--- title: "NCERT Solutions Class 8 Maths Exploring Some Geometric Themes" url: https://www.swavid.com/maths/class/8/chapter/exploring-some-geometric-themes/ncert-solutions dateModified: 2026-10-07T15:37:42+00:00 --- # NCERT Solutions Class 8 Maths Exploring Some Geometric Themes This chapter's questions cover geometric explorations of fractals, self-similarity, visualising solids, properties of nets, projections, shadows, and drawing shapes on isometric grids. Free PDF (34 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-11-exploring-some-geometric-themes-9854d5c543.pdf ## Sierpinski Carpet ### Question 1 *3 marks · Short answer* Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Carpet. **Solution** 1. Step 0 is a single filled square. 2. Step 1 is formed by dividing the square into 9 smaller squares and removing the central one, leaving 8 smaller squares. 3. Step 2 is formed by repeating the same procedure on each of the remaining 8 squares, resulting in 64 even smaller squares with central holes. **Answer:** Step 0 is a solid square, Step 1 has 8 squares surrounding a central hole, and Step 2 has 64 smaller squares with holes. > Common mistake: Confusing the number of remaining squares with the number of removed holes at each step. ### Question 2 *2 marks · Very short answer* Do you see any pattern in the number of holes and squares that remain at each step? **Solution** 1. Let $R_n$ represent the number of remaining squares and $H_n$ represent the number of holes at the $n$-th step. 2. Every remaining square at step $n$ gives rise to 8 remaining squares at step $(n+1)$, so $R_{n+1} = 8 R_n$, and the number of holes grows as $H_{n+1} = H_n + R_n$. **Answer:** The remaining squares multiply by 8 at each step ($R_{n+1} = 8 R_n$), and the new holes added equal the number of remaining squares from the previous step ($H_{n+1} = H_n + R_n$). > Common mistake: Forgetting to add the holes present in the previous step to the newly formed holes. ### Question 3 *2 marks · Very short answer* Can this be used to get a formula for $R_n$? **Solution** 1. We have $R_0 = 1$, $R_1 = 8 \times 1 = 8$, and $R_2 = 8 \times 8 = 8^2$. 2. Extending this pattern to the $n$-th step gives the general formula $R_n = 8^n$. **Answer:** $R_n = 8^n$ > Common mistake: Writing $R_n = 8n$ instead of the exponential form $8^n$. ### Question 4 *2 marks · Very short answer* Similarly, how do we find the number of holes at a given step? **Solution** 1. Every remaining square at the $n$-th step creates a new hole in the $(n+1)$-th step, and all previous holes remain. 2. This gives the relation $H_{n+1} = H_n + R_n$, leading to the sequence $H_0 = 0$, $H_1 = 1$, $H_2 = 1 + 8$, and $H_3 = 1 + 8 + 8^2$. **Answer:** The number of holes at the $n$-th step is given by the sum $H_n = 1 + 8 + 8^2 + \dots + 8^{n-1}$ for $n \ge 1$, with $H_0 = 0$. > Common mistake: Failing to account for the accumulation of all holes from the previous steps. ## Sierpinski Gasket ### Question 1 *3 marks · Proof* 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** 1. Given: An equilateral triangle ABC. 2. Let D, E, and F be the midpoints of sides AB, BC, and CA respectively. 3. By joining the midpoints D, E, and F, four smaller triangles are formed: $\triangle ADF$, $\triangle DBE$, $\triangle FEC$, and $\triangle DEF$. 4. Using the mid-point theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it, so $DE = \frac{1}{2}AC$, $EF = \frac{1}{2}AB$, and $DF = \frac{1}{2}BC$. 5. Since $\triangle ABC$ is equilateral, all its sides are equal in length ($AB = BC = CA$). 6. Therefore, the sides of the four smaller triangles are all equal to half the side length of $\triangle ABC$, making each of the four triangles equilateral. 7. Hence proved. **Answer:** Hence proved that joining the midpoints of an equilateral triangle divides it into 4 identical equilateral triangles. > Common mistake: Forgetting to state that all four triangles have equal side lengths and hence are equilateral. ## Figure it Out ### Question 1 *3 marks · Short answer* Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Triangle. **Solution** 1. Step 0: Draw a solid equilateral triangle. 2. Step 1: Join the midpoints of the sides of the equilateral triangle to divide it into 4 smaller equilateral triangles and remove the central triangle, leaving 3 smaller upright triangles. 3. Step 2: Repeat the midpoint-joining and central triangle removal procedure on each of the 3 remaining smaller triangles. **Answer:** Initial steps (Step 0, Step 1, and Step 2) of the shape sequence leading to the Sierpinski Triangle. > Common mistake: Removing more than one triangle or not joining the midpoints properly. ### Question 2 *3 marks · Short answer* Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle. **Solution** 1. At step $0$, the number of remaining triangles is $R_0 = 1$ and the number of holes is $H_0 = 0$. 2. At each step, every remaining triangle gives rise to $3$ smaller remaining triangles, so $R_n = 3^n$. 3. The number of holes satisfies the relation $H_{n+1} = H_n + R_n$, giving $H_0 = 0$, $H_1 = 1$, $H_2 = 1 + 3$, and in general $H_n = 1 + 3 + 3^2 + \dots + 3^{n-1}$. **Answer:** Remaining triangles at step $n$ is $3^n$ and holes at step $n$ is $1 + 3 + 3^2 + \dots + 3^{n-1}$. > Common mistake: Confusing the number of remaining triangles with powers of $2$ or $4$. ### Question 3 *3 marks · Short answer* Find the area of the region remaining at the $n$th step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be $1\text{ sq. unit}$. **Solution** 1. For the Sierpinski Carpet, each step multiplies the remaining area by $\frac{8}{9}$, so starting with area $1$, the remaining area at the $n$th step is $\left(\frac{8}{9}\right)^n$ sq. units. 2. For the Sierpinski Triangle, each step multiplies the remaining area by $\frac{3}{4}$, so starting with area $1$, the remaining area at the $n$th step is $\left(\frac{3}{4}\right)^n$ sq. units. **Answer:** Sierpinski Carpet area is $\left(\frac{8}{9}\right)^n$ sq. units and Sierpinski Triangle area is $\left(\frac{3}{4}\right)^n$ sq. units. > Common mistake: Using the fraction of removed area instead of the remaining area fraction. ## Figure it Out ### Question 1 *3 marks · Short answer* Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake. **Solution** 1. Step 0: Draw a plain equilateral triangle. 2. Step 1: Divide each side into 3 equal parts, raise an equilateral triangle over the middle part, and remove the middle part to form a star-like shape with 12 sides. 3. Step 2: Repeat the same bump-adding procedure on each of the 12 smaller sides of the Step 1 shape to obtain the Step 2 snowflake shape. **Answer:** A sequence showing an equilateral triangle (Step 0), a star-like 6-pointed shape (Step 1), and a more intricate snowflake shape with scaled-up bumps (Step 2). > Common mistake: Forgetting to remove the middle part of the side before raising the new equilateral triangle. ### Question 2 *3 marks · Short answer* Find the number of sides in the $n$th step of the shape sequence that leads to the Koch Snowflake. **Solution** 1. At Step 0, the starting equilateral triangle has $3$ sides, so $S_0 = 3$. 2. At each subsequent step, every side is replaced by $4$ smaller segments, multiplying the number of sides by $4$. 3. Thus, the number of sides at the $n$th step is given by the formula $S_n = 3 \times 4^n$. **Answer:** $3 \times 4^n$ > Common mistake: Writing $4 \times 3^n$ instead of $3 \times 4^n$. ### Question 3 *3 marks · Short answer* Find the perimeter of the shape at the $n$th step of the sequence. Take the starting equilateral triangle to have a sidelength of $1\text{ unit}$. **Solution** 1. The starting equilateral triangle (Step 0) has sidelength $1$ unit, so its perimeter is $P_0 = 3 \times 1 = 3$. 