Exploring Some Geometric Themes Class 8 Solutions Maths Ganita Prakash Part 2 Chapter 4

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Class 8 Maths Ganita Prakash Part 2 Chapter 4 Solutions

Class 8 Maths Exploring Some Geometric Themes Solutions

Class 8 Ganita Prakash Part 2 Chapter 4 Solutions Exploring Some Geometric Themes

NCERT (Page 72)

Question 1.
Show that by joining the mid-points of an equilateral triangle, we divide it into 4 identical equilateral triangles.
[Hint Note that the corner triangles are isosceles.]
Solution:
Consider an equilateral triangle ABC. Let D, E and F be the mid-points of sides AB, BC and AC, respectively.

In ΔADF, ∠A = 60°
AD = AF [D and F is the mid-points]
∠ADF = ∠AFD = 60° [by isosceles triangle]
So, ΔADF is an equilateral triangle.
Similarly, ΔBED and ΔECF are equilateral triangles. Therefore, by joining the mid-points of an equilateral triangle, we divide it into 4 identical equilateral triangles.

Figure it Out (NCERT Page 72)

Question 1.
Draw the initial few steps (atleast till Step 2) of the shape sequence that leads to the Sierpinski triangle.
Solution:
The Sierpinski triangle is a fractal formed by repeatedly removing a smaller triangles from a larger triangle.
Step 0: Start with a single, large, solid equilateral triangle.

Step 1 Divide the large triangle into four smaller, equilateral triangles by connecting the mid-points of its sides.
Remove the central. inverted triangle.

Step 2 Repeat the process for each of the three remaining solid triangles. Divides each into four smaller triangles and remove the central inverted triangle from each. Nine solid triangles remain in total.

This process continues infinitely to form the fractal.

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.
Solution:
Let number of remaining triangles = T_n and number of holes = H_n.
Now, we create a table as follows.

Steps	Remaining Triangles	Holes
0	1 = 3°	0
1	3 = 3‘	1
2	9 =32	1 + 3 = 4
3	27 =33	1 + 3 + 9 = 13
. . .	. . .	. . .
n	3”	(3” -1)

Hence, the total number of remaining triangle, T_n = 3
and the total number of holes, H_n = \(\frac{3^n-1}{2}latex]

Question 3.
Find the area of the region remaining at the nth 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:
The Sierpinski carpet is constructed by starting with a square, remaining a central smaller square and repeating this process on the remaining smaller squares.
So, the area is reduced to [latex]\frac{8}{9}\) of the previous step’s area.
Given, the initial area of square (A0) = 1 sq unit.

The Sierpinski triangle is constructed by starting with an equilateral triangle, removing a central smaller triangle and repeating this process on the remaining smaller triangles.
So, \(\frac{1}{4}\) of the area of each triangle being divided.

Figure it Out (NCERT Page 73)

Question 1.
Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.
Solution:
Step 0 The process begins with an equilateral triangles.

Step 1 Each side of the triangles divided into three equal segments. The middle segment is replaced by a smaller equilateral triangle pointing outwards, creating a star-like shape with 12 segments.

Step 2 The process from Step 1 is repeated for each of the new, shorter line segments, forming, a more complex, jagged shape with 48 segments.

Question 2.
Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.
Solution:
The Koch Snowflake construction starts with an
equilateral triangle, which has 3 sides. At each subsequent step, every existing line segment is divided into three equal parts and middle part is replaced by a new smaller equilateral triangle, effectively replacing one segment with four segments.
The number of sides of each step forms a sequence, where each term is 4 times the previous term.
Let S„ be the number of sides in nth step.
In Step 0 Number of sides = 3
In Step 1 Number of sides = 3 × 4 = 12
In Step 2 Number of sides = 12 × 4 = 3 × 4² = 48
In Step 3 Number of sides = 48 × 4 = 3 × 4³ = 192
The number of segments is multiplied by 4 at each stage. This can be expressed as 3 times powers of 4.
For nth step, the number of sides S_n can be represented by
S_n = 3 × 4ⁿ
Hence, the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake is 3 × 4ⁿ.

