CBSE Class 8 Mathematics Exploring Some Geometric Themes Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. Fractals occur both in nature and in human art.
(i) Explain what self-similarity means, using the fern as your example.
(ii) Describe three other objects in nature that show self-similarity, saying in each case what the “smaller copy” is.
(iii) Describe two human-made examples of fractal design mentioned in the chapter, naming the place or the artist.
Answer:
(i) A shape is self-similar when a part of it looks the same as (or very nearly the same as) the whole, only smaller. In a fern each leaf is a small copy of the whole fern, and each leaf is made of sub-leaves that are small copies of the leaf; the pattern carries on at smaller and smaller scales. Such shapes are called fractals.
(ii) A tree: the trunk carries limbs, each limb carries branches and each branch carries branchlets; the same branching pattern appears at every scale.
A coastline: a short stretch seen closely has the same ragged in-and-out shape as the whole coast on a map.
Lightning: the main flash splits into smaller forks, and each fork splits again in the same way. (Clouds and mountains behave similarly.)
(iii) The Kandariya Mahadev Temple at Khajuraho, Madhya Pradesh (about 1025 C.E.): its tall structure is made of smaller copies of the whole structure, which carry still smaller copies. Temples at Madurai, Hampi and Varanasi show similar patterns.
The Dutch artist M. C. Escher's print Smaller and Smaller, in which the same lizard pattern repeats at smaller and smaller scales. (Nigerian Fulani wedding blankets, with diamonds inside diamonds, are another example.)
In simple words: Fractals repeat their own shape in smaller parts, as in ferns, trees, coastlines, lightning, Khajuraho temple towers and Escher's prints.

Teacher's Note:
1. For each example, name both the whole and its smaller copy.
2. Human-made fractals are designed on purpose; natural ones are approximate.

Teacher's Note:
1. Fill the table one row at a time using the previous row.
2. The “remove the centre” rule gives both the 8 and the 8/9.

Question 3. Many hollow solids are manufactured by folding a flat sheet.
(i) Explain what a net of a solid is, and why a net is useful.
(ii) Describe the net of a cylinder of radius \( r \) and height \( h \), and write the two side lengths of the rectangle in the net.
(iii) Describe the net of a cone, and explain why the curved part of the net is a piece of a circle.
Answer:
(i) A net of a solid is the flat shape got by unfolding the solid onto a plane; folding it back along its fold lines gives the solid again. Nets are useful because hollow solids can be made from flat foldable material such as paper, cardboard or metal sheet, and because a problem on the surface of a solid (like finding the shortest path) can be changed into a flat problem on the net.
(ii) Open out the two circular faces and cut the curved surface once along its height. The net has two circles of radius r and one rectangle. One side of the rectangle is the height h; the other side is the circumference of the base, 2πr, because the curved surface wraps exactly once round the circle.
(iii) Slit the cone along a slant line of length l from the apex O to the base and unroll it. The net is a circle for the base together with a curved piece. Every point on the edge of the base circle is at the same distance l from O, and this stays true after unrolling. So the curved edge of the piece is part of a circle with centre O and radius l: the curved surface unrolls into a sector of a circle.
In simple words: A net is a solid opened flat; a cylinder's net is a rectangle (h by 2πr) with two circles, and a cone's curved surface opens into a sector.

Teacher's Note:
1. The rectangle's length must match the circumference so it closes up.
2. The slant height of the cone becomes the radius of the sector.

Teacher's Note:
1. Check each table row against the general rule.
2. Vertices are easiest: count base corners, then add the apex or the second polygon.

Teacher's Note:
1. In part (i), show all three angles are 60° in each small triangle.
2. More pieces does not mean more area.

Teacher's Note:
1. Multiply the two middle columns to check each perimeter.
2. A bounded shape can still have a growing boundary.

Teacher's Note:
1. The top view of both solids is a circle, so it cannot tell them apart.
2. Perpendicular dropping is the key idea of projection.

