Class 8 Maths Ganita Prakash Chapter 11 Solutions (NCERT 2026–27) – Exploring Some Geometric Themes

These Class 8 Maths Ganita Prakash Chapter 11 solutions cover Exploring Some Geometric Themes from the new NCF-2023 textbook (2026–27). This is Chapter 4 of Ganita Prakash Part II (the 11th chapter of the Class 8 course). Every “Figure it Out” set, Math Talk and reflective question is answered step by step, with verified counts of faces, edges and vertices, Euler checks, and exact shortest-path lengths.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 11 (Part II Ch 4) Exercises: Figure it Out sets, Math Talk Session: 2026–27
On this page

Chapter 11 Overview

Chapter 11, Exploring Some Geometric Themes (Chapter 4 of Ganita Prakash Part II), explores two big ideas. The first is fractals — self-similar shapes that repeat the same pattern at smaller and smaller scales, like the fern, the Sierpinski Carpet, the Sierpinski Gasket and the Koch Snowflake. The second is visualising solids — faces, edges and vertices of prisms and pyramids, building solids from nets, finding the shortest path on the surface of a cuboid by unfolding it, and representing solids on paper through projections (front, top and side views) and isometric drawing. The solutions below work through every exercise in the chapter with full, exam-ready reasoning.

Key Concepts & Definitions

Fractal: a self-similar shape that contains smaller copies of itself, repeating the same pattern at smaller and smaller scales (e.g. fern, coastline, Sierpinski Carpet, Koch Snowflake).

Face, Edge, Vertex: Faces are the flat surfaces forming the boundary of a solid; edges are the line segments where faces meet; vertices are the points where edges meet. A cube or cuboid has 6 faces, 12 edges and 8 vertices.

Prism: a solid with two congruent polygons as opposite faces and parallelograms as the side faces (triangular prism, pentagonal prism, …).

Pyramid: a solid with one polygonal base and a single apex joined to every vertex of the base (triangular pyramid = tetrahedron, square pyramid, …).

Net: a flat shape that can be folded to form a solid — obtained by ‘unfolding’ the solid onto a plane.

Projection / Views: the shadow-like image of a solid on a plane. The three standard views are the front view (on the vertical plane), top view (on the horizontal plane) and side view (on the side plane).

Isometric projection: an orientation in which the projections of all edges of a cube have equal length; drawn on a triangular (isometric) grid.

Important Formulas & Facts (Chapter 11)

Euler’s relation for a (convex) polyhedron: F − E + V = 2, where F = faces, E = edges, V = vertices.

n-sided prism: F = n + 2, E = 3n, V = 2n. Check: (n+2) − 3n + 2n = 2. ✓

n-sided pyramid: F = n + 1, E = 2n, V = n + 1. Check: (n+1) − 2n + (n+1) = 2. ✓

Sierpinski Carpet: remaining squares Rn = 8n; holes Hn = 1 + 8 + … + 8n−1 = (8n − 1)/7. Area remaining = (8/9)n.

Sierpinski Gasket (Triangle): remaining triangles Rn = 3n; holes Hn = (3n − 1)/2. Area remaining = (3/4)n.

Koch Snowflake: number of sides = 3 × 4n; perimeter = 3 × (4/3)n (starting side = 1 unit).

Shortest path on a cuboid: unfold the cuboid into a net so the path becomes a straight line; its length is found by the Baudhāyana (Pythagoras) theorem, d = √(a2 + b2).

Figure it Out — Sierpinski Gasket

From the Sierpinski Gasket section. An equilateral triangle is divided into 4 identical equilateral triangles by joining the midpoints of its sides, and the central triangle is removed; the procedure is repeated on the 3 remaining triangles, and so on. (Take Step 0 = the full triangle.)

1. Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Triangle.

