CBSE Class 8 Mathematics Exploring Some Geometric Themes MCQs Set 11

Answer: (b) repeats the same or a similar pattern at smaller and smaller scales

Explanation:
1. A fractal is a self-similar shape: parts of it look like smaller copies of the whole.
2. So the same or a similar pattern keeps showing up at smaller and smaller scales, as in a fern or the Sierpinski Carpet.
3. The other options describe ordinary shapes, not self-similarity.

Teacher's Note:
1. Self-similar means a part looks like the whole, only smaller.
2. Fractals can have straight edges too, like the Sierpinski Gasket.

Answer: (a) The fern

Explanation:
1. The chapter starts with the fern.
2. Each leaf of a fern looks like a small fern, and each sub-leaf looks like a small leaf, so the pattern repeats at smaller scales.

Teacher's Note:
1. Look at a fern frond closely to see the repeating pattern.
2. Trees, clouds and coastlines are other natural examples.

Answer: (d) The square is broken into 9 smaller squares and the central square is removed

Explanation:
1. Each square is cut into a 3 × 3 arrangement of 9 equal smaller squares.
2. The central one is removed, leaving 8 squares.
3. The same step is then repeated on each of the 8 squares that remain.

Teacher's Note:
1. A 2 × 2 split has no central square, so options (a) and (c) cannot work.
2. Remember: 9 parts, remove the middle one, 8 stay.

Answer: (d) 9

Explanation:
1. At each step every shaded triangle is split into 4 and the middle one is removed, so each one gives 3 triangles.
2. Step 0: 1 triangle; Step 1: 3 triangles; Step 2: 3 × 3 = 9 triangles.
3. In general, 3ⁿ triangles remain at Step n.

Teacher's Note:
1. Multiply by 3 at every step.
2. Do not count the white holes as triangles that remain.

Answer: (c) 4

Explanation:
1. The figure is Step 2 of the Sierpinski Gasket.
2. There is 1 large hole in the middle, made at Step 1.
3. There are 3 smaller holes, one in each corner part, made at Step 2.
4. Total holes = 1 + 3 = 4 (and 9 shaded triangles remain).

Teacher's Note:
1. Count the holes size by size: big first, then small.
2. 9 is the number of shaded triangles, not the number of holes.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is true: the Sierpinski Carpet is one of the fractals studied in the chapter.
2. R is true: the same rule (cut into 9, remove the centre) is applied to every square left.
3. Repeating the rule makes the pattern appear again at smaller scales, which is exactly why the carpet is a fractal. So R explains A.

Teacher's Note:
1. Self-similarity comes from repeating one rule again and again.
2. Check whether R gives the reason for A, not just a true fact.

Answer: (c) Both Statement I and Statement II are true

Explanation:
1. Carpet: squares remaining at Step n = 8ⁿ, so at Step 2 it is 8² = 64. Statement I is true.
2. Gasket: triangles remaining at Step n = 3ⁿ, so at Step 2 it is 3² = 9. Statement II is true.
3. So both statements are true.

Teacher's Note:
1. Carpet multiplies by 8, gasket multiplies by 3.
2. Write both rules before checking the statements.

Answer: (b) 12

Explanation:
1. A cube has 4 edges on the top face, 4 on the bottom face and 4 vertical edges joining them.
2. Total = 4 + 4 + 4 = 12 edges.
3. Edges are the line segments where two faces meet.

Teacher's Note:
1. Count edges in layers: top, bottom and upright.
2. A cube and a cuboid both have 6 faces, 12 edges and 8 vertices.

Answer: (a) the flat shape obtained by unfolding the solid onto a plane

Explanation:
1. A net is what we get when we unfold the surface of a solid flat onto a plane.
2. Folding the net back gives the solid again.
3. A shadow or a top view is only a picture of the solid; it cannot be folded into it.

Teacher's Note:
1. Think of opening a cardboard box flat.
2. Flaps used for gluing are not part of the net.

Answer: (b) 11

Explanation:
1. When nets that differ only by a rotation or a flip are treated as the same, a cube has exactly 11 different nets.
2. 43,380 is the number of nets of a dodecahedron, not of a cube.

Teacher's Note:
1. Try drawing all 11 cube nets on squared paper.
2. A regular tetrahedron has only 2 nets.

Answer: (d) top view

Explanation:
1. The horizontal plane lies below the object; looking straight down gives its projection on this plane.
2. This projection is the top view.
3. The vertical plane gives the front view and the side plane gives the side view.