Question 8. The model below is made of identical cubes glued face to face. The viewing directions are marked.
(i) How many cubes have been used?
(ii) Draw and describe the top view of the model.
(iii) Draw and describe the front view and the side view of the model.
(iv) How many more such cubes must be added to turn the model into a complete solid cuboid that is 3 cubes long, 2 cubes deep and 2 cubes high?
Answer:
(i) Three cubes lie in a row on the table, one more cube is placed behind the first of them, and one cube sits on top of that first cube. In all, 5 cubes have been used.
(ii) Top view: looking straight down we see the squares of all the cubes on the table; the cube on top lies exactly above one of them and adds no new square. The top view is an L-shape of 4 unit squares: a row of 3 squares with 1 more square attached beside the square at one end.
(iii) Front view: an L-shape of 3 unit squares: two squares side by side in the bottom row with one square on top of one of them. The other cubes of the row are hidden exactly behind the front cube, so they add nothing new.
Side view: an L-shape of 4 unit squares: a row of 3 squares with 1 square on top of the square at one end.

(iv) A cuboid 3 cubes long, 2 deep and 2 high needs 3 × 2 × 2 = 12 cubes. Since 5 are already in place, 12 − 5 = 7 more cubes must be added.
In simple words: The model has 5 cubes; its top view is an L of 4 squares, its front view an L of 3, its side view an L of 4, and 7 more cubes complete the 3 × 2 × 2 cuboid.

Teacher's Note:
1. In each view, cubes one behind another show as one square.
2. Total needed minus cubes present gives the cubes to add.

Question 9. Isometric drawing is widely used in engineering.
(i) A cube is balanced on one of its corner vertices and projected onto the floor. Explain what makes this projection isometric, and state the outline that is obtained.
(ii) Which property of projections ensures that the three families of parallel edges of a solid appear as the three families of parallel lines |, / and ∖ on isometric paper? Explain how length, depth and height are then read off the grid.
(iii) An “impossible triangle” can be drawn neatly on an isometric grid but cannot be built from real cubes. Explain why the drawing works as an illusion, using the idea that a projection does not come from a unique object.
Answer:
(i) When the cube is balanced on a corner so that the diagonal through that corner is vertical, all 12 edges make equal angles with the floor. Each edge is shortened by the same amount, so all projected edges come out equal in length; “isometric” means “equal measure” in Greek. The outline is a regular hexagon, and if the cube were made of glass all 12 edges would be seen inside it.
(ii) Projection carries parallel lines to parallel lines. A cube or cuboid has three families of parallel edges (length edges, depth edges and height edges), so in the projection they become three families of parallel lines on paper, drawn as |, / and ∖. Because the projection is isometric, one unit along each axis is drawn as the same length. So on isometric grid paper we read height along the vertical lines and length and depth along the two slanting families, simply counting grid units in each direction.
(iii) A projection does not come from a unique object: many different solids give the same picture. Each corner of the impossible triangle, taken on its own, is a correct picture of a few real cubes. But the three corners are joined using the fact that two bars far apart in depth can be drawn touching on paper. Our eye assumes that all three bars are at the same depth, and it is this wrong assumption, not the drawing, that creates the illusion. No actual set of cubes can satisfy all three corners at once.
In simple words: Equal shortening of all edges makes the projection isometric (a hexagon outline); parallel edges stay parallel; and lost depth lets an impossible triangle be drawn.

Teacher's Note:
1. Three directions on the grid stand for three directions in space.
2. Illusions come from information lost in projection.

Teacher's Note:
1. Compare decimals step by step to find the first step below 0.5.
2. Many pieces times a tiny area can still be tiny.