2. At each step, the sidelength of each segment becomes $\frac{1}{3}$ of the previous step, and the number of sides is multiplied by $4$. 3. Therefore, the perimeter at the $n$th step is given by $P_n = 3 \times \left(\frac{4}{3}\right)^n$. **Answer:** $3 \times \left(\frac{4}{3}\right)^n$ > Common mistake: Multiplying the sidelength by $\frac{4}{3}$ instead of $\frac{1}{3}$ at each step. ## Build it in Your Imagination ### Question 1 *2 marks · Very short answer* Picture your name, then read off the letters backwards. Make sure to do this by sight, not by sound—really see your name! Now try with your friend’s name. **Solution** 1. Visualise the letters of the name clearly in the mind's eye. 2. Read the visualised letters from right to left by sight to obtain the reversed name. **Answer:** The letters of the name read backwards by sight. > Common mistake: Reading the letters backwards by sound instead of visualising them. ### Question 2 *3 marks · Short answer* Cut off the four corners of an imaginary square, with each cut going between midpoints of adjacent edges. What shape is left over? How can you reassemble the four corners to make another square? **Solution** 1. Cutting off the four corners of a square between the midpoints of adjacent edges leaves a smaller square rotated by $45^\circ$ or an octagon depending on the perspective, but fundamentally a square remains. 2. Each cut-off corner is an isosceles right-angled triangle with sides equal to half the side of the square. 3. The four triangular corners can be reassembled along their hypotenuses to form another identical square. **Answer:** A smaller square is left over, and the four triangular corners can be reassembled to form another square. > Common mistake: Failing to see that the four corner right triangles combine to form a square. ### Question 3 *3 marks · Short answer* 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? **Solution** 1. Consider an equilateral triangle with sides marked into thirds. 2. Cutting off each corner of the triangle as far as the marks removes three smaller corner triangles of the same size. 3. The shape left over in the middle is a regular hexagon. **Answer:** A regular hexagon is left over. > Common mistake: Mistaking the remaining shape for a triangle or a different polygon. ### Question 4 *3 marks · Short answer* Mark the sides of a square into thirds and cut off each of its corners as far as the marks. What shape is left? **Solution** 1. Consider a square whose sides are divided into three equal parts. 2. Cutting off each of the four corners up to the one-third marks removes four right-angled triangles. 3. The shape left in the middle is a regular octagon. **Answer:** A regular octagon is left. > Common mistake: Counting the sides incorrectly and naming it a hexagon. ### Question 5 *3 marks · Short answer* A solid whose profile has a square outline **Solution** 1. A solid has a square profile when viewed from a specific direction. 2. Examples of such solids include a cube viewed directly from any of its faces. 3. A square pyramid viewed from directly above (top view) or a cylinder of equal height and diameter viewed from the side also present a square outline. **Answer:** A cube (viewed from the front, back, top, bottom, left, or right) is a solid whose profile has a square outline. > Common mistake: Thinking only a cube can have a square profile. ### Question 6 *3 marks · Short answer* A solid whose profile has a circular outline **Solution** 1. A solid has a circular profile when viewed from a specific viewpoint. 2. The most direct example of such a solid is a sphere, which appears circular from any viewpoint. 3. A cylinder viewed from the top or bottom also results in a circular outline. **Answer:** A sphere or a cylinder (viewed from the top or bottom) is a solid whose profile has a circular outline. > Common mistake: Forgetting that a cylinder viewed end-on gives a circular profile. ### Question 7 *3 marks · Short answer* A solid whose profile has a triangular outline **Solution** 1. Consider a cone or a triangular pyramid (tetrahedron). 2. When viewed from the front or side, a cone or a triangular pyramid shows a triangular outline. 3. Thus, a cone or a triangular pyramid from a side viewpoint gives a triangular profile. **Answer:** A cone viewed from the front or side results in a triangular profile. > Common mistake: Confusing a three-dimensional solid with a flat two-dimensional shape. ### Question 8 *3 marks · Short answer* A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint **Solution** 1. Consider a cylinder standing upright. 2. When viewed from the front or side, its profile is a rectangle. 3. When viewed from directly above (top view), its profile is a circle. **Answer:** A cylinder has a rectangular profile from the front and a circular profile from the top. > Common mistake: Forgetting the top view gives a circle. ### Question 9 *3 marks · Short answer* A solid with a circular profile from one viewpoint and a triangular one from another viewpoint **Solution** 1. Consider a cone standing on a circular base. 2. When viewed from directly above, the profile is a circle. 3. When viewed from the side, the profile is a triangle. **Answer:** A cone has a circular profile from the top view and a triangular profile from the side view. > Common mistake: Mixing up the viewing directions for the circle and triangle. ### Question 10 *3 marks · Short answer* A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint **Solution** 1. Consider a triangular prism resting on one of its rectangular faces. 2. When viewed from the front or side perpendicular to its rectangular face, it has a rectangular profile. 3. When viewed from the triangular base end, its profile is a triangle. **Answer:** A triangular prism has a rectangular profile from one viewpoint and a triangular profile from another viewpoint. > Common mistake: Not considering the different faces of a triangular prism. ### Question 11 *3 marks · Short answer* A solid with a trapezium shaped profile from one viewpoint and a circular one from another viewpoint **Solution** 1. Consider a frustum of a cone or a specific truncated solid. 2. Alternatively, consider a cylinder sliced at an angle, giving a trapezium profile from the side and a circular profile from the top. 3. Thus, an inclined cut cylinder gives a trapezium profile from one viewpoint and a circular profile from another. **Answer:** A cylinder cut at a slant angle has a trapezium-shaped profile from the side and a circular profile from the top. > Common mistake: Thinking standard geometric shapes without modifications can give these profiles. ### Question 12 *3 marks · Short answer* A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint **Solution** 1. Consider a pentagonal prism lying on one of its rectangular lateral faces. 2. When viewed from the side facing its pentagonal base, the profile is pentagonal. 3. When viewed from above or along its side, the profile is rectangular. **Answer:** A pentagonal prism has a pentagonal profile from the end view and a rectangular profile from the side view. > Common mistake: Confusing the base shape with the lateral face shape. ## Visualising Solids - In-text questions ### Question 1 *3 marks · Short answer* 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? **Part (i)** 1. A prism with $10$-sided polygons (decagonal prism) has $2$ decagonal bases and $10$ rectangular lateral faces, giving a total of $10 + 2 = 12$ faces. 