SOLUTION Step 0: one full equilateral triangle (no hole). Step 1: join the midpoints of the three sides to get 4 small triangles; remove the middle (inverted) one. This leaves 3 upright triangles with 1 central hole. Step 2: repeat the same step inside each of the 3 remaining triangles. Each gives 3 smaller triangles and 1 small hole, so we get 9 small upright triangles and a total of 4 holes (1 large + 3 small). Each step looks like the previous one copied at half the size in its three corners — this is the self-similarity of the fractal.

2. Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

SOLUTION Each remaining triangle splits into 4 and 1 is removed, so every remaining triangle gives 3 new remaining triangles: Rn+1 = 3Rn, with R0 = 1, giving Rn = 3n. Each remaining triangle at step n creates one new hole at step (n + 1), and old holes stay: Hn+1 = Hn + Rn, with H0 = 0. So Hn = 1 + 3 + 32 + … + 3n−1 = (3n − 1) / 2.
Step nRemaining triangles Rn = 3nHoles Hn = (3n−1)/2
010
131
294
32713
n3n(3n − 1)/2

3. Find the area of the region remaining at the nth step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit.

SOLUTION Sierpinski Gasket (Triangle): the triangle is cut into 4 equal triangles and 1 is removed, so each step keeps 3⁄4 of the previous area. Starting from area 1, the area remaining at step n = (3/4)n sq. unit. Sierpinski Carpet: the square is cut into 9 equal squares and 1 is removed, so each step keeps 8⁄9 of the area. The area remaining at step n = (8/9)n sq. unit. In both cases the remaining area → 0 as n → ∞, even though a pattern always remains — a striking property of fractals.

Figure it Out — Koch Snowflake

From the Koch Snowflake section. Start with an equilateral triangle; on each side, divide it into 3 equal parts, raise an equilateral triangle on the middle part, then remove that middle part, so each side becomes a ‘bump’. Repeat on every new side. (Take Step 0 = the starting triangle, which has 3 sides.)

1. Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.

SOLUTION Step 0: an equilateral triangle (3 straight sides). Step 1: each of the 3 sides becomes a bump — one side splits into 4 equal segments, giving a six-pointed star (Star of David) shape with 12 sides. Step 2: apply the same bump rule to every one of the 12 sides; each becomes 4 segments, giving 48 sides and a more detailed snowflake outline. Continuing forever gives the Koch Snowflake, whose boundary becomes ever more crinkled.

2. Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.

SOLUTION Each side is replaced by 4 smaller sides, so the number of sides is multiplied by 4 at every step: Sn+1 = 4Sn. With S0 = 3 (the triangle), Sn = 3 × 4n. Check: S1 = 3 × 4 = 12, S2 = 3 × 16 = 48. So number of sides at step n = 3 × 4n.

3. Find the perimeter of the shape at the nth step of the sequence. Take the starting equilateral triangle to have a sidelength of 1 unit.

SOLUTION At each step every segment is split into 3 equal parts and 4 such parts make up the new bump, so each side’s length becomes 4⁄3 of the old, while one segment’s length becomes 1⁄3 of the old. Length of one side at step n = (1/3)n, and number of sides = 3 × 4n. Perimeter Pn = (number of sides) × (length of a side) = 3 × 4n × (1/3)n = 3 × (4/3)n units. Check: P0 = 3, P1 = 3 × 4/3 = 4, P2 = 3 × 16/9 = 16/3 ≈ 5.33. The perimeter grows without bound, even though the snowflake encloses a finite area.

Figure it Out — Nets of a Cube and Cuboid

1. Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try.

SOLUTION A flat arrangement of 6 squares is a net of a cube only if, on folding, the 6 squares wrap to form all 6 faces with no two squares overlapping and no face left missing. A quick test: a valid cube net never has four squares meeting around a single point (a 2 × 2 block of squares cannot fold to a cube), and it cannot be a straight strip of more than 4 squares in a row. Apply this test to each figure (i)–(vi): figures that are simple straight strips of 6 squares, or that contain a 2 × 2 square block, are not nets of a cube; the ‘cross’, ‘T’ and ‘staircase/zig-zag of 6 squares’ arrangements that pass the test are valid cube nets. Confirm any doubtful one by cutting it out and folding.