Case Study Based Questions

Question 11. A fern in the school garden: Ananya has planted a fern at the corner of her school garden. For her science scrapbook she photographs one full frond and then, with a hand lens, one small leaflet of that frond. When she places the two photographs side by side, her partner cannot tell which is which: the leaflet looks like a smaller copy of the whole frond. Looking closer still, each leaflet is itself made of even smaller leaflets of the same shape. Their teacher tells them that such repeating structure is not a coincidence — it is common in nature, and mathematicians have invented shapes that repeat a rule forever in exactly this way.
(i) What name is given to shapes such as the fern, which show the same pattern at smaller and smaller scales?
(ii) Name two other objects in nature that show the same property, and say what the smaller copies are in each case.
(iii) Is the repetition in a real fern exact or only approximate? Give one reason.
(iv) Give one example from human art, named in the chapter, in which the same idea is used.
Answer:
(i) Such shapes are called fractals; the property they show is self-similarity.
(ii) A tree: the trunk carries limbs, a limb carries branches and a branch carries branchlets, each a smaller copy of the pattern above it. A coastline: a short stretch seen closely has the same ragged shape as the whole coast on a map. (Clouds, mountains and lightning are also correct.)
(iii) Only approximate. In a real fern the leaflets are not exact copies of the whole frond, and the repetition stops after a few levels; a mathematical fractal repeats the same rule exactly and forever.
(iv) The Kandariya Mahadev Temple at Khajuraho, whose structure is made of smaller copies of itself. (Fulani wedding blankets or M. C. Escher's Smaller and Smaller are equally correct.)
In simple words: The fern is a natural, approximate fractal, like trees and coastlines; Khajuraho's temple uses the same idea in art.

Teacher's Note:
1. Natural fractals stop after a few levels; mathematical ones never stop.
2. Always name the smaller copy for each example.

Question 12. A sweet-box maker: A shop near the market makes cuboidal sweet boxes measuring 15 cm × 10 cm × 5 cm. The worker does not join six separate pieces of cardboard. Instead he presses out one flat shape from a cardboard sheet, scores the fold lines and folds it up into the box. Along some of the edges of this flat shape he leaves narrow extra strips, which are folded inside and glued so that the box holds its form. When the shop orders a new design, the printer always asks for a drawing of this flat shape rather than a picture of the finished box.
(i) What is the mathematical name for the flat shape that is folded into the box?
(ii) How many rectangles does this flat shape contain, and how many pairs of identical rectangles are there among them?
(iii) How many edges and how many vertices does the finished box have?
(iv) Why are the narrow extra strips needed, and are they counted as part of the flat shape in mathematics?
Answer:
(i) It is the net of the box: the flat shape obtained by unfolding the cuboid onto a plane.
(ii) It has 6 rectangles, forming 3 pairs of identical rectangles: two of 15 cm × 10 cm, two of 15 cm × 5 cm and two of 10 cm × 5 cm.
(iii) The box has 12 edges and 8 vertices (and 6 faces).
(iv) The strips are flaps. They give a surface for gluing so that the faces hold together and the box stays firm. In mathematics the net is only the shape obtained by unfolding the solid, so the flaps are not part of the net.
In simple words: The flat shape is a net of 6 rectangles in 3 pairs; the box has 12 edges and 8 vertices, and the glue flaps are not part of the net.

Teacher's Note:
1. Opposite faces of a cuboid are identical rectangles.
2. Flaps help in making the box but are not counted in the net.

Question 13. A rangoli at the school exhibition: For the annual exhibition, Class 8 draws a large rangoli on the floor. They begin with an equilateral triangle of side 16 cm drawn in white chalk. They mark the midpoints of the three sides, join them, and fill the middle triangle with plain floor colour, leaving the other three coloured. They then repeat exactly the same action inside each of the three coloured triangles. The pattern that appears is the one Sierpinski described. The students stop after two rounds, because the triangles become too small to fill neatly with rangoli powder.
(i) Write the rule the students applied to go from one step to the next.
(ii) How many coloured triangles are there at Step 1, and how many at Step 2?
(iii) What is the side of each smallest coloured triangle at Step 2?
(iv) How many plain (uncoloured) triangular holes are there at Step 2?
Answer:
(i) Divide each coloured equilateral triangle into 4 identical equilateral triangles by joining the midpoints of its sides, and leave the central one uncoloured. Repeat this inside every triangle that is still coloured.
(ii) Step 1 has 3 coloured triangles, and Step 2 has 3 × 3 = 9 coloured triangles.
(iii) Joining midpoints halves the side each time: 16 cm → 8 cm at Step 1 → 4 cm at Step 2.
(iv) \( H_1 = 1 \) and \( H_2 = H_1 + 3 = \) 4: one large central hole from Step 1 and 3 smaller ones from Step 2.
In simple words: Joining midpoints and leaving the middle uncoloured gives 3 then 9 coloured triangles, the smallest of side 4 cm, with 4 holes.