2. It has $10$ edges on the top base, $10$ on the bottom base, and $10$ connecting the corresponding vertices, giving a total of $3 \times 10 = 30$ edges. 3. It has $10$ vertices on the top base and $10$ on the bottom base, giving a total of $2 \times 10 = 20$ vertices. Answer (i): $12$ faces, $30$ edges, and $20$ vertices **Part (ii)** 1. For an $n$-sided prism, the number of faces is $n + 2$. 2. The number of edges is $3n$. 3. The number of vertices is $2n$. Answer (ii): $n+2$ faces, $3n$ edges, and $2n$ vertices **Answer:** A prism with $n$-sided polygons has $n+2$ faces, $3n$ edges, and $2n$ vertices. For $10$ sides, it has $12$ faces, $30$ edges, and $20$ vertices. > Common mistake: Confusing the number of lateral faces with the total number of faces by forgetting the two bases. ### Question 2 *3 marks · Short answer* 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? **Part (i)** 1. A pyramid with a 10-sided base has 1 base face and 10 triangular side faces, making a total of $10 + 1 = 11$ faces. 2. The base has 10 edges and there are 10 edges connecting the base vertices to the apex, giving a total of $10 \times 2 = 20$ edges. 3. The base has 10 vertices and there is 1 apex vertex, giving a total of $10 + 1 = 11$ vertices. Answer (i): 11 faces, 20 edges, and 11 vertices **Part (ii)** 1. For an n-sided polygon base, the pyramid has 1 base face and n triangular side faces, resulting in $n + 1$ faces. 2. The base has n edges and there are n edges connecting the base vertices to the apex, giving a total of $2n$ edges. 3. The base has n vertices and 1 apex vertex, resulting in $n + 1$ vertices. Answer (ii): $n + 1$ faces, $2n$ edges, and $n + 1$ vertices **Answer:** For a 10-sided base pyramid: 11 faces, 20 edges, and 11 vertices. For an n-sided base pyramid: $n + 1$ faces, $2n$ edges, and $n + 1$ vertices. > Common mistake: Confusing the number of edges of the base with the total edges of the pyramid. ### Question 3 *2 marks · Very short answer* What is a net of a cube? **Solution** 1. A net of a solid is a shape obtained by unfolding the solid onto a plane. 2. A net of a cube consists of six identical squares joined together along their edges that can be folded to form the cube. **Answer:** A net of a cube is a two-dimensional arrangement of six identical squares that can be folded along its edges to form a cube. > Common mistake: Including supporting tabs or flaps as part of the formal definition of a net. ### Question 4 *2 marks · Very short answer* Visualise how it can be folded to form a cube. **Solution** 1. When folding a net of a cube like Fig. 4.1, the central square forms the base of the cube. 2. The remaining four squares attached to the base fold upwards to form the four side faces, and the top square folds over to complete the cube. **Answer:** The central square acts as the base, while the surrounding squares fold upwards and over to form the four sides and the top face of the cube. > Common mistake: Failing to visualise opposite faces correctly during the folding process. ## Figure it Out ### Question 1 *3 marks · Short answer* Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try. (i) ... (vi) **Part (i)** 1. Examine the arrangement of squares in net (i). 2. Folding this net results in overlapping faces and leaves one face uncovered. 3. Therefore, it is not a net of a cube. Answer (i): Not a net **Part (ii)** 1. Examine the arrangement of squares in net (ii) in a line of four with adjacent squares. 2. Folding this net correctly encloses a cube with six square faces without any overlap. 3. Therefore, it is a net of a cube. Answer (ii): Net of a cube **Part (iii)** 1. Examine the T-shape like arrangement of squares in net (iii). 2. Folding this net forms a closed cube successfully. 3. Therefore, it is a net of a cube. Answer (iii): Net of a cube **Part (iv)** 1. Examine the cross-like or staircase arrangement in net (iv). 2. Folding this net forms a valid cube. 3. Therefore, it is a net of a cube. Answer (iv): Net of a cube **Part (v)** 1. Examine the arrangement in net (v). 2. Folding this arrangement leads to overlapping faces and does not form a complete cube. 3. Therefore, it is not a net of a cube. Answer (v): Not a net **Part (vi)** 1. Examine the arrangement of squares in net (vi). 2. Folding this net successfully encloses a cube. 3. Therefore, it is a net of a cube. Answer (vi): Net of a cube **Answer:** Nets (ii), (iii), (iv), and (vi) can be folded to form a cube, whereas (i) and (v) cannot form a cube. > Common mistake: Assuming all arrangements of six connected squares form a cube without checking for overlapping faces. ### Question 2 *3 marks · Short answer* Find all the $11$ nets of a cube. **Solution** 1. A cube has exactly $11$ distinct net structures in total. 2. These $11$ nets can be grouped by the length of their longest row of squares (four in a row, three in a row, or two in a row). 3. By considering all rotations and flips as the same net, the $11$ possible arrangements are determined. **Answer:** There are $11$ possible nets of a cube. > Common mistake: Counting rotated or flipped versions of the same net as different nets. ### Question 3 *3 marks · Short answer* Draw a net of a cuboid having sidelengths: (i) $5\text{ cm}, 3\text{ cm},\text{ and } 1\text{ cm}$ (ii) $6\text{ cm}, 3\text{ cm},\text{ and } 2\text{ cm}$ **Part (i)** 1. A cuboid has dimensions $5\text{ cm}$ (length), $3\text{ cm}$ (breadth), and $1\text{ cm}$ (height). 2. Draw a central row of four rectangles of dimensions $5\times 3$, $3\times 1$, $5\times 3$, and $3\times 1$ joined edge to edge. 3. Attach two additional $5\times 1$ rectangles to the top and bottom of the appropriate central rectangle to complete the net. Answer (i): A net consisting of rectangles of dimensions $5\text{ cm} \times 3\text{ cm}$, $5\text{ cm} \times 1\text{ cm}$, and $3\text{ cm} \times 1\text{ cm}$. **Part (ii)** 1. A cuboid has dimensions $6\text{ cm}$ (length), $3\text{ cm}$ (breadth), and $2\text{ cm}$ (height). 2. Draw a central row of four rectangles of dimensions $6\times 3$, $3\times 2$, $6\times 3$, and $3\times 2$ joined edge to edge. 3. Attach two additional $6\times 2$ rectangles to the appropriate sides to complete the net. Answer (ii): A net consisting of rectangles of dimensions $6\text{ cm} \times 3\text{ cm}$, $6\text{ cm} \times 2\text{ cm}$, and $3\text{ cm} \times 2\text{ cm}$. **Answer:** Nets drawn with dimensions as specified. > Common mistake: Drawing adjacent faces with mismatched edge lengths. ## Visualising Solids - Nets and Pyramids ### Question 1 *3 marks · Short answer* What is a net of a regular tetrahedron? Which of the following are nets of a regular tetrahedron? **Solution** 1. 1. A net of a regular tetrahedron is formed by four identical equilateral triangles joined along their edges. 2. 2. When folded along the edges, the four triangles meet to form a regular tetrahedron (a triangular pyramid with equilateral faces). 3. 3. From the given options in the textbook, the nets consisting of four connected equilateral triangles that can fold into a closed tetrahedron are the correct ones. **Answer:** A net of a regular tetrahedron consists of four equilateral triangles joined edge-to-edge such that they fold into a closed pyramid. > Common mistake: Arranging four equilateral triangles in a straight line, which cannot fold to enclose a 3D space. ### Question 2 *2 marks · Very short answer* Are there any other possible nets? **Solution** 1. 1. A regular tetrahedron has a fixed number of distinct flat net configurations that can be folded into the solid. 2. 