euler's formula.
**Answer:**
- Vertices (V) = 10 (5 on top pentagon, 5 on bottom)
- Edges (E) = 15 (5 on top, 5 on bottom, 5 connecting top to bottom)
- Faces (F) = 7 (2 pentagonal, 5 rectangular)
Euler's formula: V – E + F = 10 – 15 + 7 = 2 ✓
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**Question 3:** Three orthogonal (perpendicular) views of a solid are: top view = square, front view = rectangle, and side view = rectangle. Identify the solid and justify your answer.
**Answer:** The solid is a **rectangular prism (cuboid)**. Justification: A cuboid has a square or rectangular base. When viewed from above, the base (a rectangle or square) appears as a square or rectangle. From the front and side, the height and width/depth appear as rectangles. A cube would have a square in all three views, but this solid has a square top and rectangular front/side, so it's a non-cubic cuboid with a square base.
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**Question 4:** A hexagonal pyramid has a hexagonal base (6 sides). Calculate the total number of faces, vertices, and edges. Is this consistent with Euler's formula?
**Answer:**
- Faces (F) = 7 (1 hexagonal base + 6 triangular sides)
- Vertices (V) = 7 (6 at the base + 1 apex)
- Edges (E) = 12 (6 forming the hexagonal base + 6 from base to apex)
Euler's formula: V – E + F = 7 – 12 + 7 = 2 ✓ Yes, it is consistent.
5-Mark Long-Answer Questions: Comprehensive Problem-Solving
**Question 1:** A company manufactures cardboard boxes shaped as rectangular prisms with dimensions length = 8 cm, width = 5 cm, and height = 6 cm. (a) Draw a net of the box. (b) Calculate the total surface area. (c) How many unit cubes of side 1 cm can fit inside the box? (d) If a diagonal line is drawn from one corner of the box to the opposite corner (space diagonal), what is its length?
**Answer:**
(a) **Net:** A rectangular prism net consists of 6 rectangles arranged in a cross or other valid unfolding:
- Two 8 × 5 cm faces (top and bottom)
- Two 8 × 6 cm faces (front and back)
- Two 5 × 6 cm faces (left and right)
(b) **Total Surface Area:**
SA = 2(lw + lh + wh)
SA = 2(8×5 + 8×6 + 5×6)
SA = 2(40 + 48 + 30)
SA = 2(118) = **236 cm²**
(c) **Number of Unit Cubes:**
Volume of box = length × width × height
V = 8 × 5 × 6 = **240 unit cubes**
(d) **Space Diagonal:**
The space diagonal d of a cuboid is given by:
d = √(l² + w² + h²)
d = √(8² + 5² + 6²)
d = √(64 + 25 + 36)
d = √125 = **5√5 cm ≈ 11.18 cm**
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**Question 2:** A triangular pyramid (tetrahedron) has four vertices: A (top/apex), B, C, and D (forming the base triangle). (a) List all faces, edges, and vertices. (b) Verify Euler's formula. (c) If each edge has length 5 cm, calculate the total length of all edges.
**Answer:**
(a) **Faces:** ABC, ABD, ACD, BCD (4 triangular faces)
**Edges:** AB, AC, AD, BC, BD, CD (6 edges)
**Vertices:** A, B, C, D (4 vertices)
(b) **Euler's Formula:**
V – E + F = 4 – 6 + 4 = 2 ✓
(c) **Total Edge Length:**
A regular tetrahedron has 6 equal edges.
Total length = 6 × 5 = **30 cm**
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**Question 3:** An octagonal prism is a prism with an octagonal (8-sided) base. (a) Draw a labelled diagram showing the top and side view. (b) Calculate the number of vertices, edges, and faces. (c) A company uses such prisms as packaging. If the octagonal base has a perimeter of 48 cm and the height of the prism is 12 cm, what is the lateral surface area (the area of the 8 rectangular sides, excluding the two octagonal bases)?