Teacher's Note:
1. The midpoint triangle has half the side of its parent.
2. Holes add up: 1 from Step 1 plus 3 from Step 2.

Question 14. Torchlight on a wall: Imran holds a solid object in front of a white wall in a dark room and shines a torch on it, keeping the beam perpendicular to the wall. A dark shape appears on the wall. He notices that when he carries the torch closer to the object the dark shape grows, and when he walks far back with the torch the shape settles down to a steady outline. His teacher explains that this steady outline is almost exactly the projection of the object on the wall, and that the Sun, being an extremely distant and powerful source, gives shadows that behave like projections.
(i) What is meant by the projection of an object on a plane?
(ii) The object is a cylinder held with its flat circular face facing the wall. What is the shape of the outline on the wall?
(iii) Why does the dark shape become larger when the torch is brought nearer to the object?
(iv) Why are the Sun's shadows on a plane practically the same as projections?
Answer:
(i) Drop a perpendicular from every point of the object to the plane; the feet of all these perpendiculars together form the projection of the object on that plane.
(ii) A circle, the same size as the circular face of the cylinder.
(iii) The torch is a nearby source, so its rays spread out. When it is brought nearer, the rays passing the edges of the object spread wider before reaching the wall, so the dark shape is scaled up.
(iv) The Sun is so far away that its rays reaching an object are practically parallel. Parallel rays falling perpendicular to a plane give a shadow of the same size and shape as the projection.
In simple words: A projection uses perpendiculars; a near torch enlarges the shadow, while far-away sunlight gives shadows like projections.

Teacher's Note:
1. Spreading rays enlarge a shadow; parallel rays do not.
2. The circular face facing the wall projects to a circle of the same size.

Question 15. A courtyard tiled like a carpet: A hostel courtyard is a square of side 2.7 m. The mason first divides it into 9 equal squares and leaves the middle one open as a small drain pit. He then divides each of the 8 remaining squares into 9 equal parts and again leaves the middle part of each one open. He stops here. The open parts are filled with gravel and the rest is covered with square stone tiles, all of the same size. The finished floor looks like the figure below, which is Step 2 of the Sierpinski Carpet.
(i) Using the rule \( R_n = 8^n \), how many stone tiles are there in the finished floor?
(ii) How many open gravel patches are there altogether?
(iii) What is the side of one stone tile?
(iv) What fraction of the courtyard area is covered by stone tiles?
Answer:
(i) The floor is at Step 2, so the number of tiles is \( R_2 = 8^2 = \) 64.
(ii) \( H_2 = H_1 + R_1 = 1 + 8 = \) 9 gravel patches: one large central patch and 8 smaller ones.
(iii) The side is divided into 3 at each step, so after two steps it is divided into 3 × 3 = 9 equal parts. Side of one tile = 2.7 m ÷ 9 = 0.3 m = 30 cm.
(iv) The courtyard is 9 × 9 = 81 small squares, of which 64 are tiled. Tiled fraction = \( \frac{64}{81} \) (about 79%), which is also \( \left(\frac{8}{9}\right)^{2} \).
In simple words: The Step 2 floor has 64 tiles of side 30 cm and 9 gravel patches, and tiles cover 64/81 of the area.

Teacher's Note:
1. Count the patches in the figure to check the rule.
2. Convert metres to centimetres for the tile side.

Question 16. A block model on a table: In a toy shop, a model is built on a table from identical wooden cubes glued face to face. It is shown below, with the three viewing directions marked. The model is solid: there is no gap hidden anywhere inside it, and no cube is floating in the air.
(i) How many cubes have been used in the model?
(ii) Describe the top view of the model.
(iii) Describe the front view of the model.
(iv) How many more cubes must be glued on to complete a full cuboid that is 2 cubes long, 2 cubes deep and 3 cubes high?
Answer:
(i) The bottom layer is a full 2 × 2 block of 4 cubes (one of them is hidden behind the others), the middle layer has 2 cubes and the top layer has 1 cube. Total = 4 + 2 + 1 = 7 cubes.
(ii) Top view: a 2 × 2 square of 4 unit squares, because the bottom layer covers a full 2 × 2 area.
(iii) Front view: an L-shaped figure of 4 unit squares: a column of 3 squares with 1 more square beside the bottom one.
(iv) A cuboid 2 long, 2 deep and 3 high needs 2 × 2 × 3 = 12 cubes. Since 7 are already used, 12 − 7 = 5 more cubes are needed.
In simple words: The model has 7 cubes (4 + 2 + 1); its top view is a 2 × 2 square, its front view an L of 4 squares, and 5 more cubes make the full cuboid.

