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). The worked explanations include original diagrams for fractals, nets, cube views and surface paths. Where a printed activity has incompatible dimensions, the issue is identified explicitly.

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
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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. Use the diagrams alongside the textbook figures, especially when a viewing direction or a hidden cube affects the answer.

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).

Sierpinski carpet activity (pages 70–71)

Divide a square into nine equal squares and remove the centre. Repeat on every retained square. At steps 0, 1 and 2, the numbers of retained small squares are 1, 8, 64; the total holes are 0, 1, 9. The retained area is 1, 8/9, 64/81 of the starting area. Here a “hole” means a removed square of any size.

Carpet steps 0, 1 and 2 leave 1, 8 and 64 small squares and contain 0, 1 and 9 holes.
Carpet steps 0, 1 and 2 leave 1, 8 and 64 small squares and contain 0, 1 and 9 holes.

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.
Three Sierpinski triangle stages show how each retained triangle produces three smaller triangles.
Three Sierpinski triangle stages show how each retained triangle produces three smaller triangles.

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.
Koch stages 0, 1 and 2 have 3, 12 and 48 sides, with perimeters 3, 4 and 16/3 units.
Koch stages 0, 1 and 2 have 3, 12 and 48 sides, with perimeters 3, 4 and 16/3 units.

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(i) No; (ii) Yes; (iii) Yes; (iv) Yes; (v) No; (vi) Yes.A correct net folds to all six different faces. In (i) and (v), two squares land on the same face and another face is missing. The absence of a 2 × 2 block is only a necessary check, not a sufficient test.Cut out each shape, crease the shared edges and check that no faces overlap. The diagrams retain the textbook numbering.
Textbook cube nets ii, iii, iv and vi fold correctly; i and v make two faces overlap.
Textbook cube nets ii, iii, iv and vi fold correctly; i and v make two faces overlap.

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.

SOLUTIONThe following two diagrams show all 11 distinct nets. A rotated or reflected copy is counted only once. Each arrangement contains six squares and folds to six different faces.Draw each on square paper, cut along the outside boundary and fold along the shared edges. A strip of four side faces must have the remaining two faces on opposite sides of that strip. A two-row 3–3 net has only one shared square-width at the join; shifting the rows by just one unit would make an invalid 2 × 2 block.
Original enumeration of all 11 cube nets, part 1. Each folds to six distinct cube faces.
Original enumeration of all 11 cube nets, part 1. Each folds to six distinct cube faces.
Original enumeration of all 11 cube nets, part 2. Each folds to six distinct cube faces.
Original enumeration of all 11 cube nets, part 2. Each folds to six distinct cube faces.

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.
Six labelled rectangular faces make each 5 by 3 by 1 cm and 6 by 3 by 2 cm cuboid net.
Six labelled rectangular faces make each 5 by 3 by 1 cm and 6 by 3 by 2 cm cuboid net.

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?

SOLUTIONIn the illustrated horizontal-line examples, the top view gives the true length. The front and side views give the two perpendicular components. If their lengths are f, t and s, then t² = f² + s². For the first line s = 0, so t = f; as the side component increases, the top segment becomes longer.More generally, a projection never exceeds the true segment length L. For three mutually perpendicular viewing planes, f² + t² + s² = 2L². A segment parallel to a viewing plane keeps its length; a perpendicular segment projects to a point.
Horizontal line examples have top-view squared length equal to the sum of squared front- and side-view lengths.
Horizontal line examples have top-view squared length equal to the sum of squared front- and side-view lengths.

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.

SOLUTIONChoose and state an orientation before drawing. The following are valid examples; turning a solid can change its views.
Solid and orientationFrontTopSide
Cube: edges along the three axesSquareSquareSquare
Cuboid: edges along the axesRectangleRectangleRectangle
Parallelepiped: rectangular base on the floor; top shifted sideways onlyParallelogramRectangleRectangle
Right circular cylinder: axis verticalRectangleCircleRectangle
Right circular cone: axis verticalIsosceles triangleCircleIsosceles triangle
Right triangular prism: triangular ends face left/right; base rests on floorRectangleRectangleTriangle
Right square pyramid: base on floor, edges parallel to axesIsosceles triangleSquare with apex projected at centreIsosceles triangle
A rectangle may be a square for particular dimensions. The example parallelepiped slants only in the front-view plane; an arbitrary slanted parallelepiped need not have these three outlines.

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.

SOLUTIONNumber each projection column from 1 at the top to 8 at the bottom. The matching row number is the same in the front, top and side columns:
ObjectProjection row
Mug8
Funnel6
Hammer7
Car1
Slide/ramp3
Chair4
Ceiling fan5
Rice cooker2
Follow the F, T and S arrows printed beside each object. For the mug, the handle is clear in the side view but is hidden behind the body in the front view. The slide is identified by its long top footprint and sloping side outline.

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.

