CBSE Class 8 Mathematics Exploring Some Geometric Themes Short Answer Questions with Answers

Short Answer Questions

Question 1. What is meant by a fractal? Using the fern, explain the idea of self-similarity, and name two other objects in nature that show the same property.
Answer:
A fractal is a self-similar geometric shape: the same or a similar pattern appears again and again at smaller and smaller scales.
In a fern, each leaf is a smaller copy of the whole fern, and each sub-leaf is a smaller copy of that leaf, and so on. This repeating of the whole in its parts is self-similarity.
Two other natural examples: a tree (trunk, limbs, branches, branchlets) and a coastline. Clouds, mountains and lightning also show it.
In simple words: A fractal repeats its own pattern in smaller and smaller parts, like a fern, a tree or a coastline.

Teacher's Note:
1. Always say what the “smaller copy” is in each example.
2. Natural fractals repeat only for a few levels, not forever.

Question 2. The Sierpinski Carpet starts from a single square.
(i) Describe the rule that is applied to go from one step to the next.
(ii) How many small squares remain, and how many holes are there, at Step 1?
Answer:
(i) Each remaining square is broken into 9 equal smaller squares and the central square is removed. The same rule is then applied to every square that remains.
(ii) At Step 1, 8 squares remain and there is 1 hole: \( R_1 = 8 \), \( H_1 = 1 \).
In simple words: Cut each square into 9, remove the middle one; after one step there are 8 squares and 1 hole.

Teacher's Note:
1. The same rule is used at every step.
2. Squares and holes are counted separately.

Question 3. Answer for the Sierpinski Gasket.
(i) Write the rule that takes the shape from Step \( n \) to Step \( n + 1 \).
(ii) State the number of triangles that remain at Step 1 and at Step 2.
Answer:
(i) Every remaining equilateral triangle is divided into 4 identical equilateral triangles by joining the midpoints of its sides, and the central one is removed. This is done to each of the triangles that remain.
(ii) Step 1: 3 triangles; Step 2: 3 × 3 = 9 triangles.
In simple words: Join midpoints, remove the middle triangle, repeat; Step 1 has 3 triangles and Step 2 has 9.

Teacher's Note:
1. Joining the midpoints gives four identical triangles.
2. The count multiplies by 3 each step.

Question 4. Describe the operation that is carried out on every side of the shape in one step of the Koch Snowflake. How many sides does the shape have at Step 1 if we begin with an equilateral triangle?
Answer:
Each side is divided into 3 equal parts, an equilateral triangle is raised on the middle part, and that middle part is then removed.
So every straight side is replaced by a “bump” of 4 smaller segments.
Starting with 3 sides, Step 1 has 3 × 4 = 12 sides.
In simple words: Every side turns into a bump of 4 pieces, so the triangle's 3 sides become 12.

Teacher's Note:
1. The middle third is removed, not kept.
2. Each of the 4 new segments is one-third of the old side.

Question 5. Define the terms face, edge and vertex of a solid, and write how many of each a cube has.
Answer:
Face: a flat surface that forms part of the boundary of the solid.
Edge: a line segment that forms a side of a face; two faces meet along it.
Vertex: a point where edges meet.
A cube has 6 faces, 12 edges and 8 vertices.
In simple words: Faces are flat surfaces, edges are their sides, vertices are corners; a cube has 6, 12 and 8 of them.

Teacher's Note:
1. Learn the three terms together with an example.
2. A cuboid has the same numbers as a cube.

Question 6. The figure shows Step 2 of the shape sequence leading to the Sierpinski Triangle.
(i) Count the shaded triangles that remain.
(ii) Count the holes.
Check your first count against the rule \( R_n = 3^n \).
Answer:
(i) 9 shaded triangles remain (3 in each corner part).
(ii) There are 4 holes: one large central hole made at Step 1 and 3 smaller holes made at Step 2.
Check: \( R_2 = 3^2 = 9 \), which matches the count. The holes also agree with \( H_2 = H_1 + R_1 = 1 + 3 = 4 \).
In simple words: Step 2 has 9 shaded triangles and 4 holes, as the rules predict.

Teacher's Note:
1. Count part by part to avoid missing triangles.
2. Holes of different sizes are all counted.

Question 7. What is a net of a solid? Name two solids that can be made by folding a net, and state how many different nets a cube has.
Answer:
A net is the flat shape obtained by unfolding a solid onto a plane; folding it along its lines gives the solid back.
Solids made from nets include cubes, cuboids, cylinders, cones, prisms, pyramids, regular tetrahedrons and octahedrons (any two).
A cube has 11 different nets, counting two as the same if one is a rotation or flip of the other.
In simple words: A net is a solid opened out flat; cubes and cones can be folded from nets, and a cube has 11 nets.

Teacher's Note:
1. A sphere is the solid that has no net.
2. Rotated or flipped nets are not counted again.

Question 8. Name the three standard views used to represent a solid on paper, and state the plane on which each one is a projection.
Answer:
Front view: projection on the vertical plane, in front of the object.
Top view: projection on the horizontal plane, below the object.
Side view: projection on the side plane, at the side of the object.
In simple words: Front view on the vertical plane, top view on the horizontal plane, side view on the side plane.