Teacher's Note:
1. “No hidden gap” tells us the hidden bottom cube is there.
2. Count layer by layer from the bottom.

Question 17. A snowflake border on cloth: A designer prints a decorative border on a shawl. She starts with an equilateral triangle of side 27 cm. On every straight edge of the figure she divides the edge into three equal parts, raises a small equilateral triangle on the middle part and then rubs out that middle part, so that each edge becomes a bump. She repeats the same operation on every edge of the new figure. After two rounds the outline is a Koch-type snowflake, and the printing machine has to trace a far longer path than the original triangle.
(i) What is the length of one small segment after Step 1 and after Step 2?
(ii) How many sides does the figure have at Step 2?
(iii) Find the perimeter of the figure at Step 2.
(iv) By what factor does the perimeter get multiplied at each step? Give a reason.
Answer:
(i) Each step divides every segment into 3 equal parts. After Step 1 a segment is 27 ÷ 3 = 9 cm, and after Step 2 it is 9 ÷ 3 = 3 cm.
(ii) Each side is replaced by 4 sides, so the count is 3 × 4ⁿ. At Step 2 it is 3 × 4² = 3 × 16 = 48 sides.
(iii) Perimeter at Step 2 = 48 × 3 cm = 144 cm. (Check: starting perimeter = 3 × 27 = 81 cm; 81 × \( \frac{4}{3} \) = 108 cm and 108 × \( \frac{4}{3} \) = 144 cm.)
(iv) By the factor \( \frac{4}{3} \). A side of length s is replaced by 4 segments of length \( \frac{s}{3} \), so the new length in its place is \( \frac{4s}{3} \). This happens to every side, so the whole perimeter is multiplied by \( \frac{4}{3} \).
In simple words: Segments become 9 cm then 3 cm, there are 48 sides, the perimeter is 144 cm, and it grows by 4/3 each step.

Teacher's Note:
1. Number of sides × length of a side gives the perimeter.
2. Check the answer a second way using the 4/3 factor.

Question 18. Drawing a mobile-tower lattice: An engineer has to send a drawing of a steel lattice for a mobile tower to a workshop in another town. A single photograph is not accepted, because two very different frames can look alike from one side. So she prepares a drawing of the frame as seen from the front, one as seen from above and one as seen from the side. She also makes a picture of the whole frame on isometric grid paper, on which the vertical lines of the grid carry the height of the frame and the two slanting families of lines carry its length and its depth.
(i) Why are three views drawn instead of one?
(ii) What does the word isometric mean, and why is an isometric grid convenient for this drawing?
(iii) Which direction on the isometric paper carries the height of the solid, and which directions carry its length and depth?
(iv) Which property of projections makes sure that parallel steel rods of the tower are drawn as parallel lines?
Answer:
(i) A single projection loses all information along the viewing direction, so two quite different frames can give the same picture. The front, top and side views are mutually perpendicular and fix the shape in all three directions, so the workshop can build exactly the right frame.
(ii) Isometric means “equal measure” in Greek. On an isometric drawing, unit lengths along all three axes of the solid are drawn equal, so the engineer can count grid units and read length, depth and height directly from one drawing.
(iii) The vertical lines of the grid carry the height. The two slanting families of lines, / and ∖, carry the length and the depth.
(iv) Projection carries parallel lines to parallel lines. So the three families of parallel rods in the tower appear as three families of parallel lines on paper, which makes the drawing easy to read.
In simple words: Three views fix the frame; isometric means equal measure; vertical lines show height, slanting lines show length and depth; projections keep parallel lines parallel.