SOLUTIONRead the six textbook solids row by row, left to right. The diagrams below give their unit-square silhouettes. Top views are drawn with the back of the object at the top of the grid; in each side view, the front is at the left.Projection superimposes cubes that lie along the same line of sight: it does not add their heights. For example, the first L-shaped arrangement has three cubes, with a two-square front, L-shaped top and two-square side. Two cubes alone cannot form an L.For drawing the exterior outline, erase grid lines between adjacent squares in the same flat face. Retain visible steps and mark hidden edges with dashes where requested.
Unit-square front, top and side silhouettes for textbook cube combinations 1 to 3, read row by row from page 95.
Unit-square front, top and side silhouettes for textbook cube combinations 1 to 3, read row by row from page 95.
Unit-square front, top and side silhouettes for textbook cube combinations 4 to 6, read row by row from page 95.
Unit-square front, top and side silhouettes for textbook cube combinations 4 to 6, read row by row from page 95.

2. Imagine eight identical cubes, glued together along faces to form the letter ‘C’. (i) This looks like an ‘C’ 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 ‘C’ from the front and a different given pattern from the top. (iii) Now, can you glue even more cubes to make it look like ‘C’ 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) The printed letter is C: three cubes across its top, one in each of the two middle rows, and three across its bottom (3 + 1 + 1 + 3 = 8). The side view is a 1 × 4 column; the top view is a 3 × 1 row.(ii) Printed-drawing inconsistency: the C is three unit cubes wide, whereas the printed A top-view pattern is four unit cubes wide. Front and top projections of the same solid must have the same left-to-right extent. Therefore, these exact patterns cannot be obtained by adding identical cubes while preserving that C.(iii) The same width conflict remains. In addition, the printed A has depth four while the F side pattern is three unit cubes across. A corrected set of dimensions is needed before constructing one solid with all three views. It would be misleading to give a cube arrangement as a solution to the incompatible printed grids.(iv) One valid alternative is a one-cube-thick L of four cubes: three in a vertical column and one next to its bottom cube. It shows L from the front, a three-square column from the side, and a two-square row from the top.
The eight-cube letter is C, not E. Its side view is a four-square column and top view a three-square row.
The eight-cube letter is C, not E. Its side view is a four-square column and top view a three-square row.

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

SOLUTIONOption (ii) matches. Its front silhouette has column heights 3 and 2. Its top footprint has two squares in each of the back two rows and only the left square in the front row. Its side silhouette has heights 1, 1, 3 from front to back.All three conditions are needed: another candidate may share a front outline but have an extra cube in its footprint.
Option ii matches the three given silhouettes: a two-column stepped front, an L-shaped footprint and an L-shaped side.
Option ii matches the three given silhouettes: a two-column stepped front, an L-shaped footprint and an L-shaped side.

4. Using identical cubes, make a solid that gives the following projections (nine drawings form three groups of Top/Front/Side views).

SOLUTIONRead the drawings in three groups: (i–iii), (iv–vi), and (vii–ix). The height plans below give one solid for each group. Each number tells you how many cubes to stack at that position; a dash means leave it empty.For (i–iii), the back-to-front rows are (0,2,2), (1,1,1), (1,0,1): 9 cubes. For (iv–vi), use (2,2), (1,1), (1,0): 7 cubes. For (vii–ix), use (2,3), (1,1), (1,0): 8 cubes.Check the footprint from above, then the maximum visible height in each front and side column. Grid lines inside a flat face may be omitted in the final outline; a dashed line represents an edge hidden by nearer cubes.
One valid height plan for projections i–iii. Rows run from back to front; zero or dash means no cube.
One valid height plan for projections i–iii. Rows run from back to front; zero or dash means no cube.
One valid height plan for projections iv–vi. Rows run from back to front; zero or dash means no cube.
One valid height plan for projections iv–vi. Rows run from back to front; zero or dash means no cube.
One valid height plan for projections vii–ix. Rows run from back to front; zero or dash means no cube.
One valid height plan for projections vii–ix. Rows run from back to front; zero or dash means no cube.

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

SOLUTIONThe stack has four triangular layers, not three square layers. From the top down they contain 1, 3, 6 and 10 cubes.Include the supporting cubes hidden behind the visible outer steps. Total = 1 + 3 + 6 + 10 = 20 cubes.
The stack has four triangular layers containing 1, 3, 6 and 10 cubes, including the hidden supporting cubes.
The stack has four triangular layers containing 1, 3, 6 and 10 cubes, including the hidden supporting cubes.

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 horizontally toward a vertical edge so that two vertical faces contribute to the outline; • 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?

SOLUTIONYes. In addition to the five planar shapes, there is a tripod (one cube meets three neighbours along three perpendicular directions) and two opposite-handed twisted chains. Each uses exactly four face-connected cubes.There are 8 shapes when rotations are identified but mirror images remain distinct. The two twists are mirror images that cannot be superimposed by rotation. If mirror images are also identified, the count is 7. The diagrams show the three additions to the five flat shapes.
A tripod and two opposite-handed twisted tetracubes add three shapes to the five planar arrangements.
A tripod and two opposite-handed twisted tetracubes add three shapes to the five planar arrangements.