Teacher's Note:
1. The three planes are mutually perpendicular.
2. Engineers use all three views together.

Question 9. For the Sierpinski Carpet, \( R_0 = 1 \), \( H_0 = 0 \), \( R_{n+1} = 8R_n \) and \( H_{n+1} = H_n + R_n \). Find \( R_3 \) and \( H_3 \), showing the steps.
Answer:
Using \( R_{n+1} = 8R_n \): \( R_1 = 8 \times 1 = 8 \), \( R_2 = 8 \times 8 = 64 \), \( R_3 = 8 \times 64 = 512 \).
Using \( H_{n+1} = H_n + R_n \): \( H_1 = 0 + 1 = 1 \), \( H_2 = 1 + 8 = 9 \), \( H_3 = 9 + 64 = 73 \).
So \( R_3 = 512 \) and \( H_3 = 73 \).
In simple words: Multiply squares by 8 and add squares to holes: 512 squares and 73 holes at Step 3.

Teacher's Note:
1. Use the previous step's R when finding the next H.
2. Make a small table to avoid mistakes.

Question 10. The starting triangle of the Sierpinski Gasket has area 1 square unit. Find the area of the region that remains at Step 1, Step 2 and Step 3, and write the rule for Step \( n \).
Answer:
Each step removes the central quarter of every remaining triangle, so \( \frac{3}{4} \) of the area is kept.
Step 1: \( \frac{3}{4} \) square unit.
Step 2: \( \frac{3}{4} \) × \( \frac{3}{4} \) = \( \frac{9}{16} \) square unit.
Step 3: \( \frac{9}{16} \) × \( \frac{3}{4} \) = \( \frac{27}{64} \) square unit.
Rule: area at Step n = \( \left(\frac{3}{4}\right)^{n} \) square units.
In simple words: Multiply by 3/4 each time: 3/4, 9/16, 27/64, and (3/4)ⁿ in general.

Teacher's Note:
1. Each smaller triangle has one-quarter of the area of its parent.
2. The area keeps shrinking towards 0.

Question 11. A Koch Snowflake is built on an equilateral triangle of side 1 unit.
(i) Find the number of sides at Step 3.
(ii) Find the perimeter of the shape at Step 3.
Answer:
Each side is replaced by 4 sides, each one-third as long.
(i) Sides at Step n = 3 × 4ⁿ, so at Step 3: 3 × 4³ = 3 × 64 = 192.
(ii) Each side at Step 3 has length \( \left(\frac{1}{3}\right)^{3} \) = \( \frac{1}{27} \) unit.
Perimeter = 192 × \( \frac{1}{27} \) = \( \frac{64}{9} \) ≈ 7.11 units.
In simple words: At Step 3 there are 192 sides of length 1/27, so the perimeter is 64/9 ≈ 7.11 units.

Teacher's Note:
1. Perimeter = number of sides × length of one side.
2. Check: 3 × (4/3)³ = 64/9.

Question 12. A prism has two congruent octagons as its opposite faces, and a pyramid has an octagonal base. Find the number of faces, edges and vertices of each solid.
Answer:
Octagonal prism: 2 octagons + 8 side faces = 10 faces; 8 + 8 + 8 = 24 edges; 8 + 8 = 16 vertices.
Octagonal pyramid: 1 base + 8 triangles = 9 faces; 8 base edges + 8 slant edges = 16 edges; 8 base vertices + 1 apex = 9 vertices.
In simple words: Octagonal prism: 10 faces, 24 edges, 16 vertices; octagonal pyramid: 9 faces, 16 edges, 9 vertices.

Teacher's Note:
1. Prism: n + 2, 3n, 2n. Pyramid: n + 1, 2n, n + 1.
2. Here n = 8 for both solids.

Question 13. Study the three flat shapes drawn below.
Which of them are nets of a cube? For the shape that is not a net of a cube, give a reason.
Answer:
Shapes (i) and (iii) are nets of a cube; shape (ii) is not.
(i) is the cross shape: the row of four squares wraps round as the four side faces, and the two extra squares fold up as the top and the bottom.
(iii) also folds into a cube: one row of three squares wraps round, and the other three squares supply the remaining faces.
(ii) is a 2 × 3 rectangle. When it is folded, two of its squares land on the same face of the cube while another face is left uncovered, so it cannot form a cube.
In simple words: The cross and the staircase fold into cubes; the 2 × 3 rectangle cannot.

Teacher's Note:
1. Six squares are needed, but not every six-square shape works.
2. Imagine folding each square to check for overlaps.

Question 14. Name a solid, and state the viewpoint, for each case.
(i) A solid whose profile is a rectangle from one viewpoint and a circle from another.
(ii) A solid whose profile is a circle from one viewpoint and a triangle from another.
Answer:
(i) A cylinder: from the side (looking across its axis) it looks like a rectangle; from directly above (along its axis) it looks like a circle.
(ii) A cone standing on its base: from the side it looks like a triangle; from directly above it looks like a circle.
In simple words: A cylinder shows a rectangle or a circle; a cone shows a triangle or a circle.