Teacher's Note:
1. Equal unit lengths on all three axes is what makes counting possible.
2. Parallel rods stay parallel in every projection.

Question 19. An ant on a packing carton: A closed cardboard carton in a godown is a cuboid measuring 16 cm × 8 cm × 4 cm. An ant sits at one corner of the carton and a grain of sugar is stuck at the corner diagonally opposite to it. The ant can only walk along the surface of the carton; it cannot fly across and it cannot go inside. The storekeeper's son flattens a spare carton into its net and finds that each path drawn on the carton becomes a path of exactly the same length on the net — so the shortest path on the carton can be found by drawing a straight line on a suitable net.
(i) Why does opening out the carton into a net help, and which theorem is then used?
(ii) One unfolding places the ant and the grain at opposite corners of a rectangle measuring 16 cm by 12 cm. Find the straight-line distance between them in this unfolding.
(iii) Another unfolding gives a rectangle measuring 20 cm by 8 cm. Find the distance in this unfolding, correct to one decimal place.
(iv) State the length of the shortest path, and say what this comparison shows about the way a cuboid is unfolded.
Answer:
(i) A path on the carton becomes a path of exactly the same length on the net, and the other way round. So the problem becomes “find the shortest path between two points on a flat sheet”, which is the straight line joining them. Its length is found with the Baudhayana (Pythagoras) theorem, since the two points are opposite corners of a rectangle.
(ii) d² = 16² + 12² = 256 + 144 = 400, so d = √400 = 20 cm.
(iii) d² = 20² + 8² = 400 + 64 = 464, so d = √464 ≈ 21.5 cm.
(iv) The shortest path is 20 cm, from the first unfolding. (A third unfolding, a 24 cm by 4 cm rectangle, gives √592 ≈ 24.3 cm, which is longer still.) This shows that the way a cuboid is unfolded matters: we must try all the unfoldings and choose the smallest distance.
In simple words: On the net the path is a straight line; the unfoldings give 20 cm, 21.5 cm and 24.3 cm, so the shortest path is 20 cm.

Teacher's Note:
1. The rectangle sides come from adding two edges of the cuboid: 8 + 4 = 12, 16 + 4 = 20, 16 + 8 = 24.
2. Always compare every unfolding before choosing.

Question 20. A fractal park: A town plans a square park of side 81 m, entirely under grass to begin with. In the first stage the park is divided into 9 equal squares and the central square is dug up and made into a flower bed. In the second stage each of the 8 remaining grassy squares is divided into 9 equal parts and the central part of each is made into a flower bed. The committee wants to know how much of the park is still grass, because the watering contract is charged on the grassed area alone. The same rule can be repeated for as many stages as the town wishes.
(i) Find the area of the flower bed made in the first stage.
(ii) How many flower beds are added in the second stage, what is the side of each, and what is their total area?
(iii) Find the total flower-bed area after the second stage, and express it as a fraction of the whole park.
(iv) Write the fraction of the park that is still grass after \( n \) stages, and evaluate it for \( n = 3 \).
Answer:
(i) Each of the 9 equal squares has side 81 ÷ 3 = 27 m. Area of the first flower bed = 27 × 27 = 729 m².
(ii) 8 flower beds are added, one in each grassy square. Each has side 27 ÷ 3 = 9 m, so each has area 81 m², and their total area = 8 × 81 = 648 m².
(iii) Total flower-bed area = 729 + 648 = 1377 m². The whole park is 81 × 81 = 6561 m², so the fraction is \( \frac{1377}{6561} \) = \( \frac{17}{81} \).
(iv) At every stage \( \frac{8}{9} \) of the grass survives, so after n stages the grassed fraction is \( \left(\frac{8}{9}\right)^{n} \).
For n = 3: \( \left(\frac{8}{9}\right)^{3} \) = \( \frac{512}{729} \) ≈ 0.70, so about 70% of the park is still grass.
In simple words: The flower beds take 729 + 648 = 1377 m² (17/81 of the park), and after n stages (8/9)ⁿ of the park is grass, about 70% for n = 3.

Teacher's Note:
1. Check: grass after 2 stages is 64/81, and 64/81 + 17/81 = 1.
2. Each new bed is one-ninth of the square it sits in.