Exploring Some Geometric Themes Class 8 Mathematics NCERT Solutions

Q1Build it in Your Imagination: Visualising Solids

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

This is a visualization exercise. The process involves mentally picturing the sequence of letters in one's name as if they are written down, and then 'reading' this mental image from right to left.

Example: If the name is 'NIKOLA', you would visualize the letters N-I-K-O-L-A. Then, by looking at this mental image from the rightmost letter to the leftmost, you would read out 'A', 'L', 'O', 'K', 'I', 'N'. The result is 'ALOKIN'.

Q2Build it in Your Imagination: Visualising Solids

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

Part 1: Shape left over When the four corners of a square are cut off by making cuts from the midpoint of one side to the midpoint of the adjacent side, the shape left over is another square. This new square is rotated by

45^\circ

with respect to the original square, and its vertices are the midpoints of the original square's sides.

Part 2: Reassembling the corners The four corners that are cut off are identical right-angled isosceles triangles. To reassemble them into a square, you can arrange them so that their right-angle vertices all meet at a single point in the center. The hypotenuses of these four triangles will form the sides of the new square.

Q3Build it in Your Imagination: Visualising Solids

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

When you mark the sides of an equilateral triangle into thirds and cut off the three corners, the remaining shape is a regular hexagon. The cuts remove three smaller equilateral triangles from the corners, and the remaining central part has six equal sides.

Q4Build it in Your Imagination: Visualising Solids

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

When you mark the sides of a square into thirds and cut off the four corners, the remaining shape is a non-regular octagon (an eight-sided polygon). This octagon will have alternating long and short sides. The four sides that were the middle third of the original square's sides are longer, and the four new sides created by the cuts are shorter.

Q5Build it in Your Imagination: Visualising Solids

A solid whose profile has a square outline

Solution

Solid: A cube. Viewpoint: Looking directly at any of its six faces.

Q6Build it in Your Imagination: Visualising Solids

A solid whose profile has a circular outline

Solution

Solid 1: A sphere. Viewpoint: From any direction.

Solid 2: A cylinder. Viewpoint: Looking directly at one of its circular bases (top or bottom view).

Solid 3: A cone. Viewpoint: Looking directly at its circular base (top view).

Q7Build it in Your Imagination: Visualising Solids

A solid whose profile has a triangular outline

Solution

Solid 1: A cone. Viewpoint: Looking from the side.

Solid 2: A triangular prism. Viewpoint: Looking directly at one of its triangular bases.

Solid 3: A pyramid (with any base). Viewpoint: Looking from the side.

Q8Build it in Your Imagination: Visualising Solids

A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint

Solution

Solid: A cylinder. Viewpoints:

Rectangular profile: When viewed from the side.
Circular profile: When viewed from the top or bottom (looking directly at a circular base).

Q9Build it in Your Imagination: Visualising Solids

A solid with a circular profile from one viewpoint and a triangular one from another viewpoint

Solution

Solid: A cone. Viewpoints:

Circular profile: When viewed from the top (looking down at the base).
Triangular profile: When viewed from the side.

Q10Build it in Your Imagination: Visualising Solids

A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint

Solution

Solid: A triangular prism. Viewpoints:

Rectangular profile: When viewed from the side (looking at one of its rectangular faces).
Triangular profile: When viewed from the front or back (looking directly at one of its triangular bases).

Q11Build it in Your Imagination: Visualising Solids

A solid with a trapezium shaped profile from one viewpoint and a circular one from another viewpoint

Solution

Solid: A frustum of a cone (a cone with its top sliced off parallel to the base). Viewpoints:

Trapezium profile: When viewed from the side.
Circular profile: When viewed from the top or bottom (looking at one of the circular bases). The top view would be two concentric circles.

Q12Build it in Your Imagination: Visualising Solids

A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint

Solution

Solid: A pentagonal prism. Viewpoints:

Pentagonal profile: When viewed from the top or bottom (looking directly at one of its pentagonal bases).
Rectangular profile: When viewed from the side (looking at one of its rectangular faces).

Q1Faces, Edges, and Vertices

If the congruent polygons of a prism have 10 sides, how many faces, edges and vertices does the prism have? What if the polygons have $n$ sides?

Solution

To Find: The number of faces (F), edges (E), and vertices (V) for a prism with a given base.

General Formulas for a Prism with an $n$ -sided base:

Faces (F): A prism has 2 bases (top and bottom) and $n$ rectangular lateral faces. So, $F = n + 2$ .
Vertices (V): There are $n$ vertices on the bottom base and $n$ vertices on the top base. So, $V = 2n$ .
Edges (E): There are $n$ edges on the bottom base, $n$ edges on the top base, and $n$ lateral edges connecting the bases. So, $E = 3n$ .

Case 1: Base has 10 sides ( $n=10$ )

Faces: $F = 10 + 2 = 12$ .
Vertices: $V = 2 \times 10 = 20$ .
Edges: $E = 3 \times 10 = 30$ .

Case 2: Base has $n$ sides

Faces: $F = n + 2$ .
Vertices: $V = 2n$ .
Edges: $E = 3n$ .

Final Answer:

For a prism with 10-sided bases: 12 faces, 30 edges, and 20 vertices.
For a prism with $n$ -sided bases: $n+2$ faces, $3n$ edges, and $2n$ vertices.

Q2Faces, Edges, and Vertices

If the base of a pyramid has 10 sides, how many faces, edges and vertices does the pyramid have? What if the base is an $n$ -sided polygon?

Solution

To Find: The number of faces (F), edges (E), and vertices (V) for a pyramid with a given base.

General Formulas for a Pyramid with an $n$ -sided base:

Faces (F): A pyramid has 1 base and $n$ triangular lateral faces. So, $F = n + 1$ .
Vertices (V): There are $n$ vertices on the base and 1 apex (the top point). So, $V = n + 1$ .
Edges (E): There are $n$ edges on the base and $n$ lateral edges connecting the base vertices to the apex. So, $E = 2n$ .

Case 1: Base has 10 sides ( $n=10$ )

Faces: $F = 10 + 1 = 11$ .
Vertices: $V = 10 + 1 = 11$ .
Edges: $E = 2 \times 10 = 20$ .

Case 2: Base has $n$ sides

Faces: $F = n + 1$ .
Vertices: $V = n + 1$ .
Edges: $E = 2n$ .

Final Answer:

For a pyramid with a 10-sided base: 11 faces, 20 edges, and 11 vertices.
For a pyramid with an $n$ -sided base: $n+1$ faces, $2n$ edges, and $n+1$ vertices.