Question 3.
Find the perimeter of the shape at the nth step of the sequence. Take the starting equilateral to have a sidelength of 1 unit.
Solution:
Given, the length of side of equilateral triangle = 1 unit
In the Koch Snowflake, the sequence starts with an equilateral triangle of sidelength 1 unit. At each subsequent step, the middle third of each side is removed and replaced with a smaller equilateral triangle pointing outwards. The initial shape has 3 sides.
At each step, every existing side is replaced by 4 smaller sides.
So, the length of each new side is one-third of the previous length.
The length of each side at step n is \(\left(\frac{1}{3}\right)^n\)
Let P_n be the total perimeter in n step and S_n be the number of sides in step n.
Now, the total perimeter at step n is the product of the number of sides and the length of each side.
P_n = S_n × \(\left(\frac{1}{3}\right)^n\)
= 3 × 4n × \(\left(\frac{1}{3}\right)^n\)
[∵ number of sides in step n = 3 × 4ⁿ]
= 3 × \(\frac{4^n}{3^n}\)
= 3 × \(\left(\frac{4}{3}\right)^n\)
Hence, the perimeter of the shape at the nth step of the sequence is P_n = 3 × \(\left(\frac{4}{3}\right)^n\).

NCERT (Page 75-77)

Question 1.
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:
The answer will vary depending on the individual’s name, e.g. The name is ‘JOHN’, reading it backwards by sight would be ‘NHOJ’.

Question 2.
Cut off the four corners of an imaginary square with each cut going between mid-points of adjacent edges. What shape is left over? How can you reassemble the four corners to make another square?
Solution:
Cutting the corners of square by connecting the mid-points of adjacent sides results in a smaller square left in the centre. The four corners cut off are congruent right angled – isosceles triangles. These four triangles can be reassembled to square by arranging them around a central point such that their right angles meet in the middle and the hypotenuses form the outer edge of the new square.

Question 3.
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 the corners of an equilateral triangle are cut off at the one-third marks along each side, the resulting shape is a regular hexagon.

Question 4.
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:
The shape left after cutting the corners of a square in this manner is an octagon. When each side of the square is divided into thirds and the corners are cut along the marks, the original four sides are reduced and four new sides are created, where the corners were removed. This results in a polygon with eight sides, specifically a regular octagon, if the square is regular and the cuts are made symmetrically.

Question 5.
Can you describe a solid and a viewpoint that would result in each of the following cases? If it helps, you can imagine the solid passing through a wall like Tom did and leaving a hole of the appropriate shape.
(i) A solid whose profile has a square outline.
(ii) A solid whose profile has a circular outline.
(iii) A solid whose profile has a triangular outline.

Solution:
(i) Cube A cube is a solid with six square faces. When viewed from a direction perpendicular to any of its faces, the profile (or shadow) will be a square.
(ii) Sphere A sphere is a perfectly round geometrical object in three-dimensional space. Regardless of the viewpoint, its profile will always be circular.
(iii) Triangular pyramid A triangular pyramid (or tetrahedron) is a solid with four triangular faces. When viewed from a direction perpendicular to any of its faces, the profile will be triangular.

Question 6.
Can you visualise solids that have the following contrasting profiles?
(i) A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint.
(ii) A solid with a circular profile from one viewpoint and a triangular one from another viewpoint.
(iii) A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint.
(iv) A solid with a trapezium shaped profile from one viewpoint and a circular one from another viewpoint.
(v) A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint.
(vi) Are there unique solids for each of the conditions, or can you come up with multiple possibilities?
Solution:
(i) Cylinder A cylinder viewed from the side (along its curved surface) appears as a rectangle. When viewed from directly above or below (along its height), it appears as a circle.
(ii) Cone A cone viewed from directly above or below (along its height) appears as a circle. When viewed from the side, it appears as a triangle.
(iii) Triangular Prism A triangular prism viewed from the side (along its length) appears as a rectangle. When viewed from the front or back (at its ends), it appears as a triangle. A square pyramid can also work.
(iv) Truncated Cone (Frustum) A truncated cone (a cone with its top cut off parallel to the base) viewed from the side appears as a trapezium. When viewed from directly above or below, it appears as a circle.
(v) Pentagonal Prism A pentagonal prism viewed from the front or back (at its ends) appears as a pentagon. When viewed from the side (along its length), it appears as a rectangle.
(vi) Generally, geometric conditions or properties (such as having a certain number of faces, edges, or a specific volume) do not uniquely define a single solid. Therefore, it is typically possible to come up with multiple solids that satisfy the same set of conditions.