2. A cube has 11 possible net structures in total. In this count, two nets are considered the same if one can be obtained from the other by a rotation or a flip. Find all the 11 nets of a cube.

SOLUTION There are exactly 11 distinct nets of a cube. They are usually grouped like this: (a) “1-4-1” type — 6 nets: a row of 4 squares, with 1 square attached above and 1 below the row. The top square can sit above any of the 4 positions and the bottom square below any of the 4, and removing repeats by rotation/flip leaves 6 different nets. (b) “2-3-1” type — 3 nets: a row of 3 squares, with 2 squares attached as a step on one side and 1 square on the other. (c) “2-2-2” staircase type — 1 net: three pairs of squares in a zig-zag staircase. (d) “3-3” type — 1 net: two rows of 3 squares offset by one square. Total = 6 + 3 + 1 + 1 = 11 nets. Drawing one example from each group and its variations on squared paper reproduces all eleven.

3. Draw a net of a cuboid having sidelengths: (i) 5 cm, 3 cm, and 1 cm (ii) 6 cm, 3 cm, and 2 cm

SOLUTION A cuboid has 3 pairs of equal rectangular faces. The easiest ‘cross’ net is a vertical strip of 4 rectangles (front, top, back, bottom) with the 2 side faces attached left and right of one of them. (i) 5 × 3 × 1: central strip of four rectangles measuring 5×1, 5×3, 5×1, 5×3 (front, top, back, bottom, all 5 cm wide); attach two 3×1 side rectangles to the left and right of one 5×3 face. Faces used: two 5×3, two 5×1 and two 3×1 rectangles. (ii) 6 × 3 × 2: similarly use two 6×3, two 6×2 and two 3×2 rectangles arranged as a cross. When folded along the edges these wrap into the cuboid. Mark the fold lines and label each rectangle so the matching faces are clear; verify by cutting and folding.

Figure it Out — Projections (Lines & Cuboids)

1. Observe the front view, top view and side view of the different lines in Fig. 4.6. Is there any relation between their lengths?

SOLUTION The projection of a line segment onto a plane is never longer than the segment itself: p ≤ l, where l is the actual length and p the projected length. The projection equals the actual length only when the line is parallel to that plane; it shrinks to a point when the line is perpendicular to the plane. So for a given slanted line the three views (front, top, side) generally have different lengths, each ≤ the true length, depending on how the line is tilted relative to that plane.

2. Find the front view, top view and side view of each of the following solids, fixing its orientation with respect to the vertical, horizontal and side planes: cube, cuboid, parallelepiped, cylinder, cone, prism, and pyramid.

SOLUTION Taking a natural ‘sitting’ orientation for each solid, the three views are:
SolidFront viewTop viewSide view
CubeSquareSquareSquare
CuboidRectangleRectangleRectangle
ParallelepipedParallelogram/rectangleParallelogramParallelogram/rectangle
Cylinder (upright)RectangleCircleRectangle
Cone (upright)TriangleCircle (with centre point)Triangle
Triangular prismRectangleRectangleTriangle
Square pyramidTriangleSquare (with diagonals)Triangle
The exact rectangles/triangles depend on the chosen orientation, but the shapes of the outlines are as above.

3. Match each of the following objects with its projections (FRONT, TOP, SIDE) — a mug, funnel, hammer, car, slide/ramp, chair, ceiling fan and rice cooker shown with their three views.

SOLUTION Match each object to the row whose three outlines fit it from the front, top and above: Car → front = car face, top = rounded rectangle, side = car silhouette. Funnel → front and side = inverted triangle/‘V’, top = circle with a small inner circle. Hammer → front = slim handle with head, top = small T/peg, side = hammer profile. Chair → front = back-rest above seat, top = square seat outline, side = the familiar chair ‘h’ shape. Ceiling fan → front and side = thin horizontal blades with central body, top = the three blades radiating from the centre. Mug → front = body with handle, top = circle with handle bump, side = body with handle. Rice cooker → front = rounded body, top = circle, side = rounded body with handle. In general, match by the top view first (a circle points to mug/funnel/cooker/fan), then separate them using the front and side outlines.