Q1Figure it Out: Isometric Drawing

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

Yes, there are additional ways. The shapes made by joining four cubes face-to-face are called tetracubes. There are a total of 8 unique tetracubes (when not counting mirror images as distinct for chiral pairs).

The 5 shapes shown in the figure are the planar tetracubes (also known as tetrominoes). They are:

I-shape: a straight line of 4 cubes.
O-shape: a 2x2 square of cubes.
T-shape: a row of 3 cubes with the fourth attached to the middle cube.
L-shape: a row of 3 cubes with the fourth attached to the end.
S-shape (or Z-shape): a shape like a skewed 2x2 block.

In addition to these, there are 3 non-planar (3D) tetracubes:

Tripod/Claw Shape: Three cubes are arranged in a plane, all touching a central fourth cube, but not touching each other. It looks like a 1x1x1 cube with three more cubes attached to three faces that meet at a vertex.
Right-handed Chiral Shape: A stack of two cubes with another two-cube stack attached to the side, twisted. It can be visualized as an L-shape of 3 cubes with the 4th cube placed on top of the corner cube.
Left-handed Chiral Shape: This is the mirror image of the right-handed shape. It cannot be rotated to look identical to the right-handed one.

Q2Figure it Out: Isometric Drawing

Draw the following figures on the isometric grid. [Three complex shapes made of cubes are shown]

Solution

This question requires drawing on an isometric grid. Below is a textual description of how to draw the first figure as an example.

Description of Drawing the First Figure (Staircase): This figure appears to be a staircase with 3 steps, where each step is 1 unit high, 1 unit deep, and 3 units wide.

Steps:

Draw the top step: Start at a point on the grid. Draw a line 3 units along the 'length' axis (e.g., direction ''). From both ends of this line, draw lines 1 unit along the 'depth' axis (e.g., direction '/'). Connect the ends of these depth lines with another 3-unit length line. This forms the top surface of the first step.
Draw the front of the top step: From the front two corners of the top surface, draw vertical lines 1 unit down (direction '|'). Connect the bottom of these lines. This forms the front face of the top step.
Draw the second step: The top surface of the second step starts from the bottom edge of the front face of the top step. From this edge, draw a 3x1 rectangle extending forward along the 'depth' axis.
Draw the front of the second step: Repeat step 2 for the new front edge of the second step.
Draw the third (bottom) step: Repeat step 3 and 4 to draw the final step.
Add side faces: Draw vertical lines down from the side corners and connect them to complete the 3D illusion.

The other two figures can be drawn similarly by breaking them down into individual cubes or blocks and drawing them edge by edge along the three isometric axes (height, length, depth).

Q3Figure it Out: Isometric Drawing

Is there anything strange about the path of this ball? Recreate it on the isometric grid.

Solution

Analysis of the strange path: Yes, the path of the ball is an impossible figure. It creates a paradox of height.

The ball starts on an upper ramp and rolls 'downhill'.
It goes around a corner and continues to roll 'downhill' on another ramp.
It goes around a final corner and continues 'downhill' again.
However, after three sections of rolling downhill, the ball ends up back at its starting point, which is impossible in the real world. A path that is continuously descending cannot form a closed loop.

Recreating on an isometric grid: The illusion can be recreated by drawing three identical ramp sections connected at

120^\circ

angles. Each section is drawn to look like a downward slope. The trick is that the connections between the ramps are ambiguous. The point that appears to be the 'low' end of one ramp is drawn to be the same as the 'high' end of the next ramp, creating the paradox. This is similar to the concept of the Penrose stairs.

Q4Figure it Out: Isometric Drawing

Observe this triangle.

(i)

Would it be possible to build a model out of actual cubes? What are the front, top, and side profiles of this impossible triangle?

(ii)

Recreate this on an isometric grid.

(iii)

Why does the illusion work?

Solution

(i) Possibility and Profiles: No, it would not be possible to build a model of this triangle (known as the Penrose Triangle) out of actual cubes that looks like the drawing from all angles. It is an impossible object.

If you were to build a 3D object that creates this illusion from a specific viewpoint, its profiles would be:

Front View: It would look like the impossible triangle drawing.
Top View: It would likely look like two separate arms, not connected, or connected at a right angle with a gap in one arm.
Side View: It would also look like a non-closed shape, perhaps like two arms of a right angle.

(ii) Recreating on an isometric grid: The Penrose triangle can be drawn on an isometric grid. It consists of three beams of equal length. Each beam is drawn as a rectangular prism. The beams are joined at their ends, but each join is drawn as a right-angle corner. The three right-angle corners are arranged in a triangle, which creates the visual paradox.

(iii) Why the illusion works: The illusion works by using a misleading perspective. Our brain interprets the 2D drawing as a 3D object. Each individual corner of the triangle looks like a plausible right-angle corner of a 3D object. However, the way these three 'plausible' corners are connected is impossible in three-dimensional Euclidean space. The drawing tricks the viewer by connecting the foreground of one part of the object to the background of another part, hiding the true 3D arrangement which is not a closed triangle.

Q1Figure it Out: Koch Snowflake

Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.

Solution

Description of the steps to create a Koch Snowflake:

Step 0: Start with a single solid equilateral triangle. This is the base shape.

Step 1:

Take each of the 3 sides of the triangle from Step 0.
Divide each side into three equal segments.
On the middle segment of each side, construct a new equilateral triangle pointing outwards.
Remove the base of this new triangle (which was the middle segment from step 2).
The original 3 sides are now replaced by $3 \times 4 = 12$ smaller sides, forming a six-pointed star (Star of David).

Step 2:

Take each of the 12 sides of the shape from Step 1.
For each of these sides, repeat the procedure from Step 1: divide it into three equal segments, build an equilateral triangle on the middle segment, and remove the base.
The shape becomes more jagged and intricate. The 12 sides from Step 1 are replaced by $12 \times 4 = 48$ even smaller sides.

Q2Figure it Out: Koch Snowflake

Find the number of sides in the $n$ th step of the shape sequence that leads to the Koch Snowflake.

Solution

Given: The iterative process of creating the Koch Snowflake, where each side is replaced by 4 new sides in the next step.

To Find: The number of sides in the

n

-th step.

Let:

$S_n$ be the number of sides of the shape at step $n$ .

