✅ SOLUTIONS
Class 5
Chapter 7: Shapes and Patterns
Math Mela · NCERT Solutions
Class 5Math Mela (NCERT)Chapter 7
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🧺 Weaving Mats
Describe the weaving patterns shown (Q2, Q3, Q4).
Solution
Q2 pattern: Row 1 — 2 over, 1 under, 2 over, 1 under... (repeat) | Row 2 — 2 under, 1 over, 2 under, 1 over... (repeat)
Q3 pattern (4 rows):
- Row 1 — 2 over, 1 under, 2 over, 1 under... (repeat)
- Row 2 — 1 under (no repeat), then 3 over, 3 under, 3 over... (repeat)
- Row 3 — 2 under, 1 over, 2 under... (repeat)
- Row 4 — 1 over (no repeat), then 3 under, 3 over... (repeat)
💡 To describe ANY weave pattern: write down the over/under sequence for one row until it starts repeating — that repeating chunk is the "pattern unit"!
🔷 Tiling and Tessellation
Do regular triangles, squares, pentagons, hexagons, octagons tessellate?
Solution
| Shape | Corner angle | Number that fit around a point | Tessellates? |
|---|---|---|---|
| Equilateral triangle | 60° | 6 (360÷60=6) | ✅ Yes |
| Square | 90° | 4 (360÷90=4) | ✅ Yes |
| Regular pentagon | 108° | Only 3 fit (324°), leaves a 36° gap | ❌ No |
| Regular hexagon | 120° | 3 (360÷120=3) | ✅ Yes |
| Regular octagon | 135° | Only 2 fit fully (270°); a 3rd leaves a gap/overlap | ❌ No |
Five squares CANNOT fit around a point: 5×90°=450°, which is MORE than 360° — they would overlap!
Tessellation with squares & triangles — are the triangles equilateral?
Solution
Yes, the triangles in a square-triangle tiling are typically equilateral (all sides and angles equal) — this is one of the well-known "semi-regular" tessellation patterns, combining two different regular shapes that still fit perfectly with no gaps around each point.
🔺 Triangles from a Rhombus
Q1. How many different types of triangles from the rhombus pieces?
Solution
A rhombus divided into 4 triangles (by its diagonals) gives triangles with 2 equal sides each — these are isosceles triangles. Folding one in half shows 2 equal angles too (matching the 2 equal sides)!
Q2 & Q3. Can you make equilateral or scalene triangles from the rhombus pieces?
Solution
Q2 — Equilateral: Generally NO with the standard rhombus triangle pieces — since all 4 pieces are identical isosceles triangles (with a specific fixed angle from the rhombus, usually not 60°), simply combining pieces from a non-special rhombus won't naturally produce a triangle with all 3 sides equal.
Q3 — Scalene (all sides different):NO — since every piece from the rhombus is an isosceles triangle with 2 equal sides, any triangle you form by combining them will still have at least 2 matching sides somewhere, so a truly scalene triangle isn't possible this way.
Try This — Cut an equilateral triangle in half. Sides equal? Scalene angles?
Solution
Cutting an equilateral triangle exactly in half (through one vertex to the midpoint of the opposite side) gives 2 right-angled triangles, each with NO two sides equal — these are scalene right triangles (unless cut in a special symmetric way).
In scalene triangles, generally NO two angles are equal — matching the fact that no two sides are equal either (equal sides ↔ equal opposite angles).
🔶 Quadrilaterals & Tangram
Q5. Measure sides of quadrilaterals A and B. What do you notice?
Solution
In BOTH shapes A and B, it is the OPPOSITE sides that are equal (not the adjacent ones) — this is what makes them parallelograms, unlike the kite where ADJACENT sides were equal.
Shape A: opposite angles are equal (a general parallelogram).
Shape B: ALL angles are equal AND are right angles — making it a rectangle (a special parallelogram)!
Shape B: ALL angles are equal AND are right angles — making it a rectangle (a special parallelogram)!
Q7. Using 3 rhombus-triangles, make 3-sided, 4-sided, 5-sided shapes.
Solution — Method
- 3-sided (triangle): Join all 3 pieces so their equal sides align, forming ONE bigger triangle.
- 4-sided (quadrilateral): Arrange the 3 pieces so one outer edge is "opened up," forming a 4-sided shape like a kite or trapezium.
- 5-sided (pentagon): Arrange the pieces in a zig-zag/fan pattern so 5 distinct outer edges are formed.
Q8. Which shapes can be made with all 4 rhombus pieces?
Solution
Using all 4 triangular pieces, you can reconstruct the original Rhombus shape itself (which has 4 sides, so it fits under "quadrilateral" options). You may also be able to form other quadrilaterals like a rectangle or parallelogram, depending on how the pieces are arranged — but forming a triangle, pentagon, hexagon, or octagon with exactly these 4 isosceles pieces is generally NOT possible, since their fixed angles don't combine evenly into those shapes.
❓ Which Shape Am I?