Figure it Out — Views of Cube Combinations

1. Draw the top view, front view and the side view of each of the following combinations of identical cubes.

SOLUTION For each combination, look straight along each axis and shade the squares you would see: Front view = the outline you see looking along the ‘Front’ arrow; Top view = the outline looking straight down; Side view = the outline looking along the ‘Side’ arrow. Each view is a flat pattern of unit squares. For an L-shaped pair of cubes, for instance, two of the views are an L of squares and the third is a single row, depending on the direction. Draw each view as squares on grid paper, counting how many cubes line up in that direction. Key idea: a stack of cubes hidden directly behind another shows only one square in that view (overlapping cubes project onto the same square).

2. Imagine eight identical cubes, glued together along faces to form the letter ‘E’. (i) This looks like an ‘E’ from the front. What does it look like from the side? From the top? (ii) Glue additional cubes to make a shape that looks like ‘E’ from the front and a different given pattern from the top. (iii) Now, can you glue even more cubes to make it look like ‘E’ from the front, a given pattern from the top, and another given pattern from the side? (iv) Can you think of other letter combinations to make with a single combination of cubes in this manner?

SOLUTION (i) A flat ‘E’ built from one layer of cubes (depth 1) shows the ‘E’ from the front. From the side it looks like a single vertical column (a tall thin rectangle), because the whole ‘E’ is only one cube deep. From the top it looks like a short horizontal row of squares (the top edge of the ‘E’). (ii) To change the top view, add cubes behind some columns of the ‘E’ (increasing depth). This does not spoil the front ‘E’ (the new cubes hide behind), but the top view now shows the new depth pattern. (iii) To control all three views at once, add cubes so that each direction’s shadow matches its target shape; this is possible because the front, top and side views constrain different directions. Build it layer by layer, checking each view as you go. (iv) Yes — for example a single cube arrangement can read as ‘L’ from the front and ‘I’ from the side, or you can design combinations giving different letters such as ‘T’, ‘H’, ‘F’ from different viewpoints. Many answers are possible.

3. Which solid corresponds to the given top view, front view, and side view? (Choose from the solids (i)–(vii) made of glued cubes.)

SOLUTION Compare each candidate solid’s three views with the given ones. The correct solid is the one whose front, top and side outlines all match simultaneously. Method: first eliminate solids whose front view is wrong, then among the rest eliminate those whose top view is wrong, and finally check the side view of the survivor. The matching solid is the cube-stack whose three projections agree with all three given drawings (the C-shaped / stepped arrangement shown); the others fail at least one of the three views.

4. Using identical cubes, make a solid that gives the following projections (nine sets of Top/Front/Side views are given).

SOLUTION For each given pair/triple of views, place cubes so that: • the number of cube-columns matches the front view across the width and height, • the depth (front-to-back) matches the side view, • the footprint on the ground matches the top view. Start with the top view as the floor plan, then raise stacks of the right height so the front and side silhouettes come out correctly. There is often more than one valid solid for the same views, because hidden cubes can sometimes be added or removed without changing any silhouette.

5. Find the number of cubes in this stack of identical cubes.

SOLUTION The stack is a step-pyramid of square layers. Counting the visible square layers from top to bottom: 1 cube on top, then a 2 × 2 layer, then a 3 × 3 layer. Total = 1 + 4 + 9 = 14 cubes (assuming the layers are full square layers, 12 + 22 + 32).

6. What are the different shapes the projection of a cube can make under different orientations?

SOLUTION Depending on how the cube is turned, its outline (projection) can be: • a square — looking straight at a face; • a rectangle — looking along an edge (two faces seen edge-on); • a regular hexagon — looking along the diagonal through opposite vertices (the isometric view). In between these, the outline can also be a non-regular hexagon. So the main outlines are square, rectangle and (regular) hexagon.

Figure it Out — Isometric Grids

1. In addition to the 5 ways shown in Fig. 4.8, are there any additional ways of gluing four cubes together along faces? Can you visualise and draw these as well?

