How to Use This Page
Work through each exercise in the Samacheer Kalvi textbook on your own first, then use the solutions below to verify your steps and answers. All questions from Chapter 6 (Information Processing — Term 1) are solved with full working. If your answer differs, trace back through the method to find where you went wrong.
Key Concepts (Quick Recap)
- Tetrominoes: Shapes made of 4 equal squares joined edge-to-edge. There are exactly 5 free tetrominoes — I-shape, L-shape, O-shape, T-shape, and S/Z-shape.
- Tiling with tetrominoes: A region of area N cells can be tiled by tetrominoes only if N is a multiple of 4. Each tetromino covers exactly 4 cells.
- Shortest route / network problems: List every possible path from start to end, add the distances along each path, and pick the minimum.
- Line of symmetry: A line that divides a figure into two mirror-image halves. Regular polygons with n sides have n lines of symmetry.
- Rotational symmetry: A figure has rotational symmetry of order n if it looks the same after a rotation of 360°/n. The number of times it matches itself in one full turn is its order.
- Pascal's triangle: A triangular array where each entry is the sum of the two entries above it. Row sums are powers of 2: 1, 2, 4, 8, 16, 32, 64, ...
Exercise 6.1 — Tetrominoes
Q1. Form a 5 × 4 square and shade it completely using tetromino shapes.
A 5 × 4 rectangle has a total of 5 × 4 = 20 cells.
Each tetromino covers exactly 4 cells. Since 20 ÷ 4 = 5, we need exactly 5 tetrominoes to fill the rectangle.
The five free tetromino shapes are:
1. I-tetromino — four squares in a straight vertical or horizontal line 2. L-tetromino — three squares in a line with one square attached at one end 3. O-tetromino — a 2 × 2 square block 4. T-tetromino — three squares in a line with one square attached at the centre 5. S/Z-tetromino — a zigzag of four squares
Since the five tetrominoes together cover 5 × 4 = 20 cells (which matches the grid area), they can tile the 5 × 4 rectangle exactly once each.
Q2. Shade a given figure using the five tetrominoes exactly once.
Step 1: Count the total number of cells in the given figure. Since each tetromino covers 4 cells, the figure must contain exactly 5 × 4 = 20 cells for all five tetrominoes to fit without overlap or gaps.
Step 2: Place the O-tetromino (2 × 2 block) first — it is the most restrictive shape because it requires a 2 × 2 space.
Step 3: Place the T-tetromino and L-tetromino next, as they have fewer orientations that fit tight corners.
Step 4: Fill the remaining cells with the I-tetromino and S-tetromino.
Step 5: Verify that all 20 cells are covered and no two tetrominoes overlap.
Exercise 6.2 — Routes and Networks
Q1. Find the shortest distance from school to playground and list all possible routes.
List every possible route from School to Playground, then calculate the total distance for each.
| Route | Path | Distance Calculation | Total Distance |
|---|---|---|---|
| 1 | School → Swimming Pool → Canteen → Playground | 9 + 2 + 10 | 21 km |
| 2 | School → Canteen → Playground | 10 + 10 | 20 km ✓ |
| 3 | School → Garden → Playground | 2 + 20 | 22 km |
| 4 | School → Garden → Teashop → Playground | 2 + 17 + 2 | 21 km |
| 5 | School → Park → Teashop → Playground | 12 + 12 + 2 | 26 km |
Comparing all five routes:
- Route 2 = 20 km (minimum)
- Routes 1 and 4 = 21 km
- Route 3 = 22 km
- Route 5 = 26 km
Q2. A postman needs to deliver letters to 5 streets. Find the shortest route starting and ending at the post office.
This is a variation of the Travelling Salesman Problem. The method is:
Step 1: Label the post office as P and the 5 streets as A, B, C, D, E.
Step 2: List all possible routes that start at P, visit each street exactly once, and return to P. The number of possible routes is (5 − 1)! = 4! = 24 routes.
Step 3: Calculate the total distance for each route by adding the distances between consecutive stops.
Step 4: Compare all 24 total distances and select the minimum.
