Figure it Out — Views of Solids Made of Cubes — Exploring Some Geometric Themes

1. 14 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 95

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

   Hint. Work on a grid: for each square of the view, ask whether any cube lies behind it in that direction.

   The five arrangements are printed diagrams, so they are not reproduced here. What follows is the method, which works on any arrangement of cubes.

   Set up coordinates. Give every cube an address (x, y, z): x across (length), y back (depth), z up (height). Then each view is a shadow on a grid: · Front view — the set of (x, z) cells for which some cube exists at that x and z, at any depth y. · Top view — the set of (x, y) cells for which some cube exists at any height z. · Side view — the set of (y, z) cells for which some cube exists at any x.

   The working rule. For each square of the grid, ask: looking along this direction, is there at least one cube in that line? If yes, shade the square. Depth is irrelevant — a stack of five cubes one behind another looks exactly like a single cube from the front.

   Worked example. Take four cubes: three in a row along x at z = 1, y = 1, and one on top of the leftmost at z = 2. · Front view: a row of three squares with one square sitting on the left end — an L shape. · Top view: a row of three squares. The stacked cube is directly above another, so it adds nothing. · Side view: one column of two squares, since the whole arrangement is only one cube deep and two cubes tall.

   Three checks that catch most errors.

   1. The width of the front view must equal the width of the top view.
   2. The height of the front view must equal the height of the side view.
   3. The depth of the top view must equal the width of the side view. If any of these disagree, a view has been drawn wrongly.

   A warning about hidden cubes. Cubes tucked behind others still count — they are part of the solid even though you cannot see them in the picture. Work from the arrangement, not from what happens to be visible in the drawing.

   ✦ Treat each view as a shadow on a grid: shade a square whenever at least one cube lies along that line of sight. Front view uses (x, z), top view (x, y), side view (y, z). Then check that front and top share a width, front and side share a height, and top and side share a depth.

2. 25 marksGanita Prakash Cl-8 Part 2, Math Talk, pages 95-96

   Eight identical cubes are glued together along faces to form the letter 'C'. (i) 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' from the top. (iii) Can you make it look like 'C' from the front, 'A' from the top and 'F' from the side? (iv) What other letter combinations could you make?

   Hint. The C is one cube thick. Ask what survives when the thickness direction is flattened.

   (i) The two views of the flat 'C'. Written on a grid, the eight-cube C is 3 wide and 4 tall and one cube thick: a top bar of 3 cubes, a single cube at the left on each of the two middle rows, and a bottom bar of 3 cubes — 3 + 1 + 1 + 3 = 8 ✓

   Side view. Looking along the width, everything collapses onto the depth-height grid. The shape is 1 cube deep and 4 cubes tall, and there is a cube at every height, so the side view is a 1 × 4 vertical bar.

   Top view. Looking down, everything collapses onto the width-depth grid. The shape is 3 wide and 1 deep, and every one of those three columns contains at least one cube, so the top view is a 3 × 1 horizontal bar.

   Both views lose the letter completely — a flat letter looks like a plain bar from the two other directions.

   (ii) 'C' from the front and 'A' from the top. The rule that makes this possible: the front view only records which (width, height) cells are occupied at some depth. So any cube added behind a cell that the C already occupies leaves the front view unchanged.

   So build the 'A' lying flat in the horizontal plane, using only those columns that the C already occupies. Since the C occupies all three widths (in the top bar and the bottom bar), and depth is completely free, an 'A' three cells wide and as deep as you like can be laid out at the height of the top bar. The front view stays a C; the top view becomes an A.

   (iii) All three at once — and how to decide it. Here is the test. A cube at position (x, y, z) is permitted only if · its front cell (x, z) belongs to the C, and · its top cell (x, y) belongs to the A, and · its side cell (y, z) belongs to the F. Take every permitted cube; call that solid S.

   S is the only candidate worth testing. Any solid with the three required views must consist entirely of permitted cubes, so it is contained in S; and S's own views are contained in C, A and F by construction. So if any solid works, S works. Build S, read off its three views, and compare.

   Doing this on a five-by-five grid with a plain block-letter C, A and F, the solid S has 12 cubes but its front view misses two cells of the C's bottom bar and its top view misses part of the A — so those particular letter shapes do not work together, while slightly different renderings of the same letters may. The test is decisive either way, and it is the point of the exercise: whether three letters are compatible depends on their exact shapes, not their names.

   (iv) Other combinations. Letters that are mostly solid bars — L, T, E, H, I — combine far more easily than letters with enclosed holes such as O, A, B or D, because a hole demands an empty cell that all three views must agree on. Good starting triples are L / T / I and T / H / I.

   ✦ (i) From the side, a 1 × 4 vertical bar; from the top, a 3 × 1 horizontal bar. (ii) Possible — add cubes behind cells the C already occupies, laying the A out flat, since extra depth never changes the front view. (iii) Test it by keeping every cube whose three shadow cells lie in C, A and F respectively; if any solid works, that one does. (iv) Bar-shaped letters such as L, T, E, H and I combine most easily; letters with holes are the hardest.

3. 34 marksGanita Prakash Cl-8 Part 2, Figure it Out, pages 96-97

   Which solid corresponds to a given top view, front view and side view? And how do you build a solid from identical cubes that gives a specified set of three projections?

   Hint. Work backwards: start from the top view to fix the footprint, then use the other two views to decide heights.

   The views are printed diagrams, so they are not reproduced here. What follows is the reconstruction method.