SOLUTION Fig. 4.8 shows the five flat (planar) tetromino shapes — the I, O (square), L, S/Z and T arrangements of four cubes lying in one layer. When the cubes may be stacked in 3-D (not just flat), there are extra three-dimensional ways of joining four cubes — for example three cubes in an L with the fourth cube placed on top, forming a non-flat ‘tripod’ / twisted shape. Counting all distinct shapes (tetracubes), there are 8 ways in total: the 5 flat ones plus 3 genuinely 3-D ones. Draw the 3 solid ones on isometric paper by stacking a cube above one end of an L or T.

2. Draw the following figures on the isometric grid. [Hint: decide for each edge whether it goes up or down along the height, and draw it along the height axis or opposite to it.]

SOLUTION Associate the three grid directions with the three axes: vertical ‘|’ = height, ‘/’ = depth, ‘\’ = length. Draw each solid edge by edge, counting units along each axis. Begin with the front-bottom edge, build the front face, then push the depth edges back by the required number of units, and finally add the top. Darken only the visible edges and leave hidden edges faint (or dashed). Following the hint, choose the up/down direction of each height edge so the steps and notches in the figures (L-shape, T-shape, staircase) come out correctly.

3. Is there anything strange about the path of this ball? Recreate it on the isometric grid. [Hint: Consider a portion of this figure that is physically realisable and identify the 3 primary directions.]

SOLUTION Yes — the picture is an optical illusion: as a whole it cannot exist as a real 3-D object, because the ball appears to roll ‘uphill’ and return to where it started, which is impossible. The trick is that the drawing joins parts that are each realisable on their own, but are connected in a way that confuses near and far. Each separate ramp or step can be drawn on the isometric grid using the three axes. To recreate it, draw each realisable portion (a ramp going up along the height and length axes) on the isometric grid, then join them where the illusion hides the depth — this reproduces the ‘strange’ looping path.

4. Observe this triangle. (i) Would it be possible to build a model out of actual cubes? What are the front, top, and side profiles of this impossible triangle? (ii) Recreate this on an isometric grid. (iii) Why does the illusion work?

SOLUTION (i) This is the famous Penrose impossible triangle. It cannot be built from actual cubes as a single connected solid, because its three arms cannot all meet at right angles in real 3-D space. Each arm on its own is a straight bar of cubes, so the front, top and side profiles are simply straight strips (rectangles) of squares — each view is an ordinary L-shaped or bar-shaped arrangement. (ii) On the isometric grid, draw three straight bars of cubes along the three axes (depth, length, height) and arrange them as an open triangle; let the ends overlap in the drawing as in the figure. (iii) The illusion works because a 2-D drawing hides depth information: at each corner the picture lets a near edge and a far edge appear to touch. Our eyes assume the bars are joined where they merely overlap on paper, so the brain ‘closes’ the triangle even though the real bars never actually meet.

Math Talk & Reflective Questions

Show that by joining the midpoints of an equilateral triangle, we divide it into 4 identical equilateral triangles. [Hint: Note that the corner triangles are isosceles.]

SOLUTION Join the midpoints of the three sides. Each midsegment is parallel to a side and equal to half of it (midpoint theorem). So the three midsegments all have the same length (half the side of the big triangle), and the central triangle they form is equilateral with side = half the original. Each corner triangle has two sides equal to half the original side, with the included angle 60° (an angle of the equilateral triangle), so it too is equilateral with side = half the original. Hence all 4 small triangles have equal sides — they are 4 identical equilateral triangles.

If the congruent polygons of a prism have 10 sides, how many faces, edges and vertices does the prism have? What if the polygons have n sides?

SOLUTION 10-sided prism (decagonal): Faces = 10 side faces + 2 polygon faces = 12 faces; Vertices = 10 on top + 10 on bottom = 20 vertices; Edges = 10 top + 10 bottom + 10 vertical = 30 edges. n-sided prism: Faces = n + 2, Vertices = 2n, Edges = 3n. Euler check: F − E + V = (n + 2) − 3n + 2n = 2. ✓

If the base of a pyramid has 10 sides, how many faces, edges and vertices does the pyramid have? What if the base is an n-sided polygon?