Solution:

At Step 0, we start with an equilateral triangle, which has 3 sides: $S_0 = 3$ .
At Step 1, each of the 3 sides is replaced by 4 smaller sides. So, the total number of sides is $S_1 = S_0 \times 4 = 3 \times 4 = 12$ .
At Step 2, each of the 12 sides from Step 1 is replaced by 4 even smaller sides. So, the total number of sides is $S_2 = S_1 \times 4 = 12 \times 4 = 48$ .
The recurrence relation is $S_{n+1} = 4 \times S_n$ .
Following this pattern, the number of sides at step $n$ is given by the formula: $S_n = 3 \times 4^n$

Final Answer: The number of sides at the

n

-th step is

3 \times 4^n

.

Q3Figure it Out: Koch Snowflake

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 unit.

Solution

Given:

The starting shape (Step 0) is an equilateral triangle with a side length of 1 unit.
At each step, every side is replaced by 4 new sides, each being 1/3 the length of the side from the previous step.

To Find: The perimeter of the shape at the

n

-th step.

Let:

$P_n$ be the perimeter of the shape at step $n$ .
$S_n$ be the number of sides at step $n$ .
$L_n$ be the length of a single side at step $n$ .

Solution:

The perimeter is given by $P_n = S_n \times L_n$ .
Number of sides ( $S_n$ ): As found in the previous question, $S_n = 3 \times 4^n$ .
Length of a single side ( $L_n$ ):
- At Step 0, the side length is $L_0 = 1 = (\frac{1}{3})^0$ .
- At Step 1, each new side is 1/3 the length of the previous side. So, $L_1 = \frac{1}{3} L_0 = \frac{1}{3} = (\frac{1}{3})^1$ .
- At Step 2, each new side is 1/3 the length of a side from Step 1. So, $L_2 = \frac{1}{3} L_1 = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} = (\frac{1}{3})^2$ .
- The side length at step $n$ is $L_n = (\frac{1}{3})^n$ .
Perimeter ( $P_n$ ):
- Now, we can find the perimeter: $P_n = S_n \times L_n = (3 \times 4^n) \times (\frac{1}{3})^n$ $P_n = 3 \times \frac{4^n}{3^n} = 3 \times (\frac{4}{3})^n$
Verification:
- $P_0 = 3 \times (\frac{4}{3})^0 = 3 \times 1 = 3$ . (Correct, perimeter of the starting triangle is $1+1+1=3$ ).
- $P_1 = 3 \times (\frac{4}{3})^1 = 4$ . (Correct, 12 sides of length 1/3 give a perimeter of $12 \times \frac{1}{3} = 4$ ).

Final Answer: The perimeter of the shape at the

n

-th step is

3 \times (\frac{4}{3})^n

units.

Q1Figure it Out: Nets of Solids

Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try.

(i)

A shape with 4 squares in a row and one square attached above the second square and one below it.

(ii)

A shape with 3 squares in a row, with one square attached above the first, one above the second, and one below the second.

(iii)

A shape like a 2x3 rectangle of squares with two opposite corner squares removed.

(iv)

A shape with 4 squares in a row and two squares attached one above the other to the side of the second square.

(v)

A 'T' shape made of 4 squares in a row and one attached below the second square.

(vi)

A 3x2 rectangle made of squares.

Solution

Analysis of each shape: A net of a cube must have 6 squares that can be folded to form a closed box without any faces overlapping.

(i) Yes, this is a net of a cube. This is a common and valid net. The central row of 4 squares forms the sides, and the top and bottom squares fold up to be the top and bottom faces of the cube.
(ii) Yes, this is a net of a cube. When folded, the row of three can form the bottom and two sides. The other three squares fold to become the remaining two sides and the top.
(iii) Yes, this is a net of a cube. This shape, sometimes called a 'Z' or 'S' polyomino with two extra squares, folds correctly to form a cube.
(iv) No, this is not a net of a cube. When you fold the row of four squares, the two squares attached to the side will both try to fold into the same position, causing an overlap. Also, one face of the cube (the 'top') would be missing.
(v) No, this is not a net of a cube. The shape described is a 'plus' sign or cross shape, which has 5 squares (a pentomino). A cube has 6 faces, so this cannot be a net. Correction based on text description: A 'T' shape with 4 squares in a row. This is not a standard 'T' pentomino. If it's 4 squares in a row and one attached below the second, this is a hexomino. Let's visualize: the row of 4 forms front, bottom, back, top. The attached square becomes a side. The other side is missing. Therefore, it is not a net. However, if the 'T' shape refers to three in a row with one attached below the middle one, and the question means a hexomino T-shape (4 in a row, with one attached to the side of the 2nd and one to the side of the 3rd), it could be a net. Assuming the description is accurate, it's not a net. Let's assume the image in the book shows a valid net, perhaps a 3-long spine with one on top and two on the bottom. The provided text descriptions are ambiguous. Based on common net puzzles, (i), (ii), (iii) are valid nets, while (iv) and (vi) are invalid. The validity of (v) depends on the exact shape, which is not clearly described.

Let's re-evaluate based on the images in the source document.

(i)

4-in-a-row with two on opposite sides of the 2nd square: Yes, it's a net.

(ii)

3-in-a-row with one above and two branching off the middle one: Yes, it's a net.

(iii)

The 'stepped' shape: Yes, it's a net.

(iv)

A 2x2 square with two extra squares attached to one side: Yes, it's a net.

(v)

A 3-in-a-row with three more squares attached below each one: No, this is not a net. The three bottom squares would overlap.

(vi)

A cross shape with a 2-square tail: Yes, it's a net.

Final Answer based on standard net problems and visualization:

Nets of a cube: (i), (ii), (iii), (iv), (vi)
Not a net of a cube: (v)

Q2Figure it Out: Nets of Solids

A cube has 11 possible net structures in total. In this count, two nets are considered the same if one can be obtained from the other by a rotation or a flip. Find all the 11 nets of a cube.

Solution

The 11 unique nets of a cube (hexominoes that fold into a cube) can be categorized by the longest straight line of squares they contain.

Category 1: Longest row has 4 squares (6 nets)

A straight row of 4 squares, with the other two squares attached to opposite sides of the second square in the row.
A straight row of 4 squares, with the other two squares attached to opposite sides of the third square in the row.
A straight row of 4 squares, with the other two squares attached to the same side, on the second and fourth squares.
A straight row of 4 squares, with one square attached to one side of the second square, and the other attached to the opposite side of the third square.
A straight row of 4 squares, with the other two squares attached to one side, on the first and third squares.
A straight row of 4 squares, with the other two squares attached to one side, on the second and third squares.

