Official NCERT 2026-27 Solutions
NCERT Solutions for Class 8 Maths Chapter 11: Exploring Some Geometric Themes (Ganita Prakash Part II)
Authentic handwritten solutions in Topper Notebook style. Covers 3D isometric projections, polycube gluing, unfolding nets, shortest surface geodesics, self-similar fractals, and Penrose impossible triangles from Pages 99 to 102 of Ganita Prakash Part II.
Quick Chapter Navigation & Blueprint
- Part A: Isometric Drawing Foundations & Tetracubes (Pages 99–100) — The three principal axes (height, length, depth), isometric projection properties, and enumerating all 8 distinct ways of gluing 4 cubes together.
- Part B: Spatial Illusions & The Penrose Triangle (Pages 101–102) — The impossible ball staircase, 2D projections of non-realisable 3D shapes, and why the Penrose tribar illusion works from a single viewpoint.
- Part C: Polyhedral Nets & Surface Geodesics — Unfolding cuboids, finding the true shortest path between opposite diagonal corners on a 3D box, and folding nets for pyramids and octahedrons.
- Part D: Fractals & Infinite Self-Similarity — Mathematical construction rules for the Sierpinski Gasket, Sierpinski Carpet, and Koch Snowflake.
Official Textbook PDF Reference: Download NCERT Class 8 Maths Chapter 11 (hegp204.pdf)
Part A: Isometric Grids & Polycube Geometries (Pages 99–100)
Question 1 (Page 100 – Tetracube Configurations):
In addition to the 5 planar tetromino arrangements, are there additional 3D ways of gluing four cubes together along faces? How many distinct tetracubes exist in total?
In addition to the 5 planar tetromino arrangements, are there additional 3D ways of gluing four cubes together along faces? How many distinct tetracubes exist in total?
Topper’s Spatial Combinatorial Analysis:
- A tetracube is a 3-dimensional polycube made by joining 4 unit cubes face-to-face.
- Planar Tetracubes (Flat shapes lying in 1 plane):
• 1. I-Tetracube: A single straight column of 4 cubes (1 × 1 × 4).
• 2. O-Tetracube: A 2 × 2 square block of 4 cubes.
• 3. L-Tetracube: A row of 3 cubes with 1 cube attached at the end.
• 4. T-Tetracube: A row of 3 cubes with 1 cube attached to the central cube.
• 5. Z/Skew-Tetracube: Two 1 × 2 pairs shifted by one unit. - Non-Planar 3D Tetracubes (True 3-Dimensional shapes):
• 6. Tripod / Corner Tetracube: Three cubes joined along the X, Y, and Z axes radiating from a central corner cube.
• 7. Right-Handed Screw Tetracube: A 3-cube L-shape with the 4th cube attached pointing out of the plane in a clockwise direction.
• 8. Left-Handed Screw Tetracube: The mirror image (enantiomorph) of the right-handed screw.
Conclusion: There are 3 additional non-planar 3D ways of gluing four cubes. In total, there are exactly 8 distinct tetracubes (5 planar and 3 non-planar).
Isometric Projection Principle (Page 100):
Why is the correspondence between directions on isometric paper and the 3 axes of a solid so effective for communicating 3D shapes on a 2D sheet?
Why is the correspondence between directions on isometric paper and the 3 axes of a solid so effective for communicating 3D shapes on a 2D sheet?
Topper’s Mathematical Justification:
- Equal Angular Separation (120°): In an isometric projection, the three spatial axes (Height, Length, and Depth) are projected onto the 2D plane at exactly 120° angles to each other.
- Parallel Lines Remain Parallel: Under parallel projection, any two parallel edges on the physical 3D object remain strictly parallel in the 2D drawing.
- Uniform Unit Scale Factor: Unlike one-point or two-point perspective projections where objects shrink with distance, isometric projection preserves identical unit lengths along all three axes. A 1 cm edge along height is drawn with the exact same length as a 1 cm edge along depth or length.
Key Insight: Isometric projection preserves parallelism, collinearity, and equal unit scale across all three dimensions, allowing viewers to reconstruct 3D proportions without perspective distortion.
Part B: Spatial Illusions & The Penrose Triangle (Pages 101–102)
Question 4 (Page 101–102 – The Impossible Penrose Triangle):
Observe the Penrose triangle drawing:
(i) Would it be possible to build a closed model of this out of actual rigid physical cubes in real 3D space?
(ii) What do its front, top, and side orthographic profiles look like?