SOLUTION 10-sided base (decagonal pyramid): Faces = 10 triangular faces + 1 base = 11 faces; Vertices = 10 base + 1 apex = 11 vertices; Edges = 10 base + 10 slant = 20 edges. n-sided pyramid: Faces = n + 1, Vertices = n + 1, Edges = 2n. Euler check: F − E + V = (n + 1) − 2n + (n + 1) = 2. ✓

What are the sidelengths of the rectangle obtained in the net of a cylinder? (Cut along the height after unrolling the curved surface.)

SOLUTION When the curved surface of a cylinder of radius r and height h is unrolled, it opens into a rectangle. One side equals the height of the cylinder, h. The other side equals the circumference of the base, 2πr (the circle ‘unrolls’ into a straight edge). So the rectangle has sidelengths h and 2πr; the net also has the two circular ends (each of radius r) attached at the top and bottom.

Find the shortest path between the ant and the laddu for the cuboid with the ant at the centre of one face, where the relevant unfolded dimensions are 24 cm and 32 cm (the laddu near the bottom edge).

SOLUTION Unfold the cuboid into a net so the ant’s path becomes a straight line. This gives a right triangle whose legs are 24 cm and 32 cm. By the Baudhāyana (Pythagoras) theorem: d2 = 242 + 322 = 576 + 1024 = 1600. d = √1600 = 40 cm. Because different unfoldings give different straight-line distances, we must try all sensible unfoldings and take the smallest; here 40 cm is that shortest length.

When is the length of the projected line equal to its actual length? What can you say about the projection of an n-sided regular polygon, and of a pair of parallel lines?

SOLUTION A projected line equals its actual length only when the line is parallel to the plane of projection; otherwise the projection is shorter (and becomes a point if the line is perpendicular to the plane). The projection of an n-sided polygon is made of the projections of its n sides, so it is again a closed figure with up to n sides — but lengths and angles can change, so a regular polygon need not project to a regular one. Parallel lines always project to parallel lines. Hence a parallelogram always projects to a parallelogram (never to a non-parallelogram quadrilateral).

Common Mistakes to Avoid

Watch out for these

  • Confusing the fractal counts: for the Sierpinski Gasket use 3n triangles (it is the Carpet that uses 8n squares).
  • Forgetting Step 0 starts the count — the Koch snowflake has 3 sides at Step 0, so sides = 3 × 4n, not 4n.
  • Miscounting a pyramid’s edges: an n-sided pyramid has 2n edges (n base + n slant), not 3n like a prism.
  • Saying a 2 × 2 block of squares is a cube net — it is not; four squares cannot meet at one point in a cube net.
  • Using the wrong unfolding for the shortest path — always compare all unfoldings and pick the smallest straight-line distance.
  • Assuming a projection (view) keeps the true length — it does so only when the edge is parallel to that plane.
  • Leaving answers like √1600 unsimplified — simplify to 40 when it is a perfect square.

Practice MCQs & Assertion–Reason

1. A fractal is best described as a shape that is:

(a) always a circle    (b) self-similar at smaller and smaller scales    (c) always two-dimensional    (d) made of only straight lines