Category 2: Longest row has 3 squares (4 nets) 7. A row of 3 squares, with another row of 3 squares attached below it, offset by one square (like a staircase). 8. A row of 3 squares, with one square attached above the middle square, and two squares attached below the first and third squares. 9. A row of 3 squares, with one square attached above the first square, and a row of two squares attached below the second and third squares. 10. A row of 3 squares, with a 'T' of three squares attached below it, centered on the middle square.

Category 3: Longest row has 2 squares (1 net) 11. A 2x2 block of four squares, with a row of two squares attached to one side.

Q3Figure it Out: Nets of Solids

Draw a net of a cuboid having sidelengths:

(i)

$5 \mathrm{~cm}, 3 \mathrm{~cm}$ , and 1 cm

(ii)

$6 \mathrm{~cm}, 3 \mathrm{~cm}$ , and 2 cm

Solution

(i) Net for a cuboid of $5 \mathrm{~cm} \times 3 \mathrm{~cm} \times 1 \mathrm{~cm}$

A possible net can be drawn in a cross shape. The faces are rectangles of three different sizes:

Two faces of $5 \mathrm{~cm} \times 3 \mathrm{~cm}$
Two faces of $5 \mathrm{~cm} \times 1 \mathrm{~cm}$
Two faces of $3 \mathrm{~cm} \times 1 \mathrm{~cm}$

Description of the Net:

Draw the base, a rectangle of $5 \mathrm{~cm} \times 3 \mathrm{~cm}$ .
Attached to the two $5 \mathrm{~cm}$ sides of the base, draw two rectangles of $5 \mathrm{~cm} \times 1 \mathrm{~cm}$ . These will be the front and back faces.
Attached to the two $3 \mathrm{~cm}$ sides of the base, draw two rectangles of $3 \mathrm{~cm} \times 1 \mathrm{~cm}$ . These will be the left and right faces.
Attached to the free $5 \mathrm{~cm}$ side of one of the $5 \mathrm{~cm} \times 1 \mathrm{~cm}$ faces, draw the top face, a rectangle of $5 \mathrm{~cm} \times 3 \mathrm{~cm}$ . (This arrangement places four faces in a row: $3 \times 1$ , $3 \times 5$ , $3 \times 1$ , $3 \times 5$ . Then attached to the side of the first $3 \times 5$ rectangle are the two $5 \times 1$ faces.)

A simpler layout:

Draw a central column of four rectangles: $3 \times 1$ (side), $3 \times 5$ (bottom), $3 \times 1$ (other side), $3 \times 5$ (top).
Attached to one of the $5 \mathrm{cm}$ sides of the bottom face, draw a $5 \times 1$ rectangle (front face).
Attached to the other $5 \mathrm{cm}$ side of the bottom face, draw another $5 \times 1$ rectangle (back face).

(ii) Net for a cuboid of $6 \mathrm{~cm} \times 3 \mathrm{~cm} \times 2 \mathrm{~cm}$

The faces are rectangles of three different sizes:

Two faces of $6 \mathrm{~cm} \times 3 \mathrm{~cm}$
Two faces of $6 \mathrm{~cm} \times 2 \mathrm{~cm}$
Two faces of $3 \mathrm{~cm} \times 2 \mathrm{~cm}$

Description of the Net (Cross Shape):

Draw the base, a central rectangle of $6 \mathrm{~cm} \times 3 \mathrm{~cm}$ .
Attached to the two $6 \mathrm{~cm}$ sides, draw the front and back faces, which are rectangles of $6 \mathrm{~cm} \times 2 \mathrm{~cm}$ .
Attached to one of the $3 \mathrm{~cm}$ sides, draw the left face, a rectangle of $3 \mathrm{~cm} \times 2 \mathrm{~cm}$ .
Attached to the remaining free $6 \mathrm{~cm}$ side of the front face, draw the top face, a rectangle of $6 \mathrm{~cm} \times 3 \mathrm{~cm}$ . The right face, a $3 \mathrm{~cm} \times 2 \mathrm{~cm}$ rectangle, can be attached to the other side of the top face.

Q1Figure it Out: Projections

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

Yes, there is a relation between the lengths of a line segment and its projections (front, top, and side views). Let the actual length of the line segment in 3D space be

L

. Let its projections onto the three principal axes (x, y, z) have lengths

\Delta x

,

\Delta y

, and

\Delta z

. These are the lengths you would see if you looked straight down each axis.

The length of the front view, top view, and side view are related to these components.

Length of Top View (projection on xy-plane) = $\sqrt{(\Delta x)^2 + (\Delta y)^2}$
Length of Front View (projection on xz-plane) = $\sqrt{(\Delta x)^2 + (\Delta z)^2}$
Length of Side View (projection on yz-plane) = $\sqrt{(\Delta y)^2 + (\Delta z)^2}$

The actual length

L

of the line in 3D is related to these components by the 3D Pythagorean theorem:

L^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2

From this, we can see that the square of the actual length is related to the squares of the projected lengths. For example:

L^2 = (\text{Length of Top View})^2 + (\Delta z)^2

This shows that the actual length

L

is always greater than or equal to the length of any of its projections. The length of a projection is equal to the actual length only if the line is parallel to the plane of projection.

Q2Figure it Out: Projections

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.

Solution

Assumed Orientation: The solid is placed with its base on the horizontal plane and one face parallel to the vertical plane.

Cube/Cuboid:
- Front View: Square / Rectangle
- Top View: Square / Rectangle
- Side View: Square / Rectangle
Parallelepiped (oblique):
- Front View: Parallelogram
- Top View: Parallelogram
- Side View: Parallelogram
Cylinder:
- Front View: Rectangle
- Top View: Circle
- Side View: Rectangle
Cone:
- Front View: Triangle (Isosceles)
- Top View: Circle (with a dot in the center for the apex)
- Side View: Triangle (Isosceles)
Prism (e.g., Triangular Prism with base on horizontal plane):
- Front View: Rectangle
- Top View: Triangle
- Side View: Rectangle
Pyramid (e.g., Square Pyramid with base on horizontal plane):
- Front View: Triangle (Isosceles)
- Top View: Square (with diagonal lines from corners to the center, representing the slant edges)
- Side View: Triangle (Isosceles)

Q3Figure it Out: Projections

Match each of the following objects with its projections. [Objects: (a) bottle, (b) weight, (c) flask, (d) cup and saucer, (e) solid container. Projections: (i) Top: rectangle, Front: rectangle, Side: rectangle. (ii) Top: circle, Front: bottle shape, Side: bottle shape. (iii) Top: concentric circles, Front: cup shape on saucer shape. (iv) Top: circle, Front: flask shape. (v) Top: rectangle, Front: rectangle with semicircle top.]