For example, if the distances are:
- P to A = 3 km, P to B = 5 km, P to C = 4 km, P to D = 6 km, P to E = 2 km
- A to B = 2 km, A to C = 4 km, A to D = 5 km, A to E = 3 km
- B to C = 1 km, B to D = 3 km, B to E = 4 km
- C to D = 2 km, C to E = 3 km
- D to E = 1 km
Sample route: P → A → B → C → D → E → P = 3 + 2 + 1 + 2 + 1 + 2 = 11 km
Check remaining routes similarly and pick the minimum total.
Exercise 6.3 — Symmetry
Q1. Draw the lines of symmetry for the following figures:
(i) Equilateral triangle
An equilateral triangle has all three sides equal and all three angles equal (60° each).
A line of symmetry passes from each vertex to the midpoint of the opposite side.
- Line 1: From vertex A to midpoint of BC
- Line 2: From vertex B to midpoint of AC
- Line 3: From vertex C to midpoint of AB
(ii) Square
A square has four equal sides and four right angles.
- Line 1: Vertical — through the midpoints of the left and right sides
- Line 2: Horizontal — through the midpoints of the top and bottom sides
- Line 3: Diagonal — from top-left to bottom-right
- Line 4: Diagonal — from top-right to bottom-left
(iii) Regular pentagon
A regular pentagon has five equal sides and five equal angles (108° each).
A line of symmetry passes from each vertex to the midpoint of the opposite side.
- Line 1: From vertex 1 to midpoint of the opposite side
- Line 2: From vertex 2 to midpoint of the opposite side
- Line 3: From vertex 3 to midpoint of the opposite side
- Line 4: From vertex 4 to midpoint of the opposite side
- Line 5: From vertex 5 to midpoint of the opposite side
(iv) Circle
A circle has infinitely many lines of symmetry. Every diameter of the circle is a line of symmetry, because a circle looks identical after reflection across any diameter.
Q2. How many lines of symmetry does a regular polygon with n sides have?
In a regular polygon, all sides are equal and all angles are equal. Each line of symmetry passes through one vertex and the midpoint of the opposite side (or the opposite vertex, if n is even).
For a regular polygon with n sides:
- There are n vertices
- Each vertex gives rise to one line of symmetry
- Therefore, there are exactly n lines of symmetry
Q3. Find the order of rotational symmetry for:
(i) Equilateral triangle
An equilateral triangle looks the same after every rotation of 360°/3 = 120°.
In one full rotation of 360°, it matches itself 3 times:
- At 0° (original position)
- At 120°
- At 240°
(ii) Square
A square looks the same after every rotation of 360°/4 = 90°.
In one full rotation of 360°, it matches itself 4 times:
- At 0° (original position)
- At 90°
- At 180°
- At 270°
(iii) Regular hexagon
A regular hexagon looks the same after every rotation of 360°/6 = 60°.
In one full rotation of 360°, it matches itself 6 times:
- At 0° (original position)
- At 60°
- At 120°
- At 180°
- At 240°
- At 300°
Exercise 6.4 — Pascal's Triangle and Patterns
Q1. Complete the next row of Pascal's triangle.
Given rows:
- Row 1: 1
- Row 2: 1, 1
- Row 3: 1, 2, 1
- Row 4: 1, 3, 3, 1
- Row 5: 1, 4, 6, 4, 1
- Row 6: 1, 5, 10, 10, 5, 1
Rule: Each number is the sum of the two numbers directly above it. The first and last numbers in every row are 1.
Row 7 is built from Row 6 (1, 5, 10, 10, 5, 1):
- 1 (first number is always 1)
- 1 + 5 = 6
- 5 + 10 = 15
- 10 + 10 = 20
- 10 + 5 = 15
- 5 + 1 = 6
- 1 (last number is always 1)
Q2. Find the sum of each row of Pascal's triangle and identify the pattern.