Match each statement to its shape.
Solution
| # | Statement | Shape |
|---|---|---|
| 1 | Right angles, sides not all equal | Rectangle |
| 2 | All sides equal, angles not all equal | Rhombus |
| 3 | Opposite angles equal, no right angles | Parallelogram |
| 4 | Two pairs of equal sides, no right angles | Kite |
| 5 | All sides equal AND meet at right angles | Square |
| 6 | Opposite angles equal, opposite sides equal | Parallelogram |
| 7 | Opposite angles equal, sides make right angles | Rectangle |
Notice statements 3 & 6 BOTH describe a Parallelogram, and statements 1 & 7 BOTH describe a Rectangle — some shapes are described more than once, just in different words!
🪁⭕ Kites & Circles
What shapes do you see in the folded paper kite?
Solution
You can see triangles (from the folded corners) and the overall kite shape (quadrilateral) itself, made up of those smaller triangles joined together.
Join the 4 endpoints of two diameters. What shape forms?
Solution
Joining the 4 endpoints of any two diameters ALWAYS forms a rectangle! This is because each diameter is equal in length (both = 2×radius) and they cross at the exact centre, so opposite sides of the resulting quadrilateral are always equal, and all four angles turn out to be 90°.
A 4-sided shape OTHER than a rectangle is not possible this way — trying with different diameter pairs always gives a rectangle (sometimes a special case — a square — if the diameters are positioned just right, e.g., perpendicular to each other).
🧊 Cube Connections
Q2. How many small cubes were removed from each big cube frame?
Solution
Each "frame" is made by drilling straight 1-cube-wide tunnels through the centre of each face, all meeting at the middle:
| Cube | Size | Method | Cubes removed |
|---|---|---|---|
| (a) | 3×3×3 (27 cubes total) | Centre cube + 6 face-centre cubes = 1+6 | 7 cubes |
| (b) | 4×4×4 (64 cubes total) | 3 tunnels of 4 cubes each, sharing overlaps at the centre | 10 cubes |
| (c) | 5×5×5 (125 cubes total) | 3 tunnels of 5 cubes each, sharing overlaps at the centre | 13 cubes |
💡 Method: For a straight 1-wide tunnel through the centre in all 3 directions (like a 3D "plus sign"), the formula is: 3×(cube size) − 3 + 1 = 3n−2. Check: for n=3: 3(3)−2=7 ✓
Q3. Nisha's 27-cube (3×3×3) painted cube — how many small cubes have 3, 2, 1, 0 faces painted?
Solution
| Position | Count | Reasoning |
|---|---|---|
| (a) 3 faces painted | 8 | The 8 corner cubes (every cube always has 8 corners) |
| (b) 2 faces painted | 12 | Edge cubes (not corners): 12 edges × 1 middle cube per edge |
| (c) 1 face painted | 6 | Face-centre cubes: 6 faces × 1 centre cube per face |
| (d) 0 faces painted | 1 | The single cube hidden right in the very centre |
Check: 8+12+6+1 = 27 ✓ (matches the total number of small cubes!)
🧩 Puzzle — Tanu's Shapes
Find the arrangement of 7 shapes using the 5 clues.
Solution
Circle → Square → Rectangle → Triangle → Square → Triangle → Hexagon
| Clue | Check |
|---|---|
| (a) Square between circle & rectangle | Circle-Square-Rectangle (positions 1,2,3) ✓ |
| (b) Rectangle between square & triangle | Square-Rectangle-Triangle (positions 2,3,4) ✓ |
| (c) Both triangles next to a square | Triangle(4)-Square(5)-Triangle(6) ✓ |
| (d) Hexagon right of the triangle | Hexagon(7) is right of Triangle(6) ✓ |
| (e) Circle left of the square | Circle(1) is left of Square(2) ✓ |
✅ All 5 clues check out perfectly with this arrangement!
🔷 Icosahedron & Dodecahedron
Answer the questions about these two solids.
Solution
| Question | Icosahedron | Dodecahedron |
|---|---|---|
| Shape of faces | Triangles | Pentagons |
| All faces same? | Yes | Yes |
| Faces meeting at a vertex | 5 | 3 |
| Same number at each vertex? | Yes | Yes |
| Number of faces | 20 ("icosa"=20) | 12 ("dodeca"=12) |
| Number of edges | 30 | 30 |
| Number of vertices (corners) | 12 | 20 |
💡 Counting edges without missing any: Count edges face-by-face (each triangle has 3 edges, each pentagon has 5), multiply by the number of faces, then DIVIDE BY 2 (since every edge is shared by exactly 2 faces). Icosahedron: 20×3÷2=30 ✓. Dodecahedron: 12×5÷2=30 ✓
Both shapes have the SAME number of edges (30) — a neat coincidence that connects to Euler's Formula: Vertices − Edges + Faces = 2 for any solid! (Icosahedron: 12−30+20=2 ✓; Dodecahedron: 20−30+12=2 ✓)