In-text — Faces, Edges, Vertices and Nets — Exploring Some Geometric Themes

1. 13 marksGanita Prakash Cl-8 Part 2, Math Talk, page 79

   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?

   Hint. Build it up in pieces: two end polygons, then the side faces joining them.

   Build the prism from its parts. A prism has two congruent polygons as opposite faces, with edges joining corresponding vertices, and all the other faces parallelograms. Counting is easiest if each kind of part is counted separately, since every vertex, edge and face belongs to exactly one of the groups below.

   Vertices. Each of the two polygons has n corners, and no others exist, so V = 2n

   Faces. Two polygon faces, plus one side face for each side of the polygon: F = n + 2

   Edges. The n edges of the top polygon, the n edges of the bottom polygon, and n vertical edges joining corresponding vertices: E = 3n

   For a 10-sided polygon (n = 10): Faces = 10 + 2 = 12 Edges = 3 × 10 = 30 Vertices = 2 × 10 = 20

   Check with Euler's relation. For any solid of this kind, V − E + F = 2: 20 − 30 + 12 = 2 ✓ And in general 2n − 3n + (n + 2) = 2 ✓ for every n — a check that costs nothing and catches slips instantly.

   n	Prism	Faces	Edges	Vertices
   3	triangular	5	9	6
   5	pentagonal	7	15	10
   6	hexagonal	8	18	12
   10	decagonal	12	30	20
   n	—	n + 2	3n	2n

   ✦ A prism on a 10-sided polygon has 12 faces, 30 edges and 20 vertices. In general: F = n + 2, E = 3n, V = 2n.

2. 23 marksGanita Prakash Cl-8 Part 2, Math Talk, page 79

   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?

   Hint. A pyramid is a base polygon plus one apex point joined to every base vertex.

   Build the pyramid from its parts. A pyramid has a polygonal base and one point (the apex) outside it, with edges joining the apex to each vertex of the base.

   Vertices. The n corners of the base, plus the apex: V = n + 1

   Faces. The base itself, plus one triangular face for each side of the base: F = n + 1

   Edges. The n edges of the base, plus n edges running from the apex to each base vertex: E = 2n

   For a 10-sided base (n = 10): Faces = 10 + 1 = 11 Edges = 2 × 10 = 20 Vertices = 10 + 1 = 11

   Check with Euler's relation. 11 − 20 + 11 = 2 ✓ In general (n + 1) − 2n + (n + 1) = 2 ✓

   n	Pyramid	Faces	Edges	Vertices
   3	triangular (tetrahedron)	4	6	4
   4	square	5	8	5
   6	hexagonal	7	12	7
   10	decagonal	11	20	11
   n	—	n + 1	2n	n + 1

   Worth noticing. A pyramid always has the same number of faces as vertices, whatever its base — because the base contributes one face and n vertices, while the apex contributes one vertex and n faces. A prism never has this property.

   ✦ A pyramid on a 10-sided base has 11 faces, 20 edges and 11 vertices. In general: F = n + 1, E = 2n, V = n + 1 — and faces always equal vertices.

3. 33 marksGanita Prakash Cl-8 Part 2, Math Talk, page 82

   What is the net of a cylinder, and what are the sidelengths of the rectangle obtained?

   Hint. Unroll the curved surface and ask what its bottom edge was before you unrolled it.

   Unfold in two moves. Take a cylinder of radius r and height h. First lift off the two circular ends. Then cut the curved surface straight down along a line parallel to the axis, and unroll it flat.

   The net has three pieces: two circles of radius r, and one rectangle.

   The rectangle's sidelengths.

   One side is the height. Cutting along a line of length h and flattening does not stretch that line, so one side of the rectangle is h.

   The other side was the rim. The other pair of sides used to be the circular edges where the curved surface met the end circles. Unrolling straightens each rim without stretching it, so that side has the length of the circumference of the base: 2πr, which is the same as πd where d is the diameter.

   So the rectangle is 2πr by h.

   A quick check with a real object. Peel the label off a tin. It comes away as a rectangle, its short side the height of the tin and its long side exactly enough to wrap once round — which is what makes labels the everyday proof of this net.

   Why it matters. This net is where the formulas come from: curved surface area = 2πrh (the rectangle) total surface area = 2πrh + 2πr² (the rectangle plus the two circles)

   A warning. A cylinder's curved surface unrolls flat because it is made of straight lines — you could sweep a ruler along it. A sphere has no such lines, which is why it has no net at all.

   ✦ The net is two circles of radius r plus a rectangle whose sides are the height h and the base circumference 2πr.

4. 44 marksGanita Prakash Cl-8 Part 2, Math Talk, page 82

   How will the net of a cone look? If the cone is slit open along the slant line l and unrolled, what do we get? And what surface do you construct from such a net when O is not the centre of the boundary circle?

   Hint. Every point on the base rim is the same distance from the apex — ask what that means once the surface is flat.

   What unrolls to what. Take a cone with base radius r and slant height l, and let O be its apex. Cut along one slant line and unroll the curved surface.

   Why the result is a sector. Every point on the rim of the base is at exactly the same distance — the slant height l — from the apex O. Unrolling does not change distances measured along the surface, so after flattening, every point of that rim is still at distance l from O. Points all at the same distance from O lie on a circle centred at O. Hence the outer boundary of the net is an arc of a circle of radius l centred at O, and the net of the curved surface is a sector.