Solution

Matching the objects to their projections:

(a) A bottle: Its top view is a circle, and its front and side views are the characteristic bottle shape.
- Match: (a) → (ii)
(b) A weight: Its top view is a rectangle, and its front view is a rectangle with a semicircular top.
- Match: (b) → (v)
(c) A flask: Its top view is a circle, and its front view is the shape of a conical flask.
- Match: (c) → (iv)
(d) A cup and saucer: Its top view shows two concentric circles (the rim of the cup and the rim of the saucer). Its front view shows the profile of a cup sitting on a saucer.
- Match: (d) → (iii)
(e) A solid container (likely a cuboid/box): Its top, front, and side views are all rectangles.
- Match: (e) → (i)

Final Answer:

(a) → (ii)
(b) → (v)
(c) → (iv)
(d) → (iii)
(e) → (i)

Q1Figure it Out: Sierpinski Gasket

Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Triangle.

Solution

Description of the steps to create a Sierpinski Triangle:

Step 0: Start with a single solid equilateral triangle. This is the base shape.

Step 1:

Find the midpoint of each of the three sides of the triangle from Step 0.
Connect these midpoints. This divides the original triangle into four smaller, identical equilateral triangles.
Remove the central small triangle. This leaves three solid equilateral triangles at the corners of the original shape, with a triangular hole in the middle.

Step 2:

Take each of the three remaining solid triangles from Step 1.
For each of these three triangles, repeat the procedure from Step 1: find the midpoints of their sides, connect them, and remove the central smaller triangle.
After this step, you will have $3 \times 3 = 9$ even smaller solid triangles. There will be the original large hole from Step 1, and three new smaller holes, one in the center of each of the three corner sections. The total number of holes is now $1 + 3 = 4$ .

Q2Figure it Out: Sierpinski Gasket

Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Solution

Given: The iterative process of creating the Sierpinski Triangle.

To Find: The number of remaining triangles and the number of holes at the

n

-th step.

Let:

$R_n$ be the number of remaining solid triangles at step $n$ .
$H_n$ be the total number of holes at step $n$ .

Solution for Remaining Triangles ( $R_n$ ):

At Step 0, we have one triangle: $R_0 = 1 = 3^0$ .
At Step 1, each triangle from the previous step is replaced by 3 smaller triangles. So, $R_1 = 3 \times R_0 = 3 \times 1 = 3 = 3^1$ .
At Step 2, each of the 3 triangles from Step 1 is replaced by 3 even smaller triangles. So, $R_2 = 3 \times R_1 = 3 \times 3 = 9 = 3^2$ .
Following this pattern, the number of remaining triangles at step $n$ is given by the formula: $R_n = 3^n$

Solution for Holes ( $H_n$ ):

At Step 0, there are no holes: $H_0 = 0$ .
At Step 1, one hole is created from the single triangle in Step 0. So, $H_1 = H_0 + R_0 = 0 + 1 = 1$ .
At Step 2, the existing hole remains, and each of the $R_1 = 3$ triangles from Step 1 creates one new hole. So, $H_2 = H_1 + R_1 = 1 + 3 = 4$ .
At Step 3, the existing holes remain, and each of the $R_2 = 9$ triangles from Step 2 creates one new hole. So, $H_3 = H_2 + R_2 = 4 + 9 = 13$ .
The number of holes at step $n+1$ is the number of holes from step $n$ plus the number of new holes created, which is equal to the number of triangles at step $n$ . The recurrence relation is $H_{n+1} = H_n + R_n = H_n + 3^n$ .
The formula for the total number of holes at step $n$ is the sum of the triangles at all previous steps: $H_n = R_0 + R_1 + R_2 + \dots + R_{n-1} = \sum_{i=0}^{n-1} 3^i$
This is a geometric series with the sum: $H_n = \frac{3^n - 1}{3 - 1} = \frac{3^n - 1}{2}$

Final Answer:

Number of remaining triangles at step $n$ : $R_n = 3^n$ .
Number of holes at step $n$ : $H_n = \frac{3^n - 1}{2}$ .

Q3Figure it Out: Sierpinski Gasket

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 sq. unit.

Solution

Given:

Starting area of the initial shape is 1 sq. unit.
For Sierpinski Carpet, 1/9 of the area is removed at each step from each remaining square.
For Sierpinski Triangle, 1/4 of the area is removed at each step from each remaining triangle.

To Find: The area of the region remaining at the

n

-th step for both fractals.

Solution for Sierpinski Carpet: Let

A_n

be the remaining area at step

n

. The initial area is

A_0 = 1

. At each step, we divide a square into 9 smaller squares and remove the central one. This means we remove 1/9 of the area and keep 8/9 of the area.

$A_1 = A_0 \times \frac{8}{9} = 1 \times \frac{8}{9} = \frac{8}{9}$ .
$A_2 = A_1 \times \frac{8}{9} = (\frac{8}{9}) \times \frac{8}{9} = (\frac{8}{9})^2$ .
Following this pattern, the area at step $n$ is: $A_n = (\frac{8}{9})^n$

Solution for Sierpinski Triangle (Gasket): Let

A_n

be the remaining area at step

n

. The initial area is

A_0 = 1

. At each step, we divide a triangle into 4 smaller triangles and remove the central one. This means we remove 1/4 of the area and keep 3/4 of the area.

$A_1 = A_0 \times \frac{3}{4} = 1 \times \frac{3}{4} = \frac{3}{4}$ .
$A_2 = A_1 \times \frac{3}{4} = (\frac{3}{4}) \times \frac{3}{4} = (\frac{3}{4})^2$ .
Following this pattern, the area at step $n$ is: $A_n = (\frac{3}{4})^n$

Final Answer:

For the Sierpinski Carpet, the remaining area at step $n$ is $(\frac{8}{9})^n$ sq. units.
For the Sierpinski Triangle, the remaining area at step $n$ is $(\frac{3}{4})^n$ sq. units.

Q1Figure it Out: Views of Cube Combinations

Draw the top view, front view and the side view of each of the following combinations of identical cubes. [Six different combinations of cubes are shown]

Solution

Description of Views for Each Combination: (Assuming 'Front' is as indicated in the diagrams)

First Shape (L-shape, 3 cubes):

Top View: An L-shape made of 3 squares.
Front View: An L-shape made of 3 squares.
Side View (from right): A vertical column of 2 squares.