Teacher's Note:
1. The same solid gives different profiles from different directions.
2. Always name the viewpoint along with the shape.

Question 15. A straight rod of length \( l \) is held above a table top, and the length of its projection on the table is \( p \). Explain why \( p \) can never be greater than \( l \), and state when \( p = l \).
Answer:
From one end of the rod, draw a line parallel to the table and drop a perpendicular to it from the other end.
This makes a right triangle in which the rod (length l) is the hypotenuse and a segment of length p (equal to the projection) is one of the other sides.
The hypotenuse is the longest side of a right triangle, so p ≤ l always.
p = l only when the rod is held parallel to the table.
In simple words: The rod is the hypotenuse and its projection a shorter side, so p ≤ l, with equality only when the rod is parallel to the table.

Teacher's Note:
1. A rod standing straight up projects to just a point.
2. Use the right-triangle picture to explain.

Question 16. What is meant by an isometric projection of a cube? What is the outline of the isometric projection obtained when a cube is balanced on one of its corner vertices?
Answer:
An isometric projection is a projection taken in a position in which all the edges of the cube project to equal lengths; “isometric” means “equal measure”.
When the cube is balanced on one corner vertex and projected down to the floor, the outline is a regular hexagon. If the cube were made of glass, all 12 projected edges would be seen as equal segments.
In simple words: In an isometric projection all edges look equal; a cube on its corner gives a regular hexagon outline.

Teacher's Note:
1. Isometric grid paper is built from this hexagon.
2. Equal lengths make counting units easy.

Question 17. For the Sierpinski Carpet, show that \( H_4 = 1 + 8 + 8^2 + 8^3 \), find its value, and explain why the holes add up in this way.
Answer:
Every square remaining at Step n makes exactly one new hole at Step n + 1, and old holes stay. So \( H_{n+1} = H_n + R_n \) with \( R_n = 8^n \).
Starting from \( H_0 = 0 \): \( H_1 = 1 \), \( H_2 = 1 + 8 \), \( H_3 = 1 + 8 + 8^2 \), and so \( H_4 = 1 + 8 + 8^2 + 8^3 \).
Value: 1 + 8 + 64 + 512 = 585.
In simple words: Each step adds one hole for every square, so the holes are 1 + 8 + 64 + 512 = 585.

Teacher's Note:
1. The holes collect the squares of all earlier steps.
2. Check: 73 + 512 = 585.

Question 18. The same ant and the same laddu on a cuboidal box give different straight-line distances when the box is unfolded in different ways. Explain why this happens, and say what it means for the problem of finding the shortest path along the surface.
Answer:
A path on the surface of the cuboid becomes a path of the same length on the net, and the other way round.
But a cuboid can be unfolded in several different ways, and each unfolding places the two points at a different pair of positions on the flat sheet, so the straight line joining them is different each time.
So we must try all the possible unfoldings, find the straight-line distance for each using the Baudhayana theorem, and take the smallest as the shortest path.
In simple words: Each unfolding gives its own straight line, so all unfoldings must be compared and the smallest chosen.

Teacher's Note:
1. One unfolding gives only one candidate.
2. A straight line that leaves the net is not allowed.

Question 19. Two different solids can have exactly the same front view. Give one such pair, and explain why engineers draw three mutually perpendicular views of an object instead of one.
Answer:
An upright cylinder and a cuboid of the same height and width both have a rectangle as their front view.
So one projection does not identify a solid, because many different objects project to the same shape.
The front, top and side views together fix the shape in all three directions, so an engineer can read the real object from the drawing and build it.
In simple words: A cylinder and a cuboid can both look like a rectangle from the front, so three views are needed.

Teacher's Note:
1. The top view separates them: a circle for the cylinder, a rectangle for the cuboid.
2. A projection loses depth information.

Question 20. Take the starting square and the starting triangle to each have area 1 square unit. Find the first step at which the region remaining in the Sierpinski Gasket falls below \( \frac{1}{2} \), and show that at that same step the Sierpinski Carpet still keeps more than two-thirds of its area.
Answer:
Gasket area at Step n = \( \left(\frac{3}{4}\right)^{n} \): \( \frac{3}{4} \) = 0.75, \( \frac{9}{16} \) = 0.5625, \( \frac{27}{64} \) ≈ 0.42.
Step 2 still keeps more than half, but Step 3 keeps \( \frac{27}{64} \) < \( \frac{1}{2} \). So Step 3 is the first such step.
Carpet area at Step 3 = \( \left(\frac{8}{9}\right)^{3} \) = \( \frac{512}{729} \).
Two-thirds of 729 is 486, so \( \frac{2}{3} \) = \( \frac{486}{729} \). Since 512 > 486, the Carpet still keeps more than two-thirds of its area at Step 3.
In simple words: The gasket drops below half at Step 3, while the carpet still has 512/729, which is more than 2/3.

Teacher's Note:
1. Make the denominators equal to compare fractions.
2. The carpet loses area much more slowly than the gasket.