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

Exploring Some Geometric Themes Class 8 Ganita Prakash Part 2 Chapter 4 NCERT Solutions Looking for the Exploring Some Geometric Themes Class 8 Ganita Prakash Part 2 Chapter 4 NCERT Solutions? You are in the right place. This chapter helps students understand important geometric concepts through diagrams, activities, and logical reasoning. Our step-by-step solutions are written in simple language to help students complete their homework, prepare for exams, and strengthen their understanding of Mathematics.

Figure it Out (Page 72)

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

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 R_n​ be the number of triangles remaining, and H_n​ be the number of holes at the nth step.
Every remaining triangle gives rise to 3 triangles in the next step. Thus,
R_{n + 1} = 3R_n
R₀ = 1
R₁ = 3 × 1 = 3
R₂ = 3 × 3 = 9
R₃ = 3 × 9 = 27
R₄ = 3 × 27 = 81 …………
In general, R_n = 3ⁿ
Also, each remaining triangle creates one new hole, while the old holes remain. Thus,
H_{n + 1} = H_n + R_n
H₀ = 0
H₁ = H₀ + R₀ = 0 + 1 = 1
H₂ = H₁ + R₁ = 1 + 3 = 4
H₃ = H₂ + R₂ = 4 + 9 = 13
H₄ = H₃ + R₃ = 13 + 27 = 40 ………..

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:
Let A_n​ be the area of the region remaining at the n^th step. The area of the starting square/triangle is 1 sq. unit.
(i) Sierpiński Triangle:
At each step, the triangle is divided into 4 equal triangles and the moddle triangle is removed, so $\frac{3}{4}$ of the area remains.
A_{n+ 1} = $\frac{3}{4}$ A_n
A₀ = 1
A₁ = $\frac{3}{4}$ A₀ = $\frac{3}{4}$ × 1 = $\frac{3}{4}$
A₂ = $\frac{3}{4}$ A₁ = $\frac{3}{4}$ × $\frac{3}{4}$ = $\frac{9}{16}$
A₃ = $\frac{3}{4}$ A₂ = $\frac{3}{4}$ × $\frac{9}{16}$ = $\frac{27}{64}$
A₄ = $\frac{3}{4}$ A₃ = $\frac{3}{4}$ × $\frac{27}{64}$ = $\frac{81}{256}$ ……..
Thus, the area remaining at the n^th step, A_n = ${\left(\frac{3}{4}\right)}^n$.

(ii) Sierpiński Square:
At each step, the square is divided into 9 equal squares and the middle square is removed, so 89​ of the area remains.
A_{n+ 1} = $\frac{8}{9}$ A_n
A₀ = 1
A₁ = $\frac{8}{9}$ A₀ = $\frac{8}{9}$
A₂ = $\frac{8}{9}$ A₁ = $\frac{8}{9}$ × $\frac{8}{9}$ = $\frac{64}{81}$
A₃ = $\frac{8}{9}$ A₂ = $\frac{8}{9}$ × $\frac{64}{81}$ = $\frac{512}{729}$
A₄ = $\frac{8}{9}$ A₃ = $\frac{8}{9}$ × $\frac{512}{729}$ = $\frac{4096}{6561}$ ……….
Thus, the area remaining at the n^th step, A_n = ${\left(\frac{8}{9}\right)}^n$.

Figure it Out (Page 73)

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

2. Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.
Solution:
Let S_n denote the number of sides at the nth step.
At each step, every side is replaced by 4 sides. Thus
S_n = 4S_n-1
S₀ = 3
S₁ = 4 × S₀ = 4 × 3 = 12
S₂ = 4 × S₁ = 4 × 12 = 48
S₃ = 4 × S₂ = 4 × 48 = 192 …….
Thus, the number of sides at the n^th step, Sn​ = 3⋅4ⁿ.

3. Find the perimeter of the shape at the n^th step of the sequence. Take the starting equilateral triangle to have a side length of 1 unit.​
Solution:
Let the perimeter at the $n^{\text{th}}$ step be $P_n$​.
At each step, each side is replaced by 4 sides, and each new side is $\frac{1}{3}$​ of the previous side.
So, the perimeter increases by a factor of $\frac{4}{3}$.
At Step 0, the equilateral triangle has side length 1 unit, so
P₀ = 3 units
P₁ = 3 × $\frac{4}{3}$ = 4 units
P₂ = 3 × $\frac{4}{3}$ × $\frac{4}{3}$ = $\frac{16}{3}$ units
P₃ = 3 × $\frac{4}{3}$ × $\frac{4}{3}$ × $\frac{4}{3}$ = $\frac{64}{9}$ units …….
Therefore, the perimeter at the n^th step, P_n = 3 × ${\left(\frac{4}{3}\right)}^n$units.

Page (75 – 77)

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:
Do it yourself.

2. 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:
Let ABCD be a square.
Let P, Q, R, and S be the midpoints of AB, BC, CD, and DA, respectively.
Join PQ, QR, RS, and SP.
When cut along the line segments PQ, QR, RS, and SP, the remaining figure PQRS is a square.

The four congruent corner triangles can be rearranged by joining them along their equal sides to form another square.

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:
Let ABC be an equilateral triangle.
Divide each side into 3 equal parts, such that AP = PQ = QB, BR = RS = SC, AU = UT = TC.
Join PU, QR and ST.
When cut along the line segments PU, QR, and ST, the remaining figure is a regular hexagon.

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:
Let ABCD be a square.
Divide each side into 3 equal parts such that AE = EF = FB, BG = GH = HC, CI = IJ = JD, DK = KL = LA.
Join LE, FB, HI, and JK.
When cut the line segments LE, FG, HI, and JK, we get a regular octagon.

