NCERT Solutions for Class 8 Maths Part 2 Chapter 4 - Exploring Some Geometric Themes

Class 8 Maths Part 2 Chapter 4 Solutions PDF Free Download - Ganita Prakash Book 2

Students can download the NCERT Solutions for Class 8 Maths Part 2 Chapter 4 Exploring Some Geometric Themes PDF by clicking the link provided below.

NCERT Solutions for Class 8 Maths Part 2 Chapter 4 Exercise Questions and Answers

Here are the NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 4 Exploring Some Geometric Themes question answers with clear and detailed solutions.

Question 1. Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Triangle.

Answer:

Step 0: Draw a single solid equilateral triangle.

Step 1: Remove the central triangle, leaving 3 smaller solid triangles surrounding 1 central triangular hole.

Step 2: Repeat the process for each of the 3 remaining triangles, remove their centres and leave 9 even smaller solid triangles.

Question 2. Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Answer:

Step 0:

Number of remaining triangles: $3^0 = 1$
New holes introduced: 0
Total cumulative holes: 0

Step 1:

Number of remaining triangles: $3^1 = 3$
New holes introduced: 1
Total cumulative holes: 1

Step 2:

Number of remaining triangles: $3^2 = 9$
New holes introduced: 3
Total cumulative holes: 1 + 3 = 4

Step 3:

Number of remaining triangles: $3^3 = 27$
New holes introduced: 9
Total cumulative holes: 4 + 9 = 13

…….

Step $n$:

Number of remaining triangles: $3^n$
New holes introduced: $3^{n-1}$, for $n \geq 1$
Total cumulative holes: $\frac{3^n-1}{2}$

Question 3. 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.

Answer:

At each step, every remaining solid triangle has its middle quarter removed. This means that exactly $\frac{3}{4}$ of the area from the previous step is remaining.
Step 0 Area: $1$
Step 1 Area: $\frac{3}{4}$
Step 2 Area: $\left(\frac{3}{4}\right)^2 = \frac{9}{16}$
.....
At the $n$th step, the remaining area is given by:
$A_n = \left(\frac{3}{4}\right)^n$ square units

At each step, every remaining solid square is divided into 9 smaller congruent squares, and the central square is removed. This means that exactly $\frac{8}{9}$ of the area from the previous step is remaining.
Step 0 Area: $1$
Step 1 Area: $\frac{8}{9}$
Step 2 Area: $\left(\frac{8}{9}\right)^2 = \frac{64}{81}$
....
At the $n$th step, the remaining area is given by:
$A_n = \left(\frac{8}{9}\right)^n\text{ sq. units}$

Question 1. Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.

Answer:

Step 0: Draw a single equilateral triangle.

Step 1: The transformation is applied once to each of the three original sides, creating a hexagon with six points, which resembles a Star of David.

Step 2: The process is repeated for every segment of the Step 1 shape, creating a more intricate, 12-pointed star.

Question 2. Find the number of sides in the $n$th step of the shape sequence that leads to the Koch Snowflake.

Answer:

Step 0: Number of sides: $N_0 = 3$.

Step 1: Number of sides: $N_1 = 3 \times 4 = 12$

Step 2: Number of sides: $N_2 = 12 \times 4 = 48$.

......

Step $n$: Number of Sides: $N_n = 3 \times 4^n$

Question 3. 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.

Answer:

The original equilateral triangle has a side length of 1 unit.

Step 0: Initial perimeter = 1 + 1 + 1 = 3 units

At each step, every side is replaced by 4 sides, and each new side is $\frac13$ of the old side.

Thus, the perimeter is multiplied by: $3\times\frac13=\frac43$

Step 1: $3\times(\frac43)^1$ units

Step 2: $3\times(\frac43)^2$ units

Step 3: $3\times(\frac43)^3$ units

Step 4: $3\times(\frac43)^4$ units

.....

Step $n$: $3\times(\frac43)^n$ units

Question 1. Which of the following are the nets of a cube? First, try to answer by

visualisation. Then, you may use cutouts and try.