   Step 1 — the top view gives the footprint. The top view shows exactly which (width, depth) positions have at least one cube somewhere above them. Those cells are your ground plan; every other cell must be completely empty.

   Step 2 — the front view gives the heights, column by column. For each width position x, the front view shows the heights at which something exists. So the tallest shaded cell in column x of the front view is the maximum height anywhere at that x.

   Step 3 — the side view constrains heights by depth. In the same way, for each depth y the side view gives the heights present at that y.

   Step 4 — build the maximal solid, then test it. Place a cube at (x, y, z) whenever all three of the following hold: (x, y) is in the top view, (x, z) is in the front view, and (y, z) is in the side view. This gives the largest solid consistent with the three views.

   This maximal solid is the one to test, for the same reason as in the letters puzzle: any solid with those three views must be contained in it, and its own views are contained in the targets. So if any solid gives the three views, the maximal one does. Read off its three views and compare with the targets.

   Step 5 — check consistency before you start. The three views must agree on shared dimensions: · front and top views must have the same width · front and side views must have the same height · top and side views must have the same depth If they do not, no solid exists and the views have been misread.

   Two things to remember. · The answer need not be unique. Several different solids can share three views — for instance a hidden cube in an interior position may or may not be present without changing any view. The maximal solid is one valid answer; a question asking for "the" solid usually wants the natural, gap-free one. · Hidden cubes count. A cube tucked behind others still belongs to the solid, and it is where the marks in this topic are usually lost.

   ✦ Use the top view as the footprint, and the front and side views to fix which heights occur at each width and each depth. Place a cube wherever all three conditions hold at once — that maximal solid works if any solid does. First check the views agree on width, height and depth, and remember the answer need not be unique.

4. 43 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 97

   Find the number of cubes in the pictured stack of identical cubes.

   Hint. Look at whether each row sits directly on the row below, or one step further back as well as one step up.

   Read the picture carefully first. The drawing shows four rows: 4 cubes across the front at the bottom, then 3, then 2, then 1 on top — ten cubes visible.

   But look at where the upper cubes sit. Every cube in the bottom row shows its whole top face, and the row above sits between them rather than squarely on them. Cubes cannot sit half on one cube and half on another, so the row above is not in the front plane at all — it is one step back as well as one step up. That is how an isometric drawing shows depth.

   So there are hidden cubes. The row of 3 rests on cubes that lie one step behind the front row — cubes we cannot see. The stack is a step pyramid with square layers, not a flat wall.

   Count layer by layer, assuming the layers are complete squares with no gaps (the usual assumption for such a stack, and the only one that leaves nothing floating):

   Layer	Size	Cubes
   bottom	4 × 4	16
   second	3 × 3	9
   third	2 × 2	4
   top	1 × 1	1
   Total		30

   Check against the picture. With each layer set one step back and one step to the side, the front row of the bottom layer shows 4 cubes, the second layer shows 3, then 2, then 1 — exactly the ten visible cubes in the drawing ✓ The other twenty are behind them.

   The general pattern. A step pyramid with square layers 1², 2², …, n² contains 1 + 4 + 9 + … + n² = n(n + 1)(2n + 1) ÷ 6 cubes. For n = 4 that is 4 × 5 × 9 ÷ 6 = 30 ✓

   Why the question is worth asking. Only ten cubes are visible, so answering 10 is the natural mistake. Counting the hidden ones is the whole point: a drawing shows the surface, and you have to reason about what must be underneath to hold it up.

   ✦ 30 cubes — layers of 16, 9, 4 and 1. Only ten are visible; the other twenty are hidden behind and beneath them, and they must be there to support the upper layers.

5. 54 marksGanita Prakash Cl-8 Part 2, Math Talk, page 97

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

   Hint. Try three positions: face-on, rolled about one edge direction, and balanced on a corner.

   Take the orientations one at a time.

   1. Face-on — a square. Hold the cube with one face parallel to the plane. The outline is a square of side equal to the cube's edge. This is the smallest possible outline.

   2. Rolled about one edge direction — a rectangle. Now turn the cube about an axis parallel to one set of edges. Those edges still project at full length, so one side of the outline stays equal to the edge; the other side is the width of the tilted square face, which is edge × (cos θ + sin θ). The outline is a rectangle, and it is widest at θ = 45°, when the factor becomes √2. So the rectangle ranges from 1 × 1 up to 1 × √2 times the edge.

   3. Balanced on a corner — a regular hexagon. Stand the cube on one vertex so that the long diagonal points straight at the viewer. Six vertices now form the outline, and by the three-fold symmetry about that diagonal all six sides and all six angles are equal — a regular hexagon. This is the isometric projection, and it is the largest outline the cube can make.

   4. Any other position — an irregular hexagon. Between these special positions the outline is a hexagon whose sides are not all equal, though opposite sides remain parallel and equal, since parallel edges of the cube project to parallel segments.

   The complete list. Square, rectangle (up to √2 : 1), regular hexagon, and non-regular hexagons.

   What is impossible, and why. A cube can never project to a triangle or a pentagon. Its edges come in three families of four parallel edges each, and parallel edges project to parallel segments — so the outline's sides always come in parallel pairs. That forces an even number of sides, ruling out triangles and pentagons at once.

   ✦ A square (face-on), a rectangle up to √2 times as long as it is wide (rolled about an edge direction), a regular hexagon (balanced on a corner — the isometric view), and irregular hexagons in between. Never a triangle or a pentagon, because parallel edges project to parallel sides, so the outline must have an even number of sides.

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.