In-text — Build it in Your Imagination — Exploring Some Geometric Themes

1. 13 marksGanita Prakash Cl-8 Part 2, Build it in Your Imagination, page 76

   Cut off the four corners of an imaginary square, with each cut going between the midpoints of adjacent edges. What shape is left over? How can you reassemble the four corners to make another square?

   Hint. Do it in your head first. Then check the areas — they should account for the whole square.

   What is left. Joining the midpoints of the four sides leaves the tilted square in the middle. It is genuinely a square: the four corner pieces are congruent right-angled isosceles triangles, so the four cut edges are equal; and each corner triangle has base angles of 45°, so at each midpoint the leftover angle is 180° − 45° − 45° = 90°. Four equal sides and four right angles.

   Its area. Each corner triangle has legs of half a side, so its area is ⅛ of the square. Four of them make ½, so the tilted square left in the middle has half the area of the original. This is exactly the halving construction from Chapter 9.

   Reassembling the four corners. The four triangles together have area ½ as well, so if they can be made into a square at all, it must be a square of area ½ — the same size as the one left behind.

   They can. Take the four right isosceles triangles and put their right-angle corners together at one point, turning each a quarter-turn from the last. The four right angles fill 360°, so they close with no gap, and the four hypotenuses form the outside. Since the hypotenuse of each triangle is exactly a side of the middle square, the assembled square is congruent to the one left over.

   A satisfying check. The original square has been split into two equal squares — the one in the middle and the one built from the corners — each of half the area, each with side equal to the original side divided by √2.

   ✦ The middle piece is a tilted square of half the area. The four corner triangles are congruent right isosceles triangles, and setting their right angles together at a point makes a second square, congruent to the first.

2. 23 marksGanita Prakash Cl-8 Part 2, Build it in Your Imagination, page 76

   Mark the sides of an equilateral triangle into thirds. Cut off each corner of the triangle, as far as the marks. What shape do you get?

   Hint. Work out the length of every edge of the leftover shape, and the angle at every new corner.

   Count the edges first. Cutting off all three corners of a triangle turns each corner into an edge, so the leftover shape has 3 original edge-pieces plus 3 cut edges — a hexagon.

   Now check whether it is regular. Take the triangle to have side 1, so each mark is 1/3 from a corner.

   The cut edges. Cutting the corner at A removes a small triangle with two sides of length 1/3 and the 60° angle of the equilateral triangle between them. That makes it isosceles with apex 60°, so its base angles are also 60° — it is equilateral with side 1/3. Therefore each cut edge has length 1/3.

   The remaining edges. Each original side of length 1 loses 1/3 at each end, leaving the middle third, of length 1/3.

   So all six sides are 1/3 — the hexagon is equilateral.

   The angles. At each new corner, the original 60° corner has been sliced away and what is left along the straight edge is 180° − 60° = 120°. All six angles are 120°.

   Six equal sides and six equal angles make it a regular hexagon.

   Check the area. The three removed corners are each 1/9 of the original triangle (side 1/3 means area 1/9), so 3/9 = 1/3 is removed and 2/3 remains. A regular hexagon of side 1/3 has area 6 × (1/9) × (area of a unit equilateral triangle) = 6/9 = 2/3 of the original ✓

   ✦ A regular hexagon, with all six sides equal to one third of the original side and all six angles 120°. It covers 2/3 of the original triangle.

3. 33 marksGanita Prakash Cl-8 Part 2, Build it in Your Imagination, page 76

   Mark the sides of a square into thirds and cut off each of its corners as far as the marks. What shape is left?

   Hint. The same method as the triangle — but check the two kinds of edge separately before calling it regular.

   Count the edges. Cutting off all four corners of a square turns each corner into an edge, so the shape has 4 original pieces plus 4 cut edges — an octagon.

   Now measure carefully. Take the square to have side 1.