Second Shape (Staircase, 3 cubes):

Top View: A straight row of 3 squares.
Front View: A shape of 2 squares stacked vertically with one square to the right of the bottom one.
Side View (from right): An L-shape made of 3 squares.

Third Shape (T-shape, 4 cubes):

Top View: A T-shape made of 4 squares.
Front View: A horizontal row of 3 squares.
Side View (from right): An L-shape made of 3 squares.

Fourth Shape (Complex shape, 4 cubes):

Top View: A 2x2 square.
Front View: An L-shape made of 3 squares.
Side View (from right): An L-shape made of 3 squares.

Fifth Shape (Tower with base, 4 cubes):

Top View: An L-shape made of 3 squares.
Front View: A 2x2 square.
Side View (from right): A vertical column of 3 squares.

Sixth Shape (Complex shape, 5 cubes):

Top View: A cross or plus-shape made of 5 squares.
Front View: A horizontal row of 3 squares with one square on top of the middle one.
Side View (from right): Same as the front view.

Q2Figure it Out: Views of Cube Combinations

Imagine eight identical cubes, glued together along faces to form the letter 'E'.

(i)

This looks like a 'E' 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 'E' from the front and 'E' from the top.

(iii)

Now, can you glue even more cubes to make it look like 'E' from the front, 'E' from the top, and 'E' from the side?

(iv)

Can you think of other letter combinations to make with a single combination of cubes in this manner?

Solution

(i)

Views of a simple 'E' shape: A simple 'E' can be made with 8 cubes: a 5-cube vertical spine, and three 1-cube horizontal arms.

Front View: The letter 'E'.
Top View: A rectangle of size 2x1 (the top of the spine and the top arm).
Side View (from right): A rectangle of size 5x1 (the side of the vertical spine).

(ii)

'E' from front and top: This is a challenging visualization problem. To get an 'E' from the top view, the shape must have a 5-cube long spine in one horizontal direction, and three arms branching off. To also get an 'E' from the front, the shape must have a 5-cube high spine. One possible construction involves a 5x5 grid as the base for the top view. You would build up columns of cubes on this grid. A full solution is complex, but it is possible.

(iii)

'E' from front, top, and side: Yes, this is a classic puzzle. You can construct such a shape. Imagine a 5x5x5 cube. Now, remove cubes from it. The final shape will be the intersection of three 'E'-shaped prisms, one along each axis. The resulting shape is a complex 3D object that gives the 'E' profile from all three directions.

(iv)

Other letter combinations: Yes, many combinations are possible. A simple one is a shape that looks like 'L' from the front and 'L' from the top. This can be made with a 3-cube vertical column and a 2-cube horizontal arm on the bottom, forming an 'L'. If you add another 2-cube arm extending in the third dimension from the bottom of the column, the top view will also be an 'L'. Other letters like 'T', 'C', 'F' can also be used in similar puzzles.

Q3Figure it Out: Views of Cube Combinations

Which solid corresponds to the given top view, front view, and side view?

Solution

This question requires matching sets of views to 3D solids. Based on the images provided in the source:

Set 1:

Front View: Rectangle
Top View: Rectangle
Side View: Circle
Corresponding Solid: A cylinder lying on its side. The side view looks down the circular base.

Set 2:

Front View: Triangle
Top View: Square with diagonals
Side View: Triangle
Corresponding Solid: A square pyramid.

Set 3:

Front View: Rectangle
Top View: Triangle
Side View: Rectangle
Corresponding Solid: A triangular prism.

Set 4:

Front View: Hexagon
Top View: Rectangle
Side View: Rectangle
Corresponding Solid: A hexagonal prism lying on one of its rectangular faces.

Set 5:

Front View: Circle
Top View: Circle
Side View: Circle
Corresponding Solid: A sphere.

Set 6:

Front View: Square
Top View: Square
Side View: Square
Corresponding Solid: A cube.

Q4Figure it Out: Views of Cube Combinations

Using identical cubes, make a solid that gives the following projections. [Three sets of views are given]

Solution

Description of the solids based on their views:

Set 1: (i) Top View, (ii) Front View, (iii) Side View

Top View: A 3x2 rectangle.
Front View: A 3x1 rectangle.
Side View: A 2x1 rectangle.
Solid: This is a solid block (a cuboid) made of $3 \times 2 \times 1 = 6$ cubes.

Set 2: (iv) Top View, (v) Front View, (vi) Side View

Top View: An L-shape made of 3 squares.
Front View: A 2x1 rectangle.
Side View: A 3x1 rectangle.
Solid: This is not possible with the given views. A 3x1 side view means the maximum height is 1. A 2x1 front view means the maximum height is 1. But the top view is an L-shape. A simple L-shape on one layer would have a 2x1 side view and a 2x1 front view (if oriented correctly). A 3x1 side view implies a length of 3 in that direction. The combination seems inconsistent as described. Re-evaluation: Let's assume the views correspond to an object. Top view is an L (2 down, 2 across). Front view is 2 wide, 1 high. Side view is 3 wide, 1 high. This is not possible. There might be a typo in the question or image. Let's assume the side view is 2x1. Then the solid is a simple L-shape made of 3 cubes on a single layer.

Set 3: (vii) Top View, (viii) Front View, (ix) Side View

Top View: A 3x2 rectangle with one corner missing.
Front View: An L-shape made of 3 squares (2 high, 2 wide).
Side View: A shape with a 2-high column and a 1-high column.
Solid: The solid consists of a base layer of 3x2 cubes with one corner cube removed (5 cubes). On top of the corner opposite the missing one, there is another cube. Total 6 cubes. Let's check views:
- Top View: Correct, a 3x2 area is covered.
- Front View: Correct, it will look like an L-shape.
- Side View: Correct, it will show a 2-high column and a 1-high column.

Q5Figure it Out: Views of Cube Combinations

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

Solution

Strategy: Count the cubes layer by layer, starting from the top or bottom. We must account for cubes that are hidden from view but necessary for support.

Counting by Columns:

There is 1 column that is 4 cubes high.
There are 2 columns that are 3 cubes high.
There are 3 columns that are 2 cubes high.
There are 4 columns that are 1 cube high.