Teacher's Note:
1. Horizontal plane means the floor-like plane.
2. Front, top and side views are the three mutually perpendicular projections.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: a cuboid has 12 edges.
2. R is false: edges are the line segments that form the sides of the faces.
3. The points where edges meet are the vertices.

Teacher's Note:
1. Faces are surfaces, edges are segments, vertices are points.
2. A wrong definition makes R false even if A is true.

Answer: (b) \( \frac{64}{81} \)

Explanation:
1. At each step 8 of the 9 equal parts stay, so \( \frac{8}{9} \) of the area is kept.
2. Area at Step n = \( \left(\frac{8}{9}\right)^{n} \).
3. At Step 2: \( \frac{8}{9} \) × \( \frac{8}{9} \) = \( \frac{64}{81} \) sq. unit.

Teacher's Note:
1. Multiply by the same fraction once for each step.
2. \( \frac{16}{18} \) is just \( \frac{8}{9} \) written differently, so it is the Step 1 area.

Answer: (d) 73

Explanation:
1. For the carpet, \( R_n = 8^n \): \( R_0 = 1, R_1 = 8, R_2 = 64 \).
2. \( H_1 = 0 + 1 = 1 \), \( H_2 = 1 + 8 = 9 \).
3. \( H_3 = 9 + 64 = \) 73.

Teacher's Note:
1. Each remaining square makes one new hole at the next step.
2. Old holes stay, so keep adding.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is true: each triangle is split into 4 identical triangles and 1 is removed, so 3 out of 4 equal parts stay.
2. Keeping \( \frac{3}{4} \) of the area at every step gives \( \left(\frac{3}{4}\right)^{n} \) after n steps, so A is true.
3. A follows directly from R, so R is the correct explanation of A.

Teacher's Note:
1. A fraction kept at every step turns into a power.
2. Compare with the carpet, which keeps 8/9 each step.

Answer: (c) (i)-(s), (ii)-(q), (iii)-(r), (iv)-(p)

Explanation:
1. (i) Carpet squares at Step 2 = 8² = 64, so (i)-(s).
2. (ii) Gasket triangles at Step 3 = 3³ = 27, so (ii)-(q).
3. (iii) Koch sides at Step 2 = 3 × 4² = 48, so (iii)-(r).
4. (iv) Gasket holes go 0, 1, 4, 13, so at Step 3 there are 13 holes: (iv)-(p).
5. The correct matching is (i)-(s), (ii)-(q), (iii)-(r), (iv)-(p).

Teacher's Note:
1. Gasket holes: add the triangles of the previous step (H_n+1 = H_n + 3ⁿ).
2. Work out every value before looking at the options.

Answer: (a) Only Statement I is true

Explanation:
1. Each side is cut into 3 equal parts and the middle part is replaced by two sides of a small equilateral triangle, giving 4 segments. Statement I is true.
2. Each new segment is \( \frac{1}{3} \) of the old side, so the length becomes 4 × \( \frac{1}{3} \) = \( \frac{4}{3} \) of the old side.
3. The perimeter is multiplied by \( \frac{4}{3} \) at each step, so it grows. Statement II is false.

Teacher's Note:
1. 4 pieces each one-third long make 4/3 of the old length.
2. The snowflake stays in a small region, yet its perimeter keeps growing.

Answer: (d) A sphere

Explanation:
1. A cone, a cylinder and an octahedron can all be folded from suitable nets.
2. A sphere has no net: no flat sheet can cover a ball without gaps, overlaps or crumpling.

Teacher's Note:
1. Try wrapping a ball in paper and notice the wrinkles.
2. Curved surfaces that bend in only one direction, like a cylinder, can be unrolled.

Answer: (a) the circumference of the base circle

Explanation:
1. The curved surface unrolls into a rectangle.
2. One side is the height; the other side is the edge of the base circle laid out flat.
3. So the other side equals the circumference of the base circle, 2πr.

Teacher's Note:
1. The rectangle wraps once around the circle.
2. Think of the label on a can unrolled.

Answer: (d) A is false, but R is true.

Explanation:
1. A is false: a regular tetrahedron has only 2 different nets.
2. R is true: a cube has 11 different nets.

Teacher's Note:
1. Do not mix up the cube and the tetrahedron counts.
2. Tetrahedron: 2 nets; cube: 11 nets; octahedron: 11 nets.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: if the segment has length l and its projection has length p, then always p ≤ l.
2. R is false: the lengths are equal when the segment is parallel to the plane.
3. A segment perpendicular to the plane projects to a single point.

Teacher's Note:
1. The segment is the hypotenuse of a right triangle; the projection is a leg.
2. Parallel keeps the length; perpendicular shrinks it to a point.