2. According to the textbook, a regular tetrahedron has only 2 possible nets. **Answer:** No, there are no other possible nets; a regular tetrahedron has only 2 possible nets. > Common mistake: Confusing the number of nets of a tetrahedron with that of a cube, which has 11 nets. ### Question 3 *3 marks · Short answer* Draw a net with appropriate measurements that can be folded into a regular tetrahedron. Verify if it works by making an actual cutout. **Solution** 1. 1. Draw four equilateral triangles, each of side length $a$ (for example, $5\text{ cm}$), joined edge-to-edge in a triangular cluster with one central triangle pointing upwards and three attached along its three edges pointing outwards. 2. 2. Add small triangular or rectangular flaps along the outer edges to help in pasting and folding. 3. 3. Cut out the shape, fold along the interior edges, and paste the flaps to verify that it forms a regular tetrahedron. **Answer:** A net made of four connected equilateral triangles of equal side length with pasting flaps. > Common mistake: Drawing triangles of differing side lengths, preventing proper closure. ### Question 4 *3 marks · Short answer* Draw a net with appropriate measurements that can be folded into a square pyramid. Verify if it works by making an actual cutout. **Solution** 1. 1. Draw a central square of side length $a$ (for example, $4\text{ cm}$) to serve as the base. 2. 2. Attach four identical isosceles triangles (or equilateral triangles depending on the pyramid) along each of the four edges of the square. 3. 3. Cut out the net, fold the triangles upwards along the edges of the square, and join their sides to form a square pyramid. **Answer:** A net consisting of one central square surrounded by four triangles on its four sides. > Common mistake: Placing triangles on opposite sides only or using wrong triangle dimensions. ### Question 5 *2 marks · Very short answer* What is the net of a cylinder? **Solution** 1. 1. The net of a cylinder is obtained by unfolding its curved lateral surface and its two circular flat bases. 2. 2. It consists of one rectangle and two identical circles attached to opposite sides of the rectangle. **Answer:** The net of a cylinder consists of a rectangle attached to two identical circular faces at its top and bottom edges. > Common mistake: Omitting the circular bases or drawing them attached to the wrong edges. ### Question 6 *2 marks · Very short answer* What are the sidelengths of the rectangle obtained? **Solution** 1. 1. When the curved surface of a cylinder of height $h$ and base radius $r$ is unrolled, it forms a rectangle. 2. 2. The height of the rectangle equals the height of the cylinder ($h$), and its length equals the circumference of the circular base ($2\pi r$). **Answer:** The sidelengths of the rectangle are the height of the cylinder ($h$) and the circumference of its circular base ($2\pi r$). > Common mistake: Taking the length of the rectangle as the radius or diameter of the circle instead of the circumference $2\pi r$. ### Question 7 *2 marks · Very short answer* How will the net of a cone look? **Solution** 1. The net of a cone consists of a circular base and a sector of a circle. 2. The sector represents the lateral surface of the cone when it is unrolled. **Answer:** The net of a cone consists of a circular base and a sector of a circle. > Common mistake: Confusing the sector with a full circle. ### Question 8 *2 marks · Very short answer* If the cone is slit open along the line $l$ and then unrolled, what will we get? **Solution** 1. When a cone is slit open along the slant height l and unrolled, the lateral surface forms a sector of a circle with centre O. 2. The base of the cone remains a circle. **Answer:** We get a sector of a circle with centre O and a circular base. > Common mistake: Forgetting that the base is also part of the net. ### Question 9 *3 marks · Short answer* What surface do you construct by using the above net, in which O is not the centre of the boundary circle? Make a physical model to help you answer this question! **Solution** 1. If O is not the centre of the boundary circle, the slant height of the cone is not uniform. 2. This construction results in an oblique cone. 3. An oblique cone is a cone where the apex is not directly above the centre of the base. **Answer:** Constructing a net where O is not the centre of the boundary circle results in an oblique cone. > Common mistake: Assuming all cones must be right circular cones. ### Question 10 *3 marks · Short answer* Draw a net with appropriate measurements that can be folded into a triangular prism. Verify if it works by making an actual cutout. **Solution** 1. A triangular prism has two congruent triangular bases and three rectangular faces. 2. Draw three rectangles of equal width joined side-by-side. 3. Attach one triangle to the top edge of the middle rectangle and one to the bottom edge of the middle rectangle. **Answer:** A net for a triangular prism consists of three rectangles and two congruent triangles attached to the rectangles. > Common mistake: Drawing the triangles on the wrong sides of the rectangles. ### Question 11 *3 marks · Short answer* Taking all the triangles in the net to be equilateral, make a cutout of the net and fold it to form an octahedron. **Solution** 1. An octahedron is a solid with 8 equilateral triangular faces. 2. The net consists of 8 equilateral triangles arranged in a specific pattern. 3. When folded, these triangles meet at the edges to form the octahedron. **Answer:** The net consists of 8 equilateral triangles arranged to fold into an octahedron. > Common mistake: Using the wrong number of triangles. ### Question 12 *3 marks · Short answer* Net of a sphere? Experiment and see if you can make a paper cutout that can perfectly wrap around a ball without leaving any wrinkles, gaps or overlaps. **Solution** 1. A sphere is a curved surface that cannot be flattened onto a plane without distortion. 2. Any attempt to wrap a flat paper cutout around a sphere will result in wrinkles, gaps, or overlaps. 3. Therefore, a sphere does not have a flat net. **Answer:** It is not possible to create a flat net for a sphere because it is a non-developable surface. > Common mistake: Thinking that a sphere can be unfolded like a polyhedron. ## Shortest Paths on a Cube ### Question 1 *3 marks · Short answer* What is the shortest path for the ant to reach the laddu? **Solution** 1. 1. Unfold the faces of the cuboid containing the ant and the laddu into a flat net. 2. 2. Join the ant and the laddu with a straight line segment on the net. 3. 3. The length of this straight line segment gives the shortest path on the surface of the cuboid. **Answer:** The shortest path is the straight line segment joining the ant and the laddu on the unfolded net of the cuboid. > Common mistake: Drawing a path directly on the 3D surface without unfolding it into a net. ### Question 2 *3 marks · Short answer* What about in the following case? **Solution** 1. 1. Open the cuboid into a net such that the faces adjacent to the edge where the laddu is placed lie flat. 2. 2. Draw a straight line between the position of the ant and the laddu on the net. 3. 3. Transform the straight line back onto the surface of the cuboid to get the shortest path. **Answer:** The shortest path is found by unfolding the relevant faces into a net and drawing a straight line between the two points. > Common mistake: Failing to choose the correct unfolding of faces that allows a straight line path. ### Question 3 *2 marks · Very short answer* If we think that a certain path is the shortest, how can we be sure that it truly is, among all the infinite possibilities? **Solution** 1. 