**Answer:**
(a) **Diagram:**
- Top view: Regular octagon with 8 sides
- Side view: Rectangle (height = 12 cm, width = perimeter ÷ 8 = 6 cm per side)
(b) **Vertices, Edges, Faces:**
- Vertices (V) = 16 (8 on top octagon, 8 on bottom)
- Edges (E) = 24 (8 on top, 8 on bottom, 8 vertical connecting)
- Faces (F) = 10 (2 octagonal + 8 rectangular)
Euler's formula check: 16 – 24 + 10 = 2 ✓
(c) **Lateral Surface Area:**
Lateral SA = perimeter of base × height
Lateral SA = 48 × 12 = **576 cm²**
Higher-Order Thinking (HOTS) & Case-Study Question
**Question:** A toy manufacturing company designs a composite toy made of a cube and a square pyramid attached on top. The cube has a side length of 4 cm, and the pyramid's apex is 3 cm above the top face of the cube.
(a) Draw a schematic 2D representation (front view) of the composite shape.
(b) Calculate the total number of faces, vertices, and edges in the composite structure. (Consider: when the pyramid sits on the cube, the touching square face is shared, so it is no longer counted as an external face.)
(c) Verify the result using Euler's formula.
(d) Calculate the total surface area of the composite toy (use only external surfaces).
(e) If the company needs to paint the toy, and paint coverage is 10 cm² per millilitre, how much paint is required?
**Solution:**
(a) **Front View (Schematic):**
```
/\\
/ \\
/____\\
| |
| |
|______|
```
The front view is a rectangle (the cube's face) with a triangle on top (the pyramid's profile).
(b) **Counting Faces, Vertices, Edges:**
**Cube alone:** 6 faces, 8 vertices, 12 edges
**Pyramid alone:** 5 faces (1 square base + 4 triangular), 5 vertices, 8 edges
**Composite (after joining):**
- The top face of the cube and the base of the pyramid merge (no longer external).
- External faces (F) = 6 (cube) – 1 (hidden top) + 5 (pyramid) – 1 (hidden base) = **9 faces**
(5 cube faces + 4 pyramid triangles)
- Vertices (V) = 8 (cube base) + 4 (pyramid apex and 3 new vertices... wait: the pyramid's base shares the cube's top 4 vertices) = 8 + 1 = **9 vertices**
(8 at the cube's base, 1 apex of the pyramid)
- Edges (E) = 12 (cube) + 8 (pyramid edges: 4 base + 4 to apex) – 4 (shared base edges are not newly counted) = 12 + 4 = **16 edges**
(12 cube edges + 4 pyramid slant edges)
(c) **Euler's Formula:**
V – E + F = 9 – 16 + 9 = 2 ✓
(d) **Total Surface Area:**
- **Cube:** 5 external faces (bottom + 4 sides) = 5 × 4² = 5 × 16 = 80 cm²
- **Pyramid:** 4 triangular faces. Each triangle has a base of 4 cm (the side of the cube's top face) and slant height = √(3² + 2²) = √13 ≈ 3.606 cm.
Area of one triangle = ½ × base × slant height = ½ × 4 × 3.606 ≈ 7.211 cm²
Four triangles ≈ 4 × 7.211 ≈ 28.844 cm²
- **Total SA ≈ 80 + 28.844 ≈ 108.84 cm²**
(e) **Paint Required:**
Paint needed = Surface area ÷ coverage = 108.84 ÷ 10 ≈ **10.88 mL ≈ 11 mL** (rounded up for practical purposes)
Master Chapter 12 with AI-Powered Daily Drills at CBSETUTOR.ai
Visualising Solid Shapes demands consistent practice—mental rotation, net folding, and Euler's formula application don't come naturally to every student. Working through 18 important questions in isolation is a good start, but mastery requires spaced repetition and instant feedback. At cbsetutor.ai, our AI tutor personalises your learning: after you attempt a question, the system identifies your exact error (e.g., miscounting vertices, misinterpreting a view), reteaches that micro-concept, and regenerates similar questions until you achieve 95% accuracy. You'll drill all question types (MCQs, 2-marks, 3-marks, and 5-marks) in random order, simulating exam pressure. The AI also tracks your weak patterns—for instance, if you consistently struggle with pentagonal prisms, it will serve more prism-type problems before moving on. Additionally, our expert educators have recorded short video walkthroughs (3–5 min each) for every question type, so if you're stuck, you can watch a worked example in your own time. Over a 60-day pre-board cycle, students using cbsetutor.ai's Chapter 12 module improve their average score from 4.2/8 marks to 7.1/8 marks. Start a free 3-day trial today—no credit card, full access to all Chapter 12 drills and explanations.