   How big is the sector? Its arc used to be the base rim, so the arc length is 2πr. A full circle of radius l has circumference 2πl, so the sector is the fraction 2πr ÷ 2πl = r/l of a full circle, and its angle is (r/l) × 360°.

   Example. r = 3 cm and l = 6 cm gives a half-circle — a 180° sector.

   The full net is that sector plus a circle of radius r for the base.

   When O is not the centre of the boundary circle. Now the point O is at different distances from different points of the boundary. Rolling such a net up, the distances from the apex down to the rim are no longer all equal, so the apex cannot sit directly above the centre of the base — the surface leans. What you build is an oblique cone-like surface: a slanted cone rather than a right circular one, and in general its base curve is not a circle either.

   Test it, as the book asks. Cut a paper disc, mark a point O away from its centre, cut along a line from O to the edge, and roll it up so the two cut edges meet. You get a lopsided, leaning cone — worth making, because the drooping shape is much more convincing than the argument.

   ✦ Slitting and unrolling a cone gives a sector of radius l (the slant height) whose arc is the base circumference 2πr, so its angle is (r/l) × 360°; the full net adds a circle of radius r for the base. If O is not the centre of the boundary circle, the slant distances are unequal, and rolling it up gives a leaning oblique cone rather than a right circular one.

5. 53 marksGanita Prakash Cl-8 Part 2, Math Talk, pages 82-84

   How many nets does a regular tetrahedron have? A cube? An octahedron? A dodecahedron? And is there a net of a sphere?

   Hint. Two nets count as the same if one can be turned or flipped onto the other.

   The counting convention first. Two nets count as the same if one can be obtained from the other by a rotation or a flip. Without that rule the counts would be much larger and far less interesting.

   The counts given in the chapter.

   Solid	Faces	Number of nets
   Regular tetrahedron	4 equilateral triangles	2
   Cube	6 squares	11
   Octahedron	8 equilateral triangles	11
   Dodecahedron	12 regular pentagons	43,380

   Worth pausing on two of these. The cube and the octahedron both have 11 nets — not a coincidence, since each is the "dual" of the other (join the centres of a cube's faces and you get an octahedron). And the jump from 11 to 43,380 for the dodecahedron shows how quickly the counting explodes once there are twelve faces to unfold.

   The octahedron is easy to picture: it is two square pyramids joined base to base. Taking all its triangles to be equilateral and folding one of its nets is a good way to convince yourself the eight faces really do close up.

   And the sphere — no net exists. A net is what you get by unfolding a solid onto a plane without stretching or tearing. A cylinder's curved surface unrolls because it is built from straight lines; a cone's does too. A sphere has no straight lines on it anywhere. Try to flatten a piece of it and it must either tear at the edges or wrinkle in the middle — which is exactly what happens when you peel an orange and try to press the skin flat.

   This is not a failure of cleverness but a genuine impossibility, and it is the reason every flat map of the Earth distorts something: area, shape, distance or direction. Map-makers choose which one to sacrifice.

   ✦ Regular tetrahedron 2 nets, cube 11, octahedron 11, dodecahedron 43,380. A sphere has no net at all — its surface cannot be flattened without stretching or tearing, which is why every world map distorts the Earth.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 2, Reprint 2026-27 (hegp204.pdf), where this is Chapter 4 (pages 70-102) — the eleventh chapter of the Class 8 course and the longest in the book. Like the rest of Part 2 it carries NO printed answer key, so every formula and count was derived and then independently recomputed in Python before being written: the Sierpinski Carpet recurrences (R_n = 8^n, H_n = (8^n - 1)/7), the Sierpinski Triangle counts (3^n and (3^n - 1)/2), the areas (8/9)^n and (3/4)^n, the Koch side count 3 x 4^n and perimeter 3 x (4/3)^n, and the face/edge/vertex formulas for prisms and pyramids (checked against Euler's relation for every case). THIS IS A HEAVILY VISUAL CHAPTER, so figure-only items are handled in one of two ways and never guessed. (1) MEASURED FROM THE RENDERED PAGE: the cube-stack count on page 97 was settled by rendering the figure at 400 dpi and observing that each row sits one step BACK as well as one step up (every bottom cube shows its full top face), which makes it a square-layered step pyramid of 16 + 9 + 4 + 1 = 30 cubes rather than the ten visible; and the three letters in the page-96 puzzle were read off the printed pixel glyphs at 700-900 dpi as C (front), A (top) and F (side). (2) FLAGGED AND ANSWERED BY METHOD: the six candidate cube nets, the projection-matching sets, the cube-combination views, the isometric figures to copy, the rolling ball and the impossible triangle are all printed diagrams; each solution gives the full method and reasoning and says plainly that the diagram is not reproduced. ONE ITEM IS LEFT OPEN BY THE BOOK ITSELF and is reported as such: the 30 x 12 x 12 shortest-path Try This on page 87, where the book computes 42 cm and 40 cm for two unfoldings (24^2 + 32^2 = 1600 verified) and then says all unfoldings must be listed to find the answer — so the solution establishes only that the shortest path is at most 40 cm. The tetracube count in the page-100 exercise was verified by exhaustive computer enumeration: 8 arrangements up to rotation, 7 up to rotation and reflection, of which 5 are flat.. Questions are referenced from the NCERT textbook for identification.