NCERT (Page 79)

Question 1.
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:
A prism is a polyhedron with two congruent and parallel bases and rectangular faces connecting the corresponding sides of the bases. The number of faces, edges and vertices of a prism are related to the number of sides (n) of its base polygon.
Faces (F) A prism has two base faces and tt lateral (rectangular) faces. ,
So, F = n + 2
If n = 10 then the total number of faces = 10 + 2 = 12.

Edges (E) There are n edges in each of the two bases and edges connecting the bases.
So, E = 3n
If n = 10 then the total number of edges = 3 × 10 = 30.

Vertices (V) There are n vertices in each of the two bases. So, V = In
If n = 10 then the total number of vertices = 2 × 10 = 20.
The general formula derived between faces, vertices and edges is given by F – E + V = 2.
This is also known as Euler’s formula for polyhedra.

Question 2.
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:
A pyramid is a three-dimensional geometric shape with a polygon as its base and the lateral surfaces as triangles v that meet at a common vertex (apex). The number of faces, edges and vertices of a pyramid are related to the number of sides (n) of its base polygon.
Faces (F) A pyramid has one base face.
So, F = n + 1
If n = 10 the total number of faces = 10 + 1 = 11.
Edges (E) The number of edges = 2n = 2 × 10 = 20.
Vertices (V) The number of vertices = n + 1 = 10 + 1 = 11.

Figure it Out (NCERT Page 80 and 81)

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.

Solution:
(ii), (iii), (iv) and (vi)
A net of a cube is a 2D-shape that can be folded to form a 3D cube. A cube has 6 square faces.
(i) This is not a net of a cube.
(ii) This is a net of a cube. This configuration is another valid arrangement that folds into a cube.
(iii) This is a net of a cube. The squares form a cube with one face as the base, four sides and one top.
(iv) This is a net of a cube. This classic “cross” shape is a common net of a cube.
(v) This is not a net of a cube. When folded, two faces will overlap and the resulting shape will not be a closed cube.
(vi) This is a net of a cube. This is a valid configuration that folds into a cube.

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.
Solution:
We know that each net of cube consists of exactly 6 connected squares.
The 11 possible net structures of cube are shown here.

Question 3.
Draw a net of a cuboid having sidelengths.
(i) 5 cm, 3 cm and 1 cm
Solution:

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

NCERT (Page 81)

Question 1.
What is a net of a regular tetrahedron? Which of the following are nets of a regular tetrahedron?

Solution:
A regular tetrahedron is a polyhedron with four faces, each of which is an equilateral triangle.
So, a net of a regular tetrahedron is a plane figure made of four equilateral triangles arranged in such a way that they can be folded along their edges to form the tetrahedron without overlapping.
The first figure consists of four equilateral triangles arranged in a larger equilateral triangle. This can be folded to form a regular tetrahedron.
The second figure cannot be folded into a tetrahedron because two faces would overlap and a side would be missing.
The third figure cannot be folded into a tetrahedron as the structure is Unear and faces would overlap or not meet correctly.
The fourth figure (a paraUelogram of four triangles with one triangle hanging off the bottom) can also be folded to form a regular tetrahedron.
Hence, the first and fourth options are nets of a regular tetrahedron.

Question 2.
Draw a net with appropriate measurements that can be folded into a regular tetrahedron. Verify if it works by making an actual cutout.
Solution:
A net for a regular tetrahedron consists of four equilateral triangles arranged in a larger equilateral triangle shape, where the central triangle shares a edge with each of the other three.
Decide on an edge length i.e. 5 cm.

Question 3.
Draw a net with appropriate measurements that can be folded into a square pyramid. Verify if it works by making an actual cutout.
Solution:
A net for a square pyramid consists of a square base and four identical triangular faces attached to each side of the square.
Draw a net [Step-by-Step]
(i) Draw a square with sides of 4 cm.
(ii) Attach on triangle to each side of the square. Each triangle will have a base of 4 cm and slant heights of 5 cm.
(iii) Ensure the triangles bases align perfectly with the squares edges.

NCERT (Page 82)

Question 1.
What are the side lengths of the rectangle obtained?
Solution:
The side lengths of the rectangle are the height of the cylinder and the circumference of its circular base.

Question 2.
How will the net of a cone look?
Solution:
A net of cone consists of one circular base and one large sector of a circle that forms the curved surface of the one.