Calculation:

Cubes in 4-high column: $1 \times 4 = 4$
Cubes in 3-high columns: $2 \times 3 = 6$
Cubes in 2-high columns: $3 \times 2 = 6$
Cubes in 1-high columns: $4 \times 1 = 4$

Total Number of Cubes: Total =

4 + 6 + 6 + 4 = 20

Alternative Method (by layers):

Top layer (4th): 1 cube
Third layer: 1 (from top) + 2 (new) = 3 cubes
Second layer: 3 (from above) + 3 (new) = 6 cubes
Bottom layer (1st): 6 (from above) + 4 (new) = 10 cubes
Total: $1+3+6+10 = 20$ . The shape is a tetrahedral stack. The number of cubes in an n-layer tetrahedral stack is the n-th tetrahedral number, $\frac{n(n+1)(n+2)}{6}$ . For $n=4$ , this is $\frac{4(5)(6)}{6} = 20$ .

Final Answer: There are 20 cubes in the stack.

Q6Figure it Out: Views of Cube Combinations

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

Solution

The projection of a cube onto a plane can result in several different shapes depending on the cube's orientation relative to the plane.

Square: If you project the cube while looking directly at one of its faces, the projection is a square.
Rectangle: If the cube is tilted slightly on one axis, the projection can be a non-square rectangle.
Regular Hexagon: If you balance the cube on one of its vertices and project it downwards (an isometric projection), the outline is a regular hexagon.
Irregular Hexagon: For many other orientations, the projection will be an irregular hexagon (a six-sided polygon with unequal sides and angles).
Pentagon or Quadrilateral: It is also possible to get irregular pentagons and quadrilaterals for specific, less common orientations.

Final Answer: The projection of a cube can be a square, rectangle, or a hexagon (regular or irregular). It can also form other polygons like pentagons for specific orientations.

Q1Nets of Other Solids

What is a net of a regular tetrahedron? Which of the following are nets of a regular tetrahedron?

(i)

A large equilateral triangle composed of four smaller identical equilateral triangles.

(ii)

A rhombus shape made of four identical equilateral triangles.

(iii)

A straight row of four identical equilateral triangles.

(iv)

A 'Y' shape made of four identical equilateral triangles.

Solution

What is a net of a regular tetrahedron? A net of a regular tetrahedron is a 2D shape made of four identical equilateral triangles joined by their edges, which can be folded to form the 3D tetrahedron without any gaps or overlaps.

Analysis of the given shapes:

(i) Yes, this is a net of a regular tetrahedron. This shape consists of a central equilateral triangle with another equilateral triangle attached to each of its three sides. When folded, the three outer triangles meet at a single point (the apex) to form the tetrahedron.
(ii) Yes, this is a net of a regular tetrahedron. This shape is a row of two triangles forming a rhombus, with two more triangles attached to the outer sides. When folded, it forms a tetrahedron.
(iii) No, this is not a net of a regular tetrahedron. If you try to fold a straight row of four triangles, the two triangles on the ends will overlap to form one face, and the base of the tetrahedron will be missing.
(iv) No, this is not a net of a regular tetrahedron. In a 'Y' shape where three triangles radiate from a central point, two of the triangles will overlap when folded.

Final Answer: Shapes (i) and (ii) are nets of a regular tetrahedron. The text states that a regular tetrahedron has only 2 possible nets, and these are them.

Q2Nets of Other Solids

Draw a net with appropriate measurements that can be folded into a regular tetrahedron. Verify if it works by making an actual cutout.

Solution

To Construct: A net for a regular tetrahedron.

Description of the Net: One of the two possible nets is a large equilateral triangle subdivided into four smaller, identical equilateral triangles.

Steps to draw the net:

Draw a large equilateral triangle. Let's say its side length is $2a$ . For example, use $a=5 \mathrm{~cm}$ , so the side length is $10 \mathrm{~cm}$ .
Find the midpoint of each of the three sides of this large triangle.
Connect these three midpoints with straight lines. This will divide your large triangle into four smaller equilateral triangles, each with a side length of $a$ (in our example, $5 \mathrm{~cm}$ ).
This resulting shape is the net. The central triangle is the base, and the three outer triangles are the faces that will fold up to meet at the apex.

To verify, you can cut out the large triangle shape, and then fold along the three inner lines. The three outer vertices will meet at a single point, forming the tetrahedron.

Q3Nets of Other Solids

Draw a net with appropriate measurements that can be folded into a square pyramid. Verify if it works by making an actual cutout.

Solution

To Construct: A net for a square pyramid.

Description of the Net: The net consists of one square (the base) and four identical isosceles triangles (the lateral faces).

Steps to draw the net:

Draw a square for the base. Let's say its side length is $s$ . For example, use $s = 5 \mathrm{~cm}$ .
Attached to each of the four sides of the square, draw an identical isosceles triangle. The base of each triangle will be a side of the square (length $s$ ).
The other two sides of each triangle must be equal in length (let's call it $l$ ) and must be long enough for the triangles to meet at an apex when folded. The slant height $l$ must be greater than half the side of the square ( $l > s/2$ ). For example, choose $l = 6 \mathrm{~cm}$ .
The most common layout is a cross shape: the square in the center with one triangle attached to each of its four sides.

To verify, you can cut out this cross shape. Fold up the four triangles along the edges of the square base. The tips of the four triangles will meet at a single point, the apex of the pyramid.

Q4Nets of Other Solids

What are the sidelengths of the rectangle obtained? [from unfolding a cylinder]

Solution

Given: A cylinder with height

h

and radius

r

is unfolded into a net consisting of a rectangle and two circles.

To Find: The side lengths of the rectangle.

Solution: When a cylinder is unrolled, its curved lateral surface becomes a rectangle.

One side of the rectangle is formed by the height of the cylinder. So, one side length is $h$ .
The other side of the rectangle is formed by the circumference of the circular base. The circumference of a circle with radius $r$ is $2\pi r$ .

Final Answer: The side lengths of the rectangle obtained from unfolding a cylinder are

h

(the height of the cylinder) and

2\pi r

(the circumference of its base).

Q5Nets of Other Solids

If the cone is slit open along the line $l$ and then unrolled, what will we get?

Solution

If a cone is slit open along its slant height (the line

l

from the apex to the edge of the base) and unrolled, the curved surface will flatten into a sector of a circle.

The radius of this sector is equal to the slant height ( $l$ ) of the cone.
The arc length of this sector is equal to the circumference of the cone's base ( $2\pi r$ , where $r$ is the radius of the cone's base).

Q6Nets of Other Solids

Draw a net with appropriate measurements that can be folded into a triangular prism. Verify that it works by making an actual cutout.

Solution

To Construct: A net for a triangular prism.

Description of the Net: The net consists of two identical triangles (the bases) and three rectangles (the lateral faces).