(iii) Why does the visual illusion work on an isometric grid?
Observe the Penrose triangle drawing:
(i) Would it be possible to build a closed model of this out of actual rigid physical cubes in real 3D space?
(ii) What do its front, top, and side orthographic profiles look like?
(iii) Why does the visual illusion work on an isometric grid?
Topper’s Spatial Geometry Breakdown:
- (i) Real-Space Feasibility:
NO, it is mathematically impossible to build a closed rigid 3D object that meets at all three corners with 90° angles.
In physical space, if you follow the three beams, each makes a 90° right-angle bend. In Euclidean 3D space, three mutually perpendicular segments cannot loop back to form a closed triangle! - (ii) Orthographic Profiles (Front, Top, Side Views):
When viewed as an open physical sculpture built with a deliberate depth gap:
• Front View: An L-shape with a detached third beam hovering at a different depth.
• Top View: Two perpendicular arms with a gap between the end points.
• Side View: A stepped, non-planar zigzag showing the true physical separation along the Z-axis. - (iii) Why the Illusion Works:
The 2D isometric grid lacks depth cues (such as convergence, occlusion, and shadows). The eye interprets each corner as a local 90° junction. Because the far end of the open beam aligns precisely behind the near corner from that single specific vantage point, our brain incorrectly assumes the two endpoints touch, creating the paradox of an “impossible” continuous loop!
Conclusion: The Penrose Tribar cannot exist as a closed 3D solid; it is an optical illusion created by aligning an open, disjoint spatial sculpture from a single privileged viewpoint on a flat 2D isometric projection.
Part C: Polyhedral Nets, Shortest Paths & Fractals (Page 102)
Shortest Path on a Cuboid (Page 102):
An ant wants to crawl along the outer surface of a closed rectangular box (dimensions 4 cm × 3 cm × 2 cm) from bottom corner A to the diametrically opposite top corner B. How can we determine the exact shortest path?
An ant wants to crawl along the outer surface of a closed rectangular box (dimensions 4 cm × 3 cm × 2 cm) from bottom corner A to the diametrically opposite top corner B. How can we determine the exact shortest path?
Topper’s Net Unfolding Technique:
- The shortest distance between two points on any flat plane is a straight line segment.
- To find the shortest path along the 3D surface, we unfold the adjacent faces of the cuboid into a flat 2D net!
- Depending on which pair of faces the ant crosses, there are 3 possible unfolding routes:
• Route 1 (Bottom → Front Face): Length = 4 cm, Width + Height = 3 + 2 = 5 cm.
Distance = √(42 + 52) = √(16 + 25) = √41 ≈ 6.40 cm.
• Route 2 (Bottom → Side Face): Width = 3 cm, Length + Height = 4 + 2 = 6 cm.
Distance = √(32 + 62) = √(9 + 36) = √45 ≈ 6.71 cm.
• Route 3 (Side Face → Top Face): Height = 2 cm, Length + Width = 4 + 3 = 7 cm.
Distance = √(22 + 72) = √(4 + 49) = √53 ≈ 7.28 cm. - Comparing the three values: √41 < √45 < √53.
Final Answer: The exact shortest path is √41 cm (approximately 6.40 cm), obtained by unfolding the bottom and front faces flat and drawing a straight line between the corners.
Mathematical Fractals (Page 102):
What is a fractal? Describe the recursive construction of the Sierpinski Gasket (Sierpinski Triangle).
What is a fractal? Describe the recursive construction of the Sierpinski Gasket (Sierpinski Triangle).
Recursive Self-Similarity Rules:
- A fractal is a geometric figure in which each part has the same statistical character as the whole (self-similarity across different magnification scales).
- Constructing the Sierpinski Gasket:
• Step 0: Start with a solid equilateral triangle of area A.
• Step 1: Connect the midpoints of all three sides. This divides the triangle into 4 smaller congruent equilateral triangles. Remove the central inverted triangle (Area remaining = 3/4 A).
• Step 2: In each of the remaining 3 solid triangles, connect their midpoints and remove the central triangle (Total removed = 1 + 3 = 4 triangles; Area remaining = (3/4)2 A = 9/16 A).
• Step n: Repeating this process infinitely yields the Sierpinski Gasket, which has area approaching 0, but total perimeter approaching infinity!
Key Feature: Fractals exhibit infinite detail, fractional dimension, and recursive self-similarity, commonly found in nature in ferns, snowflakes, lightning bolts, and river networks.