1. We can be sure by drawing the net of the cuboid and converting the 3D surface path problem into a 2D plane geometry problem. 2. 2. The straight line on the net represents the absolute shortest distance between two points. **Answer:** We can be sure by using a net of the cuboid, where the shortest path corresponds to a straight line. > Common mistake: Assuming visual approximation on the 3D figure is sufficient. ### Question 4 *2 marks · Very short answer* For example, are either of these the shortest path? **Solution** 1. 1. A path on the cuboid is the shortest only if it transforms into a straight line when the cuboid is unfolded into a net. 2. 2. Any path that bends on the net is not the shortest path. **Answer:** A path is the shortest only if it appears as a straight line on the unfolded net. > Common mistake: Choosing a curved or bent path on the net. ### Question 5 *2 marks · Very short answer* What does this show? **Solution** 1. 1. A path on the surface of the cuboid transforms to a path of the same length on the net, and vice versa. 2. 2. Straight lines on the net provide the shortest paths on the cuboid. **Answer:** It shows that paths on the cuboid and their corresponding paths on the net have the same length, converting the problem into finding a straight line on a plane. > Common mistake: Not recognizing that lengths are preserved when unfolding. ### Question 6 *3 marks · Short answer* Find the shortest path between the ant and the laddu in the following case: **Solution** 1. 1. Unfold the cuboid such that the ant and the laddu lie within a single rectangular region formed by the connected faces of dimensions $8\text{ cm}$ and $(4 + 4) = 8\text{ cm}$. 2. 2. Apply the Baudhayana Theorem (Pythagoras Theorem) on the right-angled triangle formed with base $8\text{ cm}$ and height $8\text{ cm}$. 3. 3. Calculate the distance $d = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}\text{ cm}$. **Answer:** $8\sqrt{2}\text{ cm}$ > Common mistake: Taking incorrect dimensions for the unfolded rectangle sides. ### Question 7 *2 marks · Very short answer* So what do we do now? **Solution** 1. We unfold the cuboid in a different way such that the line segment joining the ant and the laddu lies completely within the net without going outside any face. **Answer:** We unfold the cuboid differently so that the line segment between the ant and the laddu stays inside the net. > Common mistake: Using an unfolding where the straight line segment falls outside the boundary of the net. ### Question 8 *3 marks · Short answer* What is the length of the shortest path between the ant and the laddu? **Solution** 1. Given a cuboid of dimensions $30\text{ cm} \times 12\text{ cm} \times 12\text{ cm}$, the ant is at one end and the laddu is stuck to the back. 2. Using a suitable unfolding of the cuboid, the net forms a right-angled triangle with perpendicular sides of lengths $24\text{ cm}$ and $32\text{ cm}$. 3. Using the Baudhayana Theorem, the distance $d$ is given by $d^2 = 24^2 + 32^2 = 576 + 1024 = 1600$. 4. Therefore, $d = \sqrt{1600} = 40\text{ cm}$. **Answer:** $40\text{ cm}$ > Common mistake: Choosing an incorrect unfolding that does not yield the shortest straight-line distance. ## Projections and Shadows ### Question 1 *2 marks · Very short answer* Let us visualise the projection of a line. **Solution** 1. The projection of a line segment on a plane is obtained by dropping perpendiculars from all points of the line onto the plane. 2. The length of the projection is always less than or equal to the actual length of the line. **Answer:** The projection of a line on a plane is a line segment whose length is less than or equal to the actual length of the line. > Common mistake: Confusing the projection length with the actual length of the line when the line is inclined. ### Question 2 *2 marks · Very short answer* What happens to the length of a line in its projection? **Solution** 1. Let $l$ be the actual length of the line and $p$ be the length of its projection on a plane. 2. As seen from Fig. 4.3 in the textbook, the projection forms a right-angled triangle where the actual line is the hypotenuse, making the projection shorter than or equal to the actual length ($p \le l$). **Answer:** The length of the projected line is generally less than or equal to its actual length. > Common mistake: Assuming the length of the projection is always equal to the actual length of the line. ### Question 3 *2 marks · Very short answer* Can you now compare the lengths $p$ and $l$? **Solution** 1. From the right-angled triangle formed in Fig. 4.3, the actual length $l$ acts as the hypotenuse and the projection length $p$ is the adjacent side. 2. By the properties of a right-angled triangle, the hypotenuse is always greater than or equal to the other sides, so $p \le l$. **Answer:** The length of the projection $p$ is always less than or equal to the actual length $l$ ($p \le l$). > Common mistake: Writing $p \ge l$ by reversing the sides of the right-angled triangle. ### Question 4 *2 marks · Very short answer* When is the length of the projected line equal to its actual length? **Solution** 1. Consider the right-angled triangle in Fig. 4.3 where $l$ is the hypotenuse and $p$ is the base. 2. The projected length $p$ is equal to the actual length $l$ only when the angle between the line and the plane is zero, meaning the line is parallel to the plane. **Answer:** The length of the projected line is equal to its actual length when the line is parallel to the projection plane. > Common mistake: Stating that the line must be perpendicular to the plane for lengths to be equal. ### Question 5 *3 marks · Short answer* What do you think are the different possible projections of a square that we get based on its orientation? **Solution** 1. When a square is parallel to the projection plane, its projection is an identical square. 2. When it is tilted or inclined at an angle, the side lengths along the tilt appear shortened due to perspective, so its projection becomes a rectangle or a rhombus. 3. In extreme orientations where the square is perpendicular to the plane, its projection can reduce to a straight line segment. **Answer:** The projection of a square can be a square, a rectangle, a rhombus, or a straight line segment depending on its orientation. > Common mistake: Forgetting that a square can project as a line segment when viewed edge-on. ### Question 6 *3 marks · Short answer* What do you think is the projection of a parallelogram under different orientations? Can this ever be a quadrilateral that is not a parallelogram? **Solution** 1. The projection of a parallelogram under different orientations can be another parallelogram, a rectangle, or a line segment. 2. As established in the chapter, the projection of a pair of parallel lines always remains parallel. 3. Since opposite sides of a parallelogram remain parallel in parallel projection, the projected shape can never be a general quadrilateral that is not a parallelogram. **Answer:** The projection of a parallelogram is always a parallelogram (or a line segment); it can never be a quadrilateral that is not a parallelogram because parallel lines project to parallel lines. > Common mistake: Thinking that tilting a parallelogram can change it into a trapezoid or random quadrilateral. ### Question 7 *3 marks · Short answer* What can you say about the projection of an $n$-sided regular polygon? [Hint: Projection of a polygon is composed of the projections of its sides.] **Solution** 1. The projection of an $n$-sided regular polygon is composed of the projections of its individual sides. 