   The cut edges. Cutting the corner removes a right-angled isosceles triangle with legs 1/3 and the right angle of the square between them. Its hypotenuse — the cut edge — is √((1/3)² + (1/3)²) = √(2/9) = √2/3 ≈ 0.471

   The remaining edges. Each side of the square loses 1/3 at each end, leaving the middle third of length 1/3 ≈ 0.333.

   The eight sides therefore alternate 1/3, √2/3, 1/3, √2/3, … — they are not all equal.

   The angles. At each new corner the 90° corner of the square has been sliced by a 45° cut, leaving 180° − 45° = 135°. All eight angles are 135°, so the octagon is equiangular.

   The verdict. It is an octagon with all angles equal but sides of two different lengths — so it is not a regular octagon.

   Why this differs from the triangle. In the triangle the corner angle was 60°, which made the cut-off piece equilateral and the cut edge the same length as the marks. In the square the corner angle is 90°, so the cut piece is right-angled and its hypotenuse is √2 times the legs. To make the octagon regular you would have to place the marks so that 1 − 2m = m√2, giving m = 1/(2 + √2) ≈ 0.293 rather than 1/3.

   ✦ An octagon that is equiangular but not regular — all eight angles are 135°, but the sides alternate between 1/3 and √2/3 ≈ 0.471.

4. 45 marksGanita Prakash Cl-8 Part 2, Build it in Your Imagination, pages 77 and 90

   Describe a solid and a viewpoint giving each of these profiles: (5) a square outline (6) a circular outline (7) a triangular outline. Then find solids with these contrasting profiles: (8) rectangular from one viewpoint and circular from another (9) circular from one and triangular from another (10) rectangular from one and triangular from another (11) trapezium from one and circular from another (12) pentagonal from one and rectangular from another. Are the solids unique?

   Hint. A profile is the outline you see from one direction. Think of the object punching a hole through a wall.

   How to think about a profile. The profile is the outline the solid would leave if it punched a hole through a wall, moving straight ahead. So ask which direction to look from to flatten the solid into the required outline.

   Single profiles. (5) Square. A cube viewed straight at a face. Also a square-based prism seen face-on, or a square pyramid seen from directly above. (6) Circle. A sphere from any direction at all — the only solid with a circular profile from every viewpoint. Also a cylinder seen end-on, or a cone seen from directly above. (7) Triangle. A cone viewed from the side. Also a triangular prism seen end-on, or a triangular pyramid from a suitable direction.

   Contrasting pairs — the interesting ones. (8) Rectangle and circle. A cylinder. From the side it is a rectangle (height × diameter); from the end it is a circle. (9) Circle and triangle. A cone. From directly above it is a circle; from the side it is a triangle. (10) Rectangle and triangle. A triangular prism. Viewed end-on it is a triangle; viewed along a rectangular face it is a rectangle. (11) Trapezium and circle. A frustum of a cone — a cone with its top sliced off parallel to the base. From the side it is a trapezium; from above, a circle. (12) Pentagon and rectangle. A pentagonal prism. End-on it is a pentagon; from the side, a rectangle. (A house-shaped prism — a cuboid with a triangular roof — works too.)

   Are the solids unique? No — and this is the point of the section. Every one of these has many answers. A circular profile is given by a sphere, a cylinder end-on, a cone from above and a disc. A rectangular profile is given by a cuboid, a cylinder from the side, and a flat rectangular card.

   Even two different profiles do not pin a solid down: a cylinder gives a rectangle and a circle, but so does a shape like a coin standing on edge, and so does a solid made of a half-cylinder joined to a half-cuboid. A projection loses information, which is exactly why engineers record three mutually perpendicular views rather than one.

   ✦ (5) cube face-on (6) sphere, from any direction (7) cone from the side. (8) cylinder (9) cone (10) triangular prism (11) frustum of a cone (12) pentagonal prism. None of these is unique — projection loses information, which is why three views are used.

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.