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

Very Short Answer Questions

Question 1. What is meant by saying that a shape is self-similar?
Answer:
A shape is self-similar if it contains smaller copies of itself, so the same or a similar pattern appears again and again at smaller and smaller scales.

Teacher's Note:
1. Self-similarity is the key property of a fractal.
2. Zooming into a part shows a pattern like the whole.

Question 2. Name the plant used in the chapter as a natural example of a fractal, whose leaves are smaller copies of the whole plant.
Answer:
The fern. Its leaves are small copies of the whole fern, and its sub-leaves are small copies of the leaves.

Teacher's Note:
1. The fern shows self-similarity at more than one level.
2. It is the first fractal example in the chapter.

Question 3. Write any two objects in nature, other than the fern, in which self-similarity can be seen.
Answer:
Any two of: trees (trunk, limbs, branches, branchlets), clouds, coastlines, mountains and lightning.

Teacher's Note:
1. For a tree, the smaller copies are the branches and branchlets.
2. Any two correct examples earn full marks.

Question 4. In one step of the Sierpinski Carpet, a square is broken into 9 smaller squares. Which square is then removed, and how many remain?
Answer:
The central square is removed, so 8 squares remain after that step.

Teacher's Note:
1. 9 − 1 = 8 squares are left from every square.
2. This is why the count multiplies by 8.

Question 5. How many shaded triangles remain at Step 1 of the shape sequence of the Sierpinski Gasket?
Answer:
3 triangles remain, because the triangle is split into 4 and the central one is removed.

Teacher's Note:
1. 4 − 1 = 3 triangles at each split.
2. The removed middle triangle is the first hole.

Question 6. Name the temple at Khajuraho whose tall structure is made up of smaller copies of the full structure.
Answer:
The Kandariya Mahadev Temple at Khajuraho, Madhya Pradesh (built around 1025 C.E.).

Teacher's Note:
1. Its towers are made of smaller tower-like copies.
2. Temples at Madurai, Hampi, Rameswaram and Varanasi show the same idea.

Question 7. What are the edges and the vertices of a solid?
Answer:
Edges are the line segments that form the sides of the faces of a solid.
Vertices are the points where the edges meet.

Teacher's Note:
1. Faces are flat surfaces, edges are segments, vertices are corner points.
2. Two faces meet along an edge.

Question 8. How many faces, edges and vertices does a cube have?
Answer:
A cube has 6 faces, 12 edges and 8 vertices.

Teacher's Note:
1. A cuboid has the same counts.
2. Count edges as top 4, bottom 4 and 4 upright.

Question 9. What is a net of a solid?
Answer:
A net is the flat shape obtained by unfolding a solid onto a plane; folding the net back gives the solid.

Teacher's Note:
1. Gluing flaps are not part of the net.
2. A cardboard box opened flat is a net with flaps.

Question 10. Name the three mutually perpendicular projections used to represent a solid on plane surfaces.
Answer:
The front view (on the vertical plane), the top view (on the horizontal plane) and the side view (on the side plane).

Teacher's Note:
1. Three planes at right angles to one another are used.
2. Together the three views fix the shape of the solid.

Question 11. How many different nets does a regular tetrahedron have?
Answer:
A regular tetrahedron has only 2 different nets.

Teacher's Note:
1. Compare: a cube has 11 nets.
2. Both tetrahedron nets have 4 equilateral triangles.

Question 12. What is meant by the isometric projection of a cube?
Answer:
It is the projection of a cube balanced on one corner vertex, in which all the edges project to equal lengths. Its outline is a regular hexagon.

Teacher's Note:
1. Isometric means “equal measure”.
2. Tiling the plane with such hexagons gives isometric grid paper.

Question 13. Write the rule connecting \( R_{n+1} \) and \( R_n \) for the Sierpinski Carpet, and hence write \( R_n \) in terms of \( n \).
Answer:
Each remaining square gives 8 squares, so \( R_{n+1} = 8R_n \).
Starting from \( R_0 = 1 \), this gives \( R_n = 8^n \).

Teacher's Note:
1. Multiplying by 8 n times gives 8 to the power n.
2. Check: R₂ = 64 matches the Step 2 picture.

Question 14. Using \( H_{n+1} = H_n + R_n \) with \( H_0 = 0 \), find the number of holes in the Sierpinski Carpet at Step 2.
Answer:
\( H_1 = H_0 + R_0 = 0 + 1 = 1 \)
\( H_2 = H_1 + R_1 = 1 + 8 = 9 \). So there are 9 holes at Step 2.

Teacher's Note:
1. Each remaining square makes one new hole.
2. 1 big hole + 8 small holes = 9.

Question 15. Taking the area of the starting square to be 1 sq. unit, write the area of the region remaining at Step \( n \) of the Sierpinski Carpet.
Answer:
Each step keeps \( \frac{8}{9} \) of the area, so the area at Step n is \( \left(\frac{8}{9}\right)^{n} \) sq. unit.

Teacher's Note:
1. 8 of the 9 equal parts survive each step.
2. The area becomes smaller and smaller as n grows.

Question 16. How many triangles remain at Step 4 of the shape sequence of the Sierpinski Gasket?
Answer:
Triangles remaining = 3ⁿ, so at Step 4 it is 3⁴ = 81.

Teacher's Note:
1. 1, 3, 9, 27, 81: multiply by 3 each step.
2. Do not confuse with the holes, which are 40 at Step 4.