2. Since the projection of any line segment is a line segment or a point, the projection of a polygon is formed by connecting the projected segments of its sides. 3. Depending on the orientation of the polygon relative to the projection plane, the shape may appear compressed or scaled along one direction, so it forms an $n$-sided polygon or a degenerate shape if viewed edge-on. **Answer:** The projection of an $n$-sided regular polygon is an $n$-sided polygon whose shape depends on the orientation of the polygon relative to the projection plane. > Common mistake: Assuming the projection of a regular polygon is always regular. ### Question 8 *3 marks · Short answer* How would the projections of a cube and a cone look? **Solution** 1. As shown in Fig. 4.4, the projection of a cube (when viewed normally to a face) looks like a square, with internal lines representing the projected hidden edges depending on transparency. 2. As shown in Fig. 4.5, the projection of a cone (when viewed from the side) looks like a triangle with a curved base, or a circle with a point at the center when viewed from directly above. 3. Thus, the projections depend entirely on the chosen viewpoint and orientation relative to the projection plane. **Answer:** The projection of a cube appears as a square (with internal edge lines), and the projection of a cone appears as a triangle (or a circle when viewed from the apex). > Common mistake: Forgetting that solid projections can include internal lines for visible or hidden edges. ### Question 9 *3 marks · Short answer* See Figures $4.2–4.5$. In each case, see if you can visualise another object that gives the same projection. **Solution** 1. A given projection on a plane does not uniquely determine the 3D object because depth information perpendicular to the plane is lost. 2. For example, in Fig. 4.6, different lines of varying lengths and inclinations can have identical projections on a vertical plane if their lengths parallel to the projection direction differ. 3. Similarly, different cuboids of varying depths can produce the exact same rectangular or square projection on a plane. **Answer:** Another object that gives the same projection can be created by changing the depth or extending the object along the direction perpendicular to the projection plane. > Common mistake: Assuming that a 2D projection corresponds to a unique 3D object. ### Question 10 *3 marks · Short answer* Find another object that makes the same projection as that of a given cone. **Solution** 1. The standard side projection of a cone on a vertical plane is a triangle. 2. Another 3D object that produces a triangular projection from the same viewpoint is a triangular pyramid (tetrahedron) oriented with a face or edge facing the plane appropriately. 3. Any pyramid with a curved or polygonal base that tapers to a single apex aligned with the viewpoint can also yield a similar triangular outline. **Answer:** A triangular pyramid (tetrahedron) oriented correctly can produce the same triangular projection as a cone. > Common mistake: Thinking only solids with curved surfaces can project to a triangle. ### Question 11 *3 marks · Short answer* Let us see what these projections are for the objects shown in Fig. 4.6, when the planes shown are taken to be vertical planes. **Solution** 1. When the planes shown in Fig. 4.6 are taken to be vertical planes, the projections of the lines and cuboids represent their front views. 2. For the lines, the front views appear as horizontal line segments of varying lengths depending on the inclination and length of the actual segments. 3. For the cuboids, the front views appear as congruent squares or rectangles, showing that distinct 3D arrangements can share identical front views. **Answer:** The projections form horizontal line segments for the lines, and identical squares or rectangles for the cuboids, demonstrating loss of depth information. > Common mistake: Confusing front view with top or side views. ## Figure it Out ### Question 1 *3 marks · Short answer* 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** 1. Let $l$ be the actual length of a line segment and $p$ be the length of its projection on a plane. 2. By dropping perpendiculars from the endpoints of the line to the plane, we form a right-angled triangle where the length of the projection is one of the sides and the actual length is the hypotenuse. 3. Therefore, the length of the projection is always less than or equal to the actual length of the line ($p \le l$), and it equals the actual length only when the line is parallel to the projection plane. **Answer:** The length of the projection of a line is always less than or equal to its actual length ($p \le l$), and is equal to the actual length only when the line is parallel to the plane. > Common mistake: Assuming that the length of the projection is always equal to the actual length of the line regardless of its orientation. ### Question 2 *3 marks · Short answer* 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. If needed, see the next problem for clues. **Solution** 1. Cube: Front view is a square, top view is a square, and side view is a square. 2. Cylinder: Front view is a rectangle, top view is a circle, and side view is a rectangle. 3. Cone: Front view is a triangle, top view is a circle with its centre, and side view is a triangle. **Answer:** The front, top, and side views depend on the solid: a cube gives squares, a cylinder gives rectangles and a circle, and a cone gives triangles and a circle. > Common mistake: Confusing the top view with the front or side view for asymmetric solids like a cone. ### Question 3 *3 marks · Short answer* Match each of the following objects with its projections. **Solution** 1. Examine the given 3D objects in the table along their specified front, top, and side viewing directions. 2. Identify the 2D outline (profile) formed on the vertical plane (front view), horizontal plane (top view), and side plane for each object. 3. Match each object to its corresponding set of front, top, and side projections correctly. **Answer:** Each object in the table is matched to its correct set of front, top, and side orthographic projections as shown in Fig. 4.6. > Common mistake: Swapping the front view and side view directions. ## Shadows - In-text questions ### Question 1 *2 marks · Very short answer* What do you see? **Solution** 1. The shape of the shadow on a plane is quite similar to the shape of the projection of the object on that plane. 2. However, the shadow may be scaled up, stretched, or slightly distorted depending on how the object is held. **Answer:** The shadow resembles the projection of the object, though it may be scaled up, stretched, or distorted. > Common mistake: Stating that the shadow is always identical in size to the object. ### Question 2 *2 marks · Very short answer* Observe what happens to the size of the shadow as you vary the distance between your torch and your object. **Solution** 1. As the distance between the torch and the object is varied, the size of the shadow changes. 2. Moving the torch closer to the object makes the shadow larger, while moving it farther away makes the shadow smaller. **Answer:** The size of the shadow changes; moving the torch closer makes it larger, and moving it farther makes it smaller. > Common mistake: Confusing the effect of moving the torch with moving the object. ### Question 3 *2 marks · Very short answer* Why does this happen? **Solution** 1. Light travels in straight lines from a point source (the torch). 