Question 3.
What surface do you construct by using the above net in which 0 is not the centre of the boundary circle? Make a physical model to help you answer this question!
Solution:
An Oblique Cone
If the apex (point O) is not directly above the centre of the base circle, the resulting 3D-figure is an oblique cone.
The slant height and the hight will vary around the circumference.

Question 4.
Draw a net with appropriate measurements that can be folded into a triangular prism. Verify that it works by making an actual cutout.
Solution:
A net for triangular prism consists of two identical triangles (the bases) the three rectangle (the faces). The sides of the triangle must match one of the dimensions of the rectangle so that they can be joined.

NCERT (Page 86)

Question 1.
(i) Find the shortest path between the ant and the laddu in the following case:

If we unfold the cuboid as before, we get

Solution:
From figure (a), by the Baudhayana-Pythagoras theorem,
PQ² = PR² + RQ²
= (2 + 2)² + (2 + 4)²
= 4² +6²
= 16 + 36
= \(\sqrt{20}\) cm

(ii) So, what do we do now?
If the net is unfolded in the following way then we get a path on the cuboid.

Solution:
From figure (b), by the Baudhayana-Pythagoras theorem,
XY² = XZ² + YZ²
= (2 + 2)² + 22
= 4² + 2²
= 16 + 4
= \(\sqrt{20}\) cm
Thus, the way a cuboid is unfolded matters.

NCERT (Page 89-91)

Question 1.
(i) What happens to the length of a line in its projection?
Solution:
The length of the projected line is generally shorter than its actual length.

(ii) Can you now compare the lengths p and l?
Solution:
Here, p is less than 1.1 is the hypotenuse of the right-angled ∆AEB and p is a leg of the triangle. Therefore, p is always less than l.

Question 2.
When is the length of the projected line equal to its actual length?
Solution:
If the line is parallel to the projection plane, the angle between the line and the plane is 0°. In this case, the projection forms a rectangle, where the projected lengthp is equal to the actual length l.

Question 3.
What do you think are the different possible projections of a square that we get based on its orientation?
Solution:
The projections can be a line segment, a square, a rectangle or a parallelogram.
Line segment: When the square’s plane is perpendicular to the projection plane.
Square: When the square’s plane is parallel to the projection plane.
Rectangle: When the square’s plane is inclined to the projection plane but two sides are parallel to it. Parallelogram When the square’s plane is inclined to the projection plane at an angle.

Question 4.
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:
A parallelogram has two pairs of parallel sides. In a projection, parallel lines remain parallel. Since, both pairs of opposite sides stay parallel in every orientation, the projected shape must still have two pairs of parallel sides.
Hence, it is always a parallelogram and cannot become any other quadrilateral.

Question 5.
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:
We know that
(i) A polygon is made up of n-sides.
(ii) The projection of a polygon is formed by the projections of its sides.
(iii) Each side projects to a line segment, so the total number of sides in the projection remains n.
However, lengths and angles generally change under projection, so the projected polygon is not necessarily regular.
Hence, the projection of an n-sided regular polygon under different orientations is always an n-sided polygon, though it may not remain regular.

Question 6.
How would the projections of a cube and a cone look?

Solution:
The image illustrates orthographic projection, where light rays are parallel and perpendicular to the projection plane. For the cube (Fig. (i)), the projection is a two-dimensional square shape, representing one face of the cube.
For the cone (Fig. (ii)), the projection is a two-dimensional triangle shape, representing a side view of the cone.
Hence, the projection of a cube would look like a square and the projection of a cone would look like a triangle.

Question 7.
Find another object that makes the same projection as that of a cone.
Solution:
Another object that can make the same projection as a cone is a pyramid, when viewed from a suitable direction.

Figure it Out (NCERT Page 92-93)

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

Solution:
Yes, there is a clear relation between the lengths of a line and the lengths of its projections (front view, top view, and side view).
A Line appears longest in the view, where it is parallel to the plane of projection.
A line appears shorter in a view, where it is inclined to the plane. Ifa line is perpendicular to a plane of projection. its projection on that plane appears as a point.

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.
Solution:
For cube When a cube is oriented with its faces parallel to the vertical, horizontal and side planes, all three views (front, top, and side) will appear as a square.