Answer:

A valid cube net must fold so that:

• All 6 squares become the 6 faces of the cube.

• No two squares overlap when folded.

• No face is missing.

Structure 1: Not a cube net [$\because$ Overlapping face]

Structure 2: Cube net

Structure 3: Cube net

Structure 4: Cube net

Structure 5: Not a cube net [$\because$ Overlapping face]

Structure 6: Cube net

Question 2. 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. For example, the following nets are all considered the same —

Find all the 11 nets of a cube.

Answer:

Question 3. Draw a net of a cuboid having sidelengths:

(i) 5 cm, 3 cm, and 1 cm

(ii) 6 cm, 3 cm, and 2 cm

Answer:

(i)
Faces:
2 rectangles of 5 cm × 3 cm

2 rectangles of 5 cm × 1 cm

2 rectangles of 3 cm × 1 cm

(ii)
Faces:

2 rectangles of 6 cm × 3 cm

2 rectangles of 6 cm × 2 cm

2 rectangles of 3 cm × 2 cm

Question 1. Observe the front view, top view and side view of the different lines in Fig. 4.6. Is there any relation between their lengths?

Answer:

From these examples, we observe:

1. The front view, top view, and side view need not have the same length.

2. The length seen in any projection is never greater than the actual length of the line.

3. When a line is parallel to a projection plane, its projection on that plane shows the true length of the line.

4. As the line becomes more inclined to a plane, the length of its projection on that plane becomes shorter.

So, there is no fixed equality between the lengths of the front view, top view, and side view. Each projection length depends on the orientation of the line.

Question 2. 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.

Answer:

Question 3. Match each of the following objects with its projections.

Answer:

Question 1. Draw the top view, front view and the side view of each of the following combinations of identical cubes.

Answer:

Question 2．
Imagine eight identical cubes, glued together along faces to form the letter．

（i）This looks like a 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 ， from the front andfrom the top．

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

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

Answer:

(iv)

Yes, many other combinations are possible.

1. Front: L, Top: T, Side: F

2. Front: H, Top: I, Side: U

3. Front: P, Top: E, Side: F

Question 3. Which solid corresponds to the given top view, front view, and side view?

Answer: Solid (ii)

Explanation:

Since the front view has a small gap on the top right side, solids (i), (v) and (vii) cannot be possible.
In top view, there is a gap on the right corner, which means in solid front right side should be a gap at the corner.
So, solids (iii) and (vi) are not possible.
Between solid (ii) and (iv), Solid (ii) is more likely.

Hence, the correct answer is solid (ii).

Question 4. Using identical cubes, make a solid that gives the following projections.

Answer:

Question 5. Find the number of cubes in this stack of identical cubes.

Answer:

Layer 1, or the top layer, has 1 cube.
Layer 2 has 2 visible cubes, and there must be a cube directly behind this row.
So, total 2 + 1 = 3 cubes
Layer 3 has 3 cubes visible along the front edge, and the rows behind it have 2 cubes and then 1 cube to support the layers above.
So, total 3 + 2 + 1 = 6 cubes
Layer 4, or the bottom layer, has 4 cubes visible along the front edge, and the rows extending behind it contain 3 cubes, 2 cubes, and 1 cube at the very back corner.
So, total 4 + 3 + 2 + 1 = 10 cubes

Total Cubes = Sum of all 4 layer cubes = 1 + 3 + 6 + 10 = 20 cubes

Hence, the total number of cubes in the stack is 20.

Question 6. What are the different shapes the projection of a cube can make under different orientations?

Answer:

When a three-dimensional cube is projected onto a two-dimensional plane, the resulting shape changes dramatically depending on the cube's orientation relative to the projection plane.
Mathematically, the projection of a cube is always a convex polygon with anywhere from 4 to 6 sides.
Here are the specific geometric shapes a cube's projection can form under different orientations:

1. Square

• Orientation: One face of the cube is perfectly flat (parallel) against the projection plane.

• Result: You see only that single front face, creating a perfect square.