Answer: (c) Both Statement I and Statement II are true

Explanation:
1. The chapter shows that projection keeps parallel lines parallel. Statement I is true.
2. Both pairs of opposite sides of a parallelogram are parallel, so their projections are also parallel.
3. Hence the projection is again a parallelogram. Statement II is true.

Teacher's Note:
1. Statement II follows directly from Statement I.
2. This is why the edges of a cube stay parallel in a drawing.

Answer: (b) A triangular prism

Explanation:
1. The two congruent equilateral triangles become the two opposite end faces.
2. The three rectangles fold up around them, joining matching sides.
3. Two congruent triangles joined by rectangles form a triangular prism.
4. A pyramid would need its triangles to meet at a single point.

Teacher's Note:
1. Two congruent polygon faces signal a prism.
2. Count faces: 3 + 2 = 5, which matches a triangular prism.

Answer: (a) (i)-(r), (ii)-(p), (iii)-(q), (iv)-(s)

Explanation:
1. Cuboid: 6 faces, so (i)-(r).
2. Regular tetrahedron: 4 triangular faces, so (ii)-(p).
3. Square pyramid: 1 square base + 4 triangles = 5 faces, so (iii)-(q).
4. Pentagonal prism: 5 side faces + 2 pentagons = 7 faces, so (iv)-(s).
5. Answer: (i)-(r), (ii)-(p), (iii)-(q), (iv)-(s).

Teacher's Note:
1. Prism on an n-gon: n + 2 faces. Pyramid on an n-gon: n + 1 faces.
2. A tetrahedron is a triangular pyramid: 3 + 1 = 4 faces.

Answer: (b) Step 3

Explanation:
1. Squares remaining at Step n = 8ⁿ.
2. Step 2: 8² = 64, which is less than 500.
3. Step 3: 8³ = 512, which is more than 500. So it first happens at Step 3.

Teacher's Note:
1. List the powers of 8 until you pass 500.
2. First means the earliest step that works.

Answer: (a) Only Statement I is true

Explanation:
1. Each remaining square gives 8 squares, so \( R_{n+1} = 8R_n \). Statement I is true.
2. Holes follow \( H_{n+1} = H_n + R_n \), giving 0, 1, 9, 73, …
3. 9 is not 8 × 1, so Statement II is false.

Teacher's Note:
1. Test a rule on actual numbers before believing it.
2. Holes add up; squares multiply.

Answer: (c) (i)-(r), (ii)-(s), (iii)-(p), (iv)-(q)

Explanation:
1. (i) Gasket area at Step 3 = \( \left(\frac{3}{4}\right)^{3} \) = \( \frac{27}{64} \), so (i)-(r).
2. (ii) Carpet area at Step 2 = \( \left(\frac{8}{9}\right)^{2} \) = \( \frac{64}{81} \), so (ii)-(s).
3. (iii) Koch perimeter at Step 2 = 3 × \( \left(\frac{4}{3}\right)^{2} \) = \( \frac{16}{3} \), so (iii)-(p).
4. (iv) Gasket holes go 0, 1, 4, 13, 40, so Step 4 has 40 holes: (iv)-(q).
5. Answer: (i)-(r), (ii)-(s), (iii)-(p), (iv)-(q).

Teacher's Note:
1. Areas shrink by 3/4 or 8/9; the Koch perimeter grows by 4/3.
2. Holes for the gasket: 13 + 27 = 40.

Answer: (d) A is false, but R is true.

Explanation:
1. R is true: lengths of paths do not change between the cuboid and its net.
2. A is false: different unfoldings give straight segments of different lengths, and a segment may even fall outside the net.
3. So all the possible unfoldings must be checked and the smallest length taken.

Teacher's Note:
1. One unfolding gives only one candidate length.
2. The ant-and-laddu example shows different unfoldings giving different distances.

Answer: (b) Only Statement II is true

Explanation:
1. Different objects (for example, different lines or different cuboids) can give the very same projection. Statement II is true.
2. Because of this, one projection cannot tell us exactly which solid we have. Statement I is false.
3. That is why three mutually perpendicular views are used.

Teacher's Note:
1. A projection loses information along the viewing direction.
2. Front, top and side views together fix the solid.

Answer: (c) 3

Explanation:
1. Three cubes rest on the ground in an L shape, and the fourth cube sits on top of the corner cube.
2. The top view shows only the ground area covered when we look straight down.
3. The raised cube lies exactly above the corner cube, so it adds no new square.
4. So the top view is an L-shape of 3 unit squares.

Teacher's Note:
1. Cubes stacked one above another show as one square in the top view.
2. Count cubes column by column, not one by one.