2. As the light rays spread out from the source, objects placed closer intercept rays that are further apart, casting a larger shadow. **Answer:** This happens because light travels in straight lines and spreads out from the source, enlarging the intercepted shadow area when closer. > Common mistake: Forgetting that light rays diverge from a point source. ## Figure it Out ### Question 1 *3 marks · Short answer* Draw the top view, front view and the side view of each of the following combinations of identical cubes. **Part (i)** 1. Observe the first combination of cubes from the front, top, and side. 2. The front view shows the outline facing the vertical plane. 3. The top view shows squares corresponding to the plan from above, and the side view shows the profile from the side. Answer (i): Front view, top view, and side view as determined by the projections of the given shape. **Answer:** The top view, front view, and side view are determined by projecting each combination of cubes onto the horizontal, vertical, and side planes respectively. > Common mistake: Confusing the viewing direction for the top view and side view. ### Question 2 *3 marks · Short answer* Imagine eight identical cubes, glued together along faces to form the letter ‘C’. (i) This looks like a ‘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 ‘C’ from the top. (iii) Now, can you glue even more cubes to make it look like ‘C’ from the front, ‘C’ from the top, and ‘C’ from the side? (iv) Can you think of other letter combinations to make with a single combination of cubes in this manner? **Part (i)** 1. A shape formed by 8 cubes resembling the letter C is viewed from different directions. 2. From the side, it shows a rectangular profile of width equal to a single cube and height equal to the full structure. 3. From the top, it shows the horizontal bars of the letter C. Answer (i): Side view is a rectangle of height equal to the letter and width of one cube; top view shows the two parallel horizontal bars. **Part (ii)** 1. To make a shape that looks like C from both front and top, add cubes to fill the gaps appropriately. 2. Ensure the front projection remains the letter C and the top projection also forms the letter C. Answer (ii): Add cubes in the appropriate positions to satisfy both front and top projections. **Part (iii)** 1. Glue additional cubes such that the projections from front, top, and side all form the letter C. 2. This requires extending the cube structure symmetrically along the depth and width axes. Answer (iii): A symmetrical 3D block structure whose orthogonal projections all resemble the letter C. **Part (iv)** 1. Consider other block letters like L, T, or H formed by connected cubes. 2. Visualise their combinations and check their projections. Answer (iv): Other letters such as L and T can similarly be formed using appropriate cube combinations. **Answer:** Analysis of views and extensions for cube structures forming the letter C. > Common mistake: Failing to account for the depth of cubes when considering side and top views. ### Question 3 *3 marks · Short answer* Which solid corresponds to the given top view, front view, and side view? **Solution** 1. Examine the given front view, top view, and side view of the solid. 2. Compare the dimensions and layout of the squares in the given projections with the options (i) to (vii). 3. Identify that the solid corresponds to option (iv) as its orthogonal projections exactly match the given views. **Answer:** (iv) > Common mistake: Matching only one view instead of checking all three mutually perpendicular views. ### Question 4 *3 marks · Short answer* Using identical cubes, make a solid that gives the following projections. **Solution** 1. Analyze the given top view, front view, and side view panels for each sub-part. 2. Determine the number of cubes and their arrangement needed in each layer to satisfy all three views simultaneously. 3. Assemble identical cubes to form the solid corresponding to the projections. **Answer:** Solids constructed using identical cubes that match the given projections for panels (i) to (ix). > Common mistake: Placing cubes in higher layers without verifying if they obstruct the required front or side views. ### Question 5 *3 marks · Short answer* Find the number of cubes in this stack of identical cubes. **Solution** 1. Count the number of cubes layer by layer from top to bottom as shown in the figure. 2. The top layer has $1 \times 1 = 1$ cube. 3. The second layer has $2 \times 2 = 4$ cubes. 4. The third layer has $3 \times 3 = 9$ cubes. 5. The bottom layer has $4 \times 4 = 16$ cubes. 6. Total number of cubes = $1 + 4 + 9 + 16 = 30$ cubes. **Answer:** 30 cubes > Common mistake: Counting only the visible cubes or forgetting to include the cubes in the hidden lower layers. ### Question 6 *3 marks · Short answer* What are the different shapes the projection of a cube can make under different orientations? **Solution** 1. When a cube is projected onto a plane, its apparent 2D shape depends on how it is oriented relative to the projection plane. 2. Orienting it face-on gives a square projection, balancing it on an edge can give a rectangular or polygonal shape, and balancing it perfectly on a corner vertex gives a regular hexagon as its isometric projection. **Answer:** A square, various rectangles, and a regular hexagon (isometric projection). > Common mistake: Thinking a cube can only project as a square. ## Isometric Projections ### Question 1 *3 marks · Short answer* Construct a model of a cube and use your hands to keep it balanced on one corner vertex. Can you try to understand why all the projected edges have equal length? **Solution** 1. When a cube is balanced on one of its corner vertices, all three visible pairs of parallel edges meet at equal angles of $120^\circ$ around the projected center. 2. Due to the symmetrical orientation of the cube with respect to the projection plane, each edge is inclined at the same angle to the plane. 3. Consequently, the lengths of the projections of all the edges are equal, forming a regular hexagon. **Answer:** All projected edges have equal length because the cube's edges make equal angles with the projection plane due to its symmetrical balance on a corner vertex. > Common mistake: Thinking the physical edges change in length instead of recognising that equal inclination causes equal projected lengths. ### Question 2 *3 marks · Short answer* Imagine these are cubes, not squares. Draw each of these on your isometric paper (you can find it at the end of the book). **Solution** 1. Identify the three principal axes on the isometric grid: height (vertical lines), depth (lines sloping down-left), and length (lines sloping down-right). 2. Draw each of the five Tetris shapes cube by cube along the chosen axes, counting the unit lengths carefully. 3. Darken the visible edges and erase any hidden lines to complete the 3D representation of each shape. **Answer:** The five Tetris shapes are drawn by representing each constituent square as a cube oriented along the isometric axes. > Common mistake: Drawing edges along incorrect grid directions. ### Question 3 *3 marks · Short answer* For example, you can draw a $1 \times 1 \times 1$ cube as follows. How would you draw a $2 \times 2 \times 2$ cube? Feel free to add shading, if it helps you visualise the solid. **Solution** 1. Start from a corner vertex on the isometric grid and draw edges of length 2 units along the height, length, and depth axes. 2. Complete the remaining parallel edges to form the outer boundary of the $2 \times 2 \times 2$ cubical box. 3. Add internal grid lines to show the individual $1 \times 1 \times 1$ cubes and shade appropriately to enhance visualisation. **Answer:** A $2 \times 2 \times 2$ cube is drawn by scaling the unit lengths to 2 units along each of the three isometric axes. > Common mistake: Drawing edges of length 1 unit instead of 2 units. ### Question 4 *2 marks · Very short answer* Why is this correspondence between directions on isometric paper and axes of the solid so effective for communicating the shape of the solid? **Solution** 1. Parallel lines on a solid always project to parallel lines on the isometric paper. 