Question 17. How many sides does the Koch Snowflake shape have at Step 3?
Answer:
Number of sides = 3 × 4ⁿ, so at Step 3 it is 3 × 4³ = 3 × 64 = 192 sides.

Teacher's Note:
1. Every side becomes 4 sides at each step.
2. 3, 12, 48, 192 is the sequence.

Question 18. Name the Dutch artist whose print Smaller and Smaller shows the same pattern of lizards at smaller and smaller scales.
Answer:
M. C. Escher.

Teacher's Note:
1. Escher used fractal-like repetition in many prints.
2. The lizards shrink towards the centre of the print.

Question 19. A prism has two congruent octagons as its opposite faces. How many faces and how many edges does it have?
Answer:
Faces = 8 + 2 = 10; edges = 3 × 8 = 24.

Teacher's Note:
1. Prism on an n-gon: n + 2 faces and 3n edges.
2. It also has 2 × 8 = 16 vertices.

Question 20. How many faces, edges and vertices does a pyramid with a hexagonal base have?
Answer:
Faces = 6 + 1 = 7, edges = 2 × 6 = 12, vertices = 6 + 1 = 7.

Teacher's Note:
1. Pyramid on an n-gon: n + 1 faces, 2n edges, n + 1 vertices.
2. The extra vertex is the apex.

Question 21. Which flat shapes make up the net of a cylinder?
Answer:
One rectangle (the unrolled curved surface) and two circles (the circular faces). The rectangle's sides are the height and the circumference of the base.

Teacher's Note:
1. The rectangle's length must match the circle's circumference.
2. Two equal circles are needed for a closed cylinder.

Question 22. A cone is slit open along a slant line and unrolled. What is the shape of the unrolled curved surface?
Answer:
It is part of a circle (a sector) whose centre is the apex, because every point of the base circle is the same slant distance from the apex.

Teacher's Note:
1. The slant height becomes the radius of the sector.
2. The base circle is added separately in the net.

Question 23. Name a solid whose profile is a circle when seen from directly above and a triangle when seen from the side.
Answer:
A cone standing on its base.

Teacher's Note:
1. Seen from the side, its slant edges and base make a triangle.
2. A cylinder would look like a rectangle from the side.

Question 24. Why does the shadow cast by sunlight on a wall look almost the same as the projection of the object on that wall?
Answer:
The Sun is extremely far away, so its rays are practically parallel. When they fall perpendicular to the wall, the shadow is practically the same as the perpendicular projection.

Teacher's Note:
1. A nearby torch makes the shadow larger than the projection.
2. A distant source behaves like a projection.

Question 25. Taking the sidelength of the starting equilateral triangle to be 1 unit, find the perimeter of the Koch Snowflake shape at Step 2.
Answer:
The perimeter is multiplied by \( \frac{4}{3} \) at each step.
Step 2 perimeter = 3 × \( \left(\frac{4}{3}\right)^{2} \) = \( \frac{16}{3} \) ≈ 5.33 units.

Teacher's Note:
1. Check: 48 sides each of length 1/9 give 48/9 = 16/3.
2. The perimeter grows at every step.

Question 26. At which step does the area remaining in the Sierpinski Gasket first become less than half the area of the starting triangle?
Answer:
Area at Step n = \( \left(\frac{3}{4}\right)^{n} \).
Step 2: \( \frac{9}{16} \) = 0.5625 (more than half); Step 3: \( \frac{27}{64} \) ≈ 0.42 (less than half). So it first happens at Step 3.

Teacher's Note:
1. Half is 0.5, so compare decimals.
2. 9/16 is still more than 8/16.

Question 27. At Step 2, which loses area faster — the Sierpinski Carpet or the Sierpinski Gasket? Support your answer with the two fractions.
Answer:
Carpet at Step 2: \( \frac{64}{81} \) ≈ 0.79 of the area; Gasket at Step 2: \( \frac{9}{16} \) = 0.5625 of the area.
So the Sierpinski Gasket loses area faster.

Teacher's Note:
1. The gasket removes 1 part in 4; the carpet only 1 part in 9.
2. Compare fractions by turning them into decimals.

Question 28. How does drawing a net help in finding the shortest path between two points on the surface of a cuboid?
Answer:
Any surface path becomes a path of the same length on the net, so the shortest path is the straight line joining the two points on a suitable net. All unfoldings must be tried, since they give different lengths.

Teacher's Note:
1. On a flat sheet the shortest path is a straight line.
2. Its length is found with the Baudhayana theorem.

Question 29. In one unfolding of a cuboid, the ant and the laddu lie at the two ends of the hypotenuse of a right triangle whose legs are 24 cm and 32 cm. Find this distance.
Answer:
By the Baudhayana theorem, d² = 24² + 32² = 576 + 1024 = 1600.
So d = √1600 = 40 cm.

Teacher's Note:
1. 24 : 32 : 40 is 3 : 4 : 5 multiplied by 8.
2. The hypotenuse is the longest side.

Question 30. A closed loop of three beams, drawn on an isometric grid, looks like a triangle but cannot be built with real cubes. Give one reason why such a drawing is possible on paper.
Answer:
A drawing on paper loses depth information, so beams that are actually far apart in depth can be drawn as if they meet at a corner. The picture is an optical illusion; no real arrangement of cubes gives that loop.

Teacher's Note:
1. A projection does not come from a unique object.
2. Our eye wrongly assumes all the beams are at the same depth.