2. In the Sierpinski Gasket, the number of remaining triangles at step n is:

(a) 8n    (b) 4n    (c) 3n    (d) 2n

3. The number of sides of the Koch Snowflake at step n is:

(a) 4n    (b) 3 × 4n    (c) 3n    (d) 3 + 4n

4. A cuboid has how many faces, edges and vertices?

(a) 6, 12, 8    (b) 8, 12, 6    (c) 6, 8, 12    (d) 4, 6, 4

5. Euler’s relation for a polyhedron is:

(a) F + E + V = 2    (b) F − E + V = 2    (c) F + E − V = 2    (d) F − E − V = 2

6. A pentagonal prism has how many edges?

(a) 10    (b) 12    (c) 15    (d) 20

7. A square pyramid has how many faces?

(a) 4    (b) 5    (c) 6    (d) 8

8. The total number of distinct nets of a cube is:

(a) 6    (b) 8    (c) 11    (d) 12

9. When the curved surface of a cylinder of radius r and height h is unrolled, the rectangle has sides:

(a) r and h    (b) 2πr and h    (c) πr2 and h    (d) 2r and h

10. If a cuboid is unfolded into a right triangle with legs 24 cm and 32 cm, the shortest surface path is:

(a) 28 cm    (b) 40 cm    (c) 56 cm    (d) 1600 cm

Answer key: 1-(b), 2-(c), 3-(b), 4-(a), 5-(b), 6-(c), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The Sierpinski Carpet keeps (8/9)n of the original area at step n.

Reason: At each step the square is divided into 9 equal squares and the central one is removed.

A-R 2. Assertion: A triangular pyramid (tetrahedron) has 4 faces, 6 edges and 4 vertices.

Reason: For an n-sided pyramid, F = n + 1, E = 2n and V = n + 1.

A-R 3. Assertion: Every arrangement of 6 squares is a net of a cube.

Reason: A cube has 6 faces.

A-R 4. Assertion: The shortest path between two points on a cuboid’s surface can be found by unfolding it into a net.

Reason: A path on the surface transforms to a path of the same length on the net, and a straight line on the net is the shortest.

A-R 5. Assertion: The projection of a pair of parallel lines is always a pair of parallel lines.

Reason: A projection can never change the length of any line segment.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(C).

Quick Revision Summary

  • Fractals are self-similar shapes (fern, Sierpinski Carpet/Gasket, Koch Snowflake) repeating at smaller scales.
  • Sierpinski Gasket: triangles = 3n, holes = (3n−1)/2, area = (3/4)n. Sierpinski Carpet: area = (8/9)n.
  • Koch Snowflake: sides = 3 × 4n, perimeter = 3 × (4/3)n — finite area but infinite perimeter.
  • Faces, edges, vertices: cube/cuboid have 6, 12, 8. Euler: F − E + V = 2.
  • n-sided prism: F = n+2, E = 3n, V = 2n. n-sided pyramid: F = n+1, E = 2n, V = n+1.
  • A net folds into a solid; a cube has 11 nets, a regular tetrahedron has 2, an octahedron has 11.
  • Shortest path on a cuboid = straight line on a suitable unfolded net (use Pythagoras, e.g. √(242+322) = 40).
  • Solids are drawn using front, top and side views (projections), and isometric drawing on a triangular grid.

How to score full marks in this chapter

State the rule before computing — e.g. write “each triangle → 3 triangles, so Rn = 3n” before tabulating. For faces/edges/vertices, always finish with the Euler check F − E + V = 2. For shortest-path problems, draw the unfolded net, mark the right triangle, and simplify surds (√1600 = 40). For projection and net questions, label your diagram clearly and verify a net by mentally folding it.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 11 about?

Chapter 11, Exploring Some Geometric Themes (Chapter 4 of Ganita Prakash Part II), covers two themes: fractals (self-similar shapes like the Sierpinski Carpet/Gasket and Koch Snowflake) and visualising solids (faces, edges, vertices, nets, shortest paths on cuboids, projections and isometric drawing).

What is Euler’s relation for solids?

For a polyhedron, F − E + V = 2, where F is the number of faces, E the number of edges and V the number of vertices. For example a cube gives 6 − 12 + 8 = 2.

How many nets does a cube have?

A cube has exactly 11 distinct nets (counting rotations and flips as the same). A regular tetrahedron has 2 nets and an octahedron has 11.

Are these Class 8 Maths Ganita Prakash Chapter 11 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash textbook (Part II) for 2026–27, with questions reproduced verbatim and original step-by-step answers.

← Chapter 10 All Ganita Prakash Chapters Chapter 12 →
Scroll to Top