2. The three families of parallel edges on the solid project to the three specific grid orientations corresponding to height, length, and depth, while preserving unit distances. **Answer:** Because parallel lines project to parallel lines and unit distances along the three principal axes project equally on the grid. > Common mistake: Failing to mention the preservation of parallel lines and equal unit lengths. ### Question 5 *3 marks · Short answer* Can you try drawing the other tetris shapes on isometric paper? **Solution** 1. Select an appropriate orientation for each remaining Tetris shape along the depth, length, and height axes. 2. Count the unit edge lengths on the isometric grid for each component cube. 3. Connect the edges to form the composite 3D block and darken the visible outlines. **Answer:** All remaining Tetris shapes are successfully drawn by aligning their cube units with the height, length, and depth axes of the isometric grid. > Common mistake: Miscounting the number of unit cubes when changing orientation. ## Figure it Out ### Question 1 *3 marks · Short answer* 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** 1. Yes, in addition to the $5$ planar shapes shown in Fig. 4.8, there are additional ways to glue four cubes together along their faces in three dimensions. 2. These are called non-planar tetracubes or L-trominoes with a cube attached out of the plane, giving rise to chiral forms. 3. There are $3$ more such distinct non-planar arrangements, making a total of $8$ free tetracubes in all. **Answer:** Yes, there are 3 additional non-planar ways, making a total of 8 free tetracubes. > Common mistake: Counting rotated or flipped versions as distinct new shapes. ### Question 2 *3 marks · Short answer* Draw the following figures on the isometric grid. **Solution** 1. Observe the given figures and identify the three primary directions: length, depth, and height corresponding to the axes of the isometric grid. 2. Draw the visible edges cube by cube, moving along the grid lines ($|$, $/$, and $\$) according to the dimensions of the solid. 3. Ensure that parallel lines remain parallel and hidden interior lines are omitted to produce the correct 3D representation. **Answer:** The figures are accurately reproduced on the isometric grid by aligning their edges with the height, length, and depth axes. > Common mistake: Drawing edges along incorrect isometric axes, distorting the shape of the solid. ### Question 3 *3 marks · Short answer* Is there anything strange about the path of this ball? Recreate it on the isometric grid. **Solution** 1. The strange aspect of the ball's path is that it appears to continuously climb or descend in an endless loop while remaining on a seemingly flat or rectangular structure, which is physically impossible in three-dimensional space. 2. To recreate it on the isometric grid, break down the figure into smaller physically realisable portions and identify the local three primary directions. 3. The illusion is created by cleverly misrepresenting the connections between corners at different depths, tricking the brain into perceiving an inconsistent 3D geometry. **Answer:** The path forms an impossible perpetual loop due to conflicting depth cues in the 2D drawing. > Common mistake: Trying to construct a physical model of the impossible object directly without noticing the conflicting depth cues. ### Question 4 *3 marks · Short answer* 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? **Part (i)** 1. State that it is physically impossible to build a 3D model out of actual cubes because the apparent three-way connections contain contradictory depth information. 2. State that the front, top, and side profiles each appear as ordinary L-shaped or rectangular configurations of squares when viewed orthogonally. Answer (i): Not possible to build; profiles appear as orthogonal projections of connected square arms. **Part (ii)** 1. Choose an isometric grid and align three rectangular blocks of cubes to meet at the corners. 2. Draw the visible faces of the cubes along the height, length, and depth axes to match the impossible perspective loop. Answer (ii): Recreated on the isometric grid following the stepped cuboid layout shown in the figure. **Part (iii)** 1. Explain that the illusion works because the 2D drawing uses consistent shading and parallel lines that trick the brain into assuming all meeting corners are right-angled in 3D space. 2. Conclude that local consistency misleads the global interpretation, making incompatible spatial relationships appear continuous. Answer (iii): Local consistency of individual corners tricks the brain into assuming a globally impossible 3D structure. **Answer:** The impossible triangle is an optical illusion created by exploiting our brain's tendency to interpret 2D projections as 3D objects with consistent depth, which cannot physically exist as a 3D model. > Common mistake: Stating that the impossible triangle can be physically constructed in three-dimensional Euclidean space. ## Frequently asked questions ### How many total questions are there in NCERT Solutions for Class 8 Maths Chapter 11 Exploring Some Geometric Themes? This chapter for the 2026-27 session contains a total of 79 questions distributed across various sections like Sierpinski Carpet, Visualising Solids, and Isometric Projections. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page. ### Which topics and concepts are covered in the Class 8 Maths Chapter 11 solutions? The solutions cover important concepts such as fractal patterns like the Sierpinski Carpet and Koch Snowflake, faces, edges, and vertices of solids, and shortest paths using the Baudhayana Theorem. Additional topics include projections, shadows, and isometric drawings to help strengthen spatial visualization. ### What are the hardest question types in this chapter and how should I approach them? Questions involving shortest paths on 3D surfaces and complex projections of solids are generally considered challenging by students. To approach them, use nets to flatten surfaces or draw clear isometric projections as explained in SwaVid's detailed solutions. ### How can I write answers for full marks in Class 8 Maths Chapter 11 questions? To score full marks, you should write clear steps, state relevant geometric properties or theorems clearly, and draw neat diagrams for nets and projections. Following SwaVid's structured step-by-step solutions on this page will help you format your answers correctly. ### Is the free PDF for Class 8 Maths Chapter 11 Exploring Some Geometric Themes available based on the new NCERT book? Yes, the complete chapter solutions aligned with the new NCERT book for the 2026-27 session are available as a free PDF. You can easily access and download SwaVid's comprehensive solutions directly from this page. ## Related pages - [Class 8 Maths chapters](https://www.swavid.com/maths/class/8) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student