For cuboid Similarly, for a cuboid with faces aligned to the planes, the views will be

• Front view Rectangle or square, depending on dimensions
• Top view Rectangle or square, depending on dimensions
• Side view Rectangle or square, depending on dimensions

For parallelepiped A parallelepiped a 3D-figure with six faces that are parallelograms, with faces aligned will have

• Front view Parallelogram or rectangle
• Top view Parallelogram or rectangle
• Side view Parallelogram or rectangle

For cylinder When a cylinder is oriented vertically (axis perpendicular to the horizontal plane).

• Front view Rectangle
• Top view Circle
• Side view Rectangle

For cone When a cone is oriented vertically (base on the horizontal plane).

• Front view Triangle
• Top view Circle
• Side view Triangle

For prism For a right prism standing on its base (e.g. a triangular prism).

• Front view Rectangle (representing a face)
• Top view Polygon (the shape of its base e.g. a triangle)
• Side view Rectangle (representing a face)

For pyramid For a right pyramid standing on its base (e.g. a square pyramid).

• Front view Triangle (representing a face)
• Top view Polygon (the shape of its base e.g. a square, with lines to the centre point).
• Side view Triangle

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

Solution:
(i) → (h), (ii) → (f), (iii) → (g), (iv) → (a), (v) → (c), (vi) → (d), (vii) → (e), (viii) → (b)

NCERT (Page 94)

Question 1.
(i) Observe what happens to the size of the shadow as you vary the distance between your torch and your object.
Solution:
The size of the shadow increases as the torch is moved closer to the object and decreases as the torch is moved further away.

(ii) Why does this happen?
Solution:
This happens because light travels in straight lines. When the torch is closer to the object, the light rays spread out more before hitting the object, creating a larger area of blocked light on the screen. When the torch is further away, the light rays are more focused, when they reach the object, resulting in a smaller shadow. This is also related to the angle at which the light hits the object and the screen.

Figure it Out (NCERT Page 95-97)

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

Solution:

Question 2.
Which solid corresponds to the given top view, front view, and side view?

Solution:
(ii)

Question 3.
Using identical cubes, make a solid that gives the following projections.

Solution:
Do Yourself

Question 4.
Find the number of cubes in this stack of identical cubes.

Solution:
The total number of cubes is found by counting the cubes in each layer and adding them together.
The top layer has 1 cube.
The second layer has 2 + 1 i.e. 3 cubes.
The third layer has 3 + 2 + 1 i.e. 6 cubes.
The fourth layer has 4 + 3 + 2 + 1 i.e. 10 cubes.
Hence, the total number of cubes =1 + 3 + 6 +10 = 20

Question 5.
What are the different shapes the projection of a cube can make under different orientations?
Solution:
The shape of the orthogonal projection of a cube depends on its orientation relative to the projection plane.
A square is formed, when the cube is viewed face on.
A rectangle is formed, when the cube is viewed edge on.
A hexagon is formed, when the cube is viewed vertex on.

Question 6.
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:
Do yourself

Figure it Out (NCERT Page 100-102)

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?

Solution:
Do yourself

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.]
Solution:

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.]
Solution:
Yes, the path is strange.
The path shown is an optical illusion and cannot exist in a physically realisable 3D world. The arrows on the grid indicate a loop that goes both up and down simultaneously in a way that is impossible in a three-dimensional space.

The red ball appears to be at a lower level than the blue ball but by following the arrows, one would move from the blue ball’s position, around the path and come to the red ball’s position, now both the balls appearing at the same level. This creates a logical inconsistency in the perceived height, similar to a penrose staircase illusion.

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?
Solution:
No, it would not be possible to build a model out of actual cubes.
The image depicts a penrose triangle also known as an impossible triangle, which is an optical illusion and an impossible object in three-dimensional Euclidean space. It can only exist in two-dimensional representations or as a 3D-object that appears as such only from a very specific view point.

The front, top and side profiles would appear as simple triangles or L-shapes, as each individual 2D projection is valid. The illusion only works, when viewing the entire object from a perspective that merges the different planes.

(ii) Recreate this on an isometric grid.
Solution:

(iii) Why does the illusion work?
Solution:
The illusion works due to a conflict between the 2D representation and our brain’s interpretation of a 3D-object.

• The image provides strong monocular depth cues (like perspective and shading) that suggest the object is a single, solid 3D structure made of cubes.
• However, when tracing the edges of the triangle, one finds a continuous loop, where each corner appears to be a right angle, which is geometrically impossible for a single object in 3D-space.
• The brain attempts to reconcile these conflicting cues, resulting in the perception of an impossible, infinitely looping object.