2. Rectangle

• Orientation: The cube is rotated along one axis. Two faces are visible, but one set of edges remains completely horizontal or vertical.

• Result: The footprint stretches or squashes in one direction, forming a rectangle.

3. Parallelogram

• Orientation: The cube is tilted dynamically. A face is slanted relative to the projection plane.

• Result: The right angles skew into acute and obtuse angles, while opposite sides stay parallel.

4. Rhombus

• Orientation: A specific tilted position where the cube is angled symmetrically.

• Result: A slanted diamond-like shape where all four outer boundary edges happen to have the exact same length.

5. Hexagon

• Orientation: The cube is tipped to look directly down one of its corners (along a body diagonal), making three faces equally visible.

• Result: The outer silhouette reaches its maximum possible complexity, forming a symmetrical 6-sided hexagon.

Question 1. 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?

Answer:

Question 2. Draw the following figures on the isometric grid.

[Hint: It may be useful to determine whether the edge to be currently drawn - say, along the height - goes from down to up or up to down. Accordingly, draw the line segment on the grid either in the direction of the height axis or opposite to it.]

Answer:

Question 3. Is there anything strange about the path of this ball? Recreate it on the isometric grid.

[Hint: Consider a portion of this figure that is physically realisable and identify the 3 primary directions.]

Answer:

The strange thing is that the ball is trapped in an impossible loop.

• If you follow the arrows, the ball goes downstairs at every single step.

• But after rolling all the way around the loop, it somehow ends up right back at the top where it started!

In real life, you cannot keep walking downstairs and magically end up back at the top of the staircase. It breaks the laws of physics!

Question 4. 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?

Answer:

4 (i):

No, it is not possible to build this as a single, solid, closed triangle in 3D space.

This is called a Penrose Triangle.

Front, top, and side views are as follows:

4 (ii):

4 (iii):

Our eyes get tricked because of how the blocks are drawn:

• In small pieces: If you look at just one corner or one step at a time, it looks completely normal and real.

• As a whole picture: When you look at the entire shape together, the front steps and the back steps are connected in a way that creates a trick. The artist hid the depth, forcing a staircase that should go down to connect back to the top floor on a flat piece of paper.

Key Concepts Explained in Exploring Some Geometric Themes - Ganita Prakash Book 2

Topics you will learn in NCERT Solutions for Class 8 Maths Part 2 Chapter 4 Exploring Some Geometric Themes include:

• 4.1 Fractals

• 4.2 Visualising Solids

NCERT Solutions for Class 8 Maths Chapter Wise

We at Careers360 compiled all the latest NCERT Class 8 Maths solutions based on latest textbook in one place for easy student reference. The following links will allow you to access them.

NCERT Books and NCERT Syllabus

Before planning a study schedule, always analyse the latest syllabus. Here are the links to the latest NCERT syllabus and some of the important books that will help students in this cause.

• NCERT Books Class 8 Maths

• NCERT Syllabus Class 8 Maths

• NCERT Books Class 8

• NCERT Syllabus Class 8

Chapter Summary: NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 4

Number of Exercises: 6

Total number of questions: 22

Expert Review of NCERT Solutions for Class 8 Maths Part 2 Chapter 4

Here are the concepts and topics that are related to the latest NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 4 in higher classes.

Class 9-10

• Fractals

• Koch Snowflake

• Sierpinski Triangle

• Visualising solids and three-dimensional shapes

• Nets of solids

• Surface area and volume concepts

• Symmetry and geometric patterns

• Spatial visualization and geometric reasoning

Class 11-12

• Three-Dimensional Geometry

• Coordinate geometry of space

• Surface area and volume of solids

• Geometric transformations and symmetry

• Mathematical patterns and recursive structures

• Visualisation of complex geometric objects

JEE (Main & Advanced)

• Three-Dimensional Geometry

• Surface area and volume problems

• Coordinate geometry in space

• Symmetry and geometric transformations

• Recursive and pattern-based reasoning

• Advanced problem-solving involving geometric structures and patterns