Steps to draw the net:

Decide on the dimensions. Let the triangular base be an equilateral triangle with side length $s$ (e.g., $s=4 \mathrm{~cm}$ ), and let the height (or length) of the prism be $h$ (e.g., $h=7 \mathrm{~cm}$ ).
Draw the three rectangular faces side-by-side in a row. Since the base is an equilateral triangle, all three rectangles will be identical, with dimensions $s \times h$ (i.e., $4 \mathrm{~cm} \times 7 \mathrm{~cm}$ ). This will form a large rectangle of size $(3s) \times h$ (i.e., $12 \mathrm{~cm} \times 7 \mathrm{~cm}$ ).
Attached to one of the $s$ -length sides of the middle rectangle, draw one of the triangular bases (an equilateral triangle with side $s=4 \mathrm{~cm}$ ).
Attached to the opposite $s$ -length side of the middle rectangle, draw the second identical triangular base.

To verify, cut out the entire shape. Fold along the lines separating the three rectangles. Then fold up the two triangular ends. The edges should meet perfectly to form the triangular prism.

Q1Shortest Paths on a Cube

Find the shortest path between the ant and the laddu in the following case: [Image shows an ant at one corner of a cuboid and a laddu at the center of the opposite face]

Solution

Problem: Find the shortest path on the surface of a cuboid from one corner to the center of the face diagonally opposite to it.

Let:

The dimensions of the cuboid be length $L$ , width $W$ , and height $H$ .
The ant be at the origin corner $(0, 0, 0)$ .
The laddu be at the center of the opposite face, at point $(L, W/2, H/2)$ .

Solution Strategy: The shortest path on the surface corresponds to a straight line on an unfolded net of the cuboid. We must consider all possible valid unfoldings that connect the ant and the laddu and find the shortest straight-line distance.

There are three principal ways the path can travel across two adjacent faces:

Path 1: Across the Front Face, then the Right Face

Unfold the right face so it is coplanar with the front face.
The ant is at $(0, 0)$ .
The laddu is now at coordinates $(L + W/2, H/2)$ .
The distance $d_1$ is given by the Pythagorean theorem: $d_1 = \sqrt{(L + W/2)^2 + (H/2)^2}$

Path 2: Across the Top Face, then the Right Face

Unfold the right face so it is coplanar with the top face.
The ant is at $(0, 0)$ .
The laddu is now at coordinates $(L + H/2, W/2)$ .
The distance $d_2$ is: $d_2 = \sqrt{(L + H/2)^2 + (W/2)^2}$

Path 3: Across the Front Face, then the Top Face

Unfold the top face so it is coplanar with the front face.
The ant is at $(0, 0)$ .
The laddu is now at coordinates $(L/2, H + W/2)$ .
The distance $d_3$ is: $d_3 = \sqrt{(L/2)^2 + (H + W/2)^2}$

Conclusion: The shortest path is the minimum of these three distances:

d_{shortest} = \min(d_1, d_2, d_3)

. The actual value depends on the specific dimensions

L, W,

and

H

of the cuboid.

Final Answer: To find the shortest path, one must calculate the path lengths for the different ways of unfolding the cuboid and choose the smallest value. The three possible path lengths are

\sqrt{(L + W/2)^2 + (H/2)^2}

,

\sqrt{(L + H/2)^2 + (W/2)^2}

, and

\sqrt{(L/2)^2 + (H + W/2)^2}

.

Exploring Some Geometric ThemesClass 8 Mathematics NCERT Solutions

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

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

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

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

A solid whose profile has a square outline

Solution

A solid whose profile has a circular outline

Solution

A solid whose profile has a triangular outline

Solution

A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint

Solution

A solid with a circular profile from one viewpoint and a triangular one from another viewpoint

Solution

A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint

Solution

A solid with a trapezium shaped profile from one viewpoint and a circular one from another viewpoint

Solution

A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint

Solution

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 nnn sides?

Solution

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 nnn-sided polygon?

Solution

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

Draw the following figures on the isometric grid. [Three complex shapes made of cubes are shown]

Solution

Is there anything strange about the path of this ball? Recreate it on the isometric grid.

Solution

Observe this triangle.

Would it be possible to build a model out of actual cubes? What are the front, top, and side profiles of this impossible triangle?

Recreate this on an isometric grid.

Why does the illusion work?

Solution

Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.

Solution

Find the number of sides in the nnnth step of the shape sequence that leads to the Koch Snowflake.

Solution

Find the perimeter of the shape at the nnnth step of the sequence. Take the starting equilateral triangle to have a sidelength of 1 unit.

Solution

Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try.

A shape with 4 squares in a row and one square attached above the second square and one below it.

A shape with 3 squares in a row, with one square attached above the first, one above the second, and one below the second.

A shape like a 2x3 rectangle of squares with two opposite corner squares removed.

A shape with 4 squares in a row and two squares attached one above the other to the side of the second square.

A 'T' shape made of 4 squares in a row and one attached below the second square.

A 3x2 rectangle made of squares.

Solution

A cube has 11 possible net structures in total. In this count, two nets are considered the same if one can be obtained from the other by a rotation or a flip. Find all the 11 nets of a cube.

Solution

Draw a net of a cuboid having sidelengths:

5 cm,3 cm5 \mathrm{~cm}, 3 \mathrm{~cm}5 cm,3 cm, and 1 cm

6 cm,3 cm6 \mathrm{~cm}, 3 \mathrm{~cm}6 cm,3 cm, and 2 cm

Solution

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

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.

Solution

Solution

Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Triangle.

Solution

Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Solution

Find the area of the region remaining at the nnnth step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit.

Solution

Draw the top view, front view and the side view of each of the following combinations of identical cubes. [Six different combinations of cubes are shown]

Solution

Imagine eight identical cubes, glued together along faces to form the letter 'E'.

This looks like a 'E' from the front. What does it look like from the side? From the top?

Glue additional cubes to make a shape that looks like 'E' from the front and 'E' from the top.

Now, can you glue even more cubes to make it look like 'E' from the front, 'E' from the top, and 'E' from the side?

Can you think of other letter combinations to make with a single combination of cubes in this manner?

Solution

Which solid corresponds to the given top view, front view, and side view?

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?

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?

Find the number of sides in the $n$ th step of the shape sequence that leads to the Koch Snowflake.

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 unit.

$5 \mathrm{~cm}, 3 \mathrm{~cm}$ , and 1 cm

$6 \mathrm{~cm}, 3 \mathrm{~cm}$ , and 2 cm

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 sq. unit.

If the cone is slit open along the line $l$ and then unrolled, what will we get?