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.
Original isometric teaching drawings of the L-shaped solid, raised-middle solid and three-step staircase on page 101.
Original isometric teaching drawings of the L-shaped solid, raised-middle solid and three-step staircase on page 101.

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.]

SOLUTIONYes. The drawing pretends to join a closed route even though following its arrows includes upward height changes without enough downward changes to return to the starting height. This is an impossible staircase, not a physically consistent closed track.Trace a realisable section using the three grid directions. Label the height before and after each vertical step. When the path supposedly rejoins its beginning, the heights disagree. To copy the illusion, reproduce its two-dimensional outlines; do not interpret every apparent junction as a real connection.

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) A connected model of ordinary cubes cannot have all the joins shown in this impossible triangle. Its three apparent perpendicular arms are locally plausible but globally inconsistent. Therefore there is no single consistent set of front, top and side orthographic profiles for the claimed solid.A partial arm can be projected as a strip or L-shaped portion, but those local drawings cannot all be views of the complete impossible object. A specially arranged disconnected model may imitate the picture from one viewpoint; it is not the connected cube model claimed by the drawing.(ii) To recreate the picture, use the three isometric grid directions and copy the visible bands and cube divisions, allowing the apparent final join to conceal the depth conflict.(iii) The illusion works because a flat picture hides depth. Our interpretation treats overlapping edges as physical joins even though the positions cannot agree in three dimensions.

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.
Joining the three midpoints of an equilateral triangle makes four congruent equilateral triangles.
Joining the three midpoints of an equilateral triangle makes four congruent 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.
Cylinder net with two radius-r circles and a rectangle whose width is circumference 2 pi r and height is h.
Cylinder net with two radius-r circles and a rectangle whose width is circumference 2 pi r and height is h.

Find the shortest ant-to-laddu path on the 30 cm × 12 cm × 12 cm box (page 87). The ant is at the horizontal midpoint of one end face, 1 cm below its top; the laddu is at the horizontal midpoint of the opposite end face, 1 cm above its bottom.

SOLUTIONThe illustrated five-face unfolding gives perpendicular endpoint separations 32 cm and 24 cm. The straight segment stays within the unfolded faces, so it is a valid surface path.Its length is √(32² + 24²) = √1600 = 40 cm. The simpler route straight over one long face is 1 + 30 + 11 = 42 cm; going around one side gives √(42² + 10²) ≈ 43.17 cm.Unfoldings must be checked for both length and validity: a straight segment that leaves the net does not represent a route on the box. Comparing the possible face routes gives the minimum of 40 cm for these endpoints.
A valid five-face unfolding gives a 24 by 32 cm right triangle and a 40 cm ant-to-laddu path.
A valid five-face unfolding gives a 24 by 32 cm right triangle and a 40 cm ant-to-laddu path.

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 project to parallel lines when their images are distinct non-degenerate lines. They can also project to coincident lines, or to points when their direction is perpendicular to the plane. A non-degenerate projection of a parallelogram is a parallelogram; an edge-on projection collapses to a segment.

Further textbook activities

Ant on the 8 cm × 4 cm × 4 cm box (page 86)

Unfold the end face directly beside the face containing the laddu. The ant is 2 cm from the shared edge and the laddu is 2 cm along the adjacent face, giving a 4 cm horizontal separation. Their vertical separation is 2 cm. Thus the shortest adjacent-face route has length √(4² + 2²) = 2√5 cm ≈ 4.47 cm. The straight line must remain within the two unfolded rectangles.

Unfolding the adjacent faces of the 8 by 4 by 4 cm cuboid gives a straight path of 2 root 5 cm.
Unfolding the adjacent faces of the 8 by 4 by 4 cm cuboid gives a straight path of 2 root 5 cm.

Cutting corners (page 76)

Joining adjacent side-midpoints of a square removes four equal corner triangles and leaves a smaller square with half the original area. The four removed triangles can be paired and rearranged to make another square of the same area. For an equilateral triangle, cutting corners at the one-third points leaves a regular hexagon. Doing the same to a square leaves an octagon that is not regular: if the original side is a, four edges have length a/3 and the four cut edges have length a√2/3.

Nets of other solids (pages 81–84)

A regular tetrahedron has two distinct nets: an equilateral triangle subdivided into four equal triangles, and a zigzag strip of four equal equilateral triangles. Among the four candidate drawings on page 81, the first and third are valid; the second does not join all triangles along edges, and the fourth has too many triangles.

A square pyramid can be made from a square with an equal isosceles triangle attached to each of its four sides. For example, use a 4 cm square and four triangles with sides 4 cm, 5 cm, 5 cm. A triangular prism needs three rectangles in a strip plus two matching triangular ends.

A cone of base radius r and slant height l unfolds to a circle of radius r and a sector of radius l. The sector arc is 2πr, so its angle is 360° × r/l. A sphere has no exact flat paper net that folds smoothly into its surface without stretching, tearing or overlap; globe gores are approximations.

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 convex 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: If two parallel lines have distinct, non-degenerate line projections on a plane, those projections are parallel.

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 convex 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 original worked explanations and diagrams.

Textbook reference: NCERT Ganita Prakash, Chapter 11 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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