5. A solid whose profile has a square outline
Solution:
Cube.

6. A solid whose profile has a circular outline.
Solution:
Sphere.

7. A solid whose profile has a triangular outline.
Solution:
Triangular pyramid.

8. A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint.
Solution:
Cylinder.

9. A solid with a circular profile from one viewpoint and a triangular one from another viewpoint.
Solution:
Cone.

10. A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint.
Solution:
Prism.

11. A solid with a trapezium-shaped profile from one viewpoint and a circular one from another viewpoint.
Solution:
A frustum, which is a truncated cone.

12. A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint.
Solution:
Pentagonal base prism.

Page 79

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:
(i) For a prism whose bases are 10-sided polygons:
Faces = Number of side faces + 2 bases
= 10 + 2 = 12
Edges = Edges of top base + edges of bottom base + vertical edges
= 10 + 10 + 10 = 30
Vertices = Vertices of top base + vertices of bottom base
= 10 + 10 = 20
(ii) For a prism whose bases are n-sided polygons:
Faces = n + 2
Edges = 3n
Vertices = 2n

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:
(i) For a pyramid whose base has 10 sides:
Faces: Number of triangular side faces + Base
= 10 + 1 = 11
Edges = Edges of the base + edges joining the apex to the base vertices
= 10 + 10 = 20
Vertices = Vertices of the base + Apex
= 10 + 1 = 11
(ii) For a pyramid with an n-sided base:
Vertices = n + 1
Faces = n + 1
Edges = 2n

Figure it Out (Page 80)

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:
(i) No
(ii) Yes
(iii) Yes
(iv) Yes
(v) No
(vi) Yes

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 —

Solution:

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

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

Page 81

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

Solution:
Figure 1 – Net of a regular tetrahedron
Figure 2 – Net of a regular tetrahedron
Figure 3 – Not a net of a regular tetrahedron
Figure 4 – Net of a regular tetrahedron

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:

(i) Draw an equilateral triangle ABC of side length 6 cm.
(ii) Mark the midpoints D, E, and F of sides AB, BC, and CA, respectively.
(iii) Join D, E, and F. This divides the large equilateral triangle into four congruent equilateral triangles.
(iv) The central triangle DEF becomes the base of the tetrahedron.
(v) Fold the three outer triangles upward along the sides DE, EF, and FD.
(vi) Join the free edges of the folded triangles. The shape formed is a regular tetrahedron.
Verification: Make a cutout of the net, fold along the indicated lines, and check that all four faces are congruent equilateral triangles. This confirms that the solid formed is a regular tetrahedron.

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:

(i) Draw a square ABCD of side 5 cm.
(ii) On side AB, construct an isosceles triangle ABE such that AE = BE = 5 cm.
(iii) Similarly, construct isosceles triangles on sides BC, CD, and DA.
(iv) The resulting figure consists of a square with four congruent equilateral triangles attached to its sides.
(v) Fold along the sides of the square. The four triangular faces meet at a common vertex to form a square pyramid.
Verification: Fold the four triangles upward along the edges of the square. The triangles join at the top to form a square pyramid with a square base of side 6 cm.

4. What is the net of a cylinder? What are the side lengths of the rectangle obtained?
Solution:
Consider a cylinder of radius r and height h. If the curved surface of the cylinder is cut along its height and unfolded, the net consists of two equal circles and one rectangle. This is the net of a cylinder.

Length of the rectangle = Circumference of the base = $2\pi r$
Breadth of the rectangle = Height of the cylinder = h

5. How will the net of a cone look?
Solution:
If the circular base is unfolded and a cut is made along the slant height of the cone, then the net of the cone consists of:
• a sector of a circle of radius l (slant height), and
• a circle of radius r attached to the arc.
The length of the arc of the sector is equal to the circumference of the base of the cone.
Therefore,
Length of the arc of the sector = Circumference of the base = 2πr.

6. Draw a net with appropriate measurements that can be folded into a triangular prism. Verify that it works by making an actual cutout.
Solution:
Draw three rectangles in a row with the dimensions 6 cm × 3 cm.
On the two opposite sides of the second rectangle, construct two congruent equilateral triangles with side 3 cm.
The figure obtained is the required net of a triangular prism.

Verification:
Cut the net along the outer boundary and fold along the edges. The three rectangles form the lateral faces, and the two triangles form the two ends of the prism.
Hence, the net folds into a triangular prism.

Figure it Out (Page 92)

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?

Solution:
Relation: As the top view of the line becomes more oblique, its side-view length increases while front-view length remains unchanged.

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:

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

Solution:

Figure it Out (Page 95)

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

Solution:

2. Imagine eight identical cubes, glued together along faces to form the letter ‘C’.

Solution:
(i)

(ii)

(iii) Not possible.

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

Solution:
Solid (ii).

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

Solution:

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

Solution:
Total cubes in the stack = 1 + 3 + 6 + 10 = 20 cubes.

6. What are the different shapes the projection of a cube can make under different orientations?
Solution:
Five different shapes can be observed as projections of a cube under different orientations:
(a) Square – when one face of the cube is parallel to the plane of projection.
(b) Rectangle – when the cube is tilted so that two faces are visible.
(c) Parallelogram – when a face is inclined to the plane of projection.
(d) Rhombus – a special case of a parallelogram when all sides are equal.
(e) Hexagon – when the cube is oriented such that three faces are equally visible.

Figure it Out (Page 100 – 101)

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:

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:

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.]

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?

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