Calculate the sum of each row:
| Row | Entries | Sum |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 1, 1 | 1 + 1 = 2 |
| 3 | 1, 2, 1 | 1 + 2 + 1 = 4 |
| 4 | 1, 3, 3, 1 | 1 + 3 + 3 + 1 = 8 |
| 5 | 1, 4, 6, 4, 1 | 1 + 4 + 6 + 4 + 1 = 16 |
| 6 | 1, 5, 10, 10, 5, 1 | 1 + 5 + 10 + 10 + 5 + 1 = 32 |
| 7 | 1, 6, 15, 20, 15, 6, 1 | 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 |
The sums are: 1, 2, 4, 8, 16, 32, 64, ...
These are powers of 2: 2⁰, 2¹, 2², 2³, 2⁴, 2⁵, 2⁶, ...
So the sum of Row n = 2^(n−1)
Q3. Identify the pattern and find the next two numbers in each sequence:
(i) 2, 4, 6, 8, 10, ...
Observe the pattern:
- 4 − 2 = 2
- 6 − 4 = 2
- 8 − 6 = 2
- 10 − 8 = 2
The common difference is 2. This is an arithmetic sequence with first term 2 and common difference 2.
Next term: 10 + 2 = 12 Following term: 12 + 2 = 14
(ii) 1, 4, 9, 16, 25, ...
Observe the pattern:
- 1 = 1²
- 4 = 2²
- 9 = 3²
- 16 = 4²
- 25 = 5²
These are perfect squares. The nth term is n².
Next term: 6² = 36 Following term: 7² = 49
(iii) 1, 1, 2, 3, 5, 8, 13, ...
This is the Fibonacci sequence. Each term is the sum of the two preceding terms:
- 1 + 1 = 2
- 1 + 2 = 3
- 2 + 3 = 5
- 3 + 5 = 8
- 5 + 8 = 13
Next term: 8 + 13 = 21 Following term: 13 + 21 = 34
Multiple Choice Questions (MCQs)
Q1. How many lines of symmetry does an equilateral triangle have?
(a) 1 (b) 2 (c) 3 (d) 4
An equilateral triangle has 3 equal sides and 3 equal angles. Each vertex is paired with the midpoint of the opposite side to form a line of symmetry.
Q2. The shortest path between two points is:
(a) Curved (b) Zigzag (c) Straight line (d) L-shaped
By definition, the shortest distance between any two points is the straight line segment connecting them.
Q3. How many tetromino shapes are there (free tetrominoes)?
(a) 3 (b) 4 (c) 5 (d) 6
The five free tetromino shapes are: I, L, O, T, and S (or Z). These are the only distinct shapes that can be formed from 4 squares joined edge-to-edge, counting rotations and reflections as the same shape.
Q4. The order of rotational symmetry of a square is:
(a) 1 (b) 2 (c) 3 (d) 4
A square matches itself after rotations of 90°, 180°, 270°, and 360° — a total of 4 times in one full turn.
Q5. The sum of the 6th row of Pascal's triangle is:
(a) 16 (b) 32 (c) 64 (d) 128
Sum of Row n = 2^(n−1). For n = 6: 2^(6−1) = 2⁵ = 32.
True or False
1. An equilateral triangle has 3 lines of symmetry.
2. A circle has only one line of symmetry.
3. The sum of each row in Pascal's triangle is a power of 2.
4. A square has 2 lines of symmetry.
5. The Fibonacci sequence starts with 0, 1, 1, 2, 3, 5, ...
Fill in the Blanks
1. An equilateral triangle has ___ lines of symmetry.
2. The sum of the 5th row of Pascal's triangle is ___.
3. A 5 × 4 rectangle can be tiled using ___ tetrominoes.
4. The order of rotational symmetry of a square is ___.
5. In Pascal's triangle, each number is the sum of the two numbers ___ it.
Common Mistakes
1. Confusing lines of symmetry with order of rotational symmetry — Lines of symmetry are mirror lines; rotational symmetry order is the number of times a shape matches itself in a full 360° turn. For example, an equilateral triangle has 3 lines of symmetry AND rotational symmetry of order 3, but these are different concepts.
2. Missing routes when listing all possible paths in a network — Always systematically list every possible path before calculating distances. It is easy to overlook a route, especially in networks with many nodes.
3. Forgetting that each tetromino covers exactly 4 cells — When tiling, the total area must be a multiple of 4. If it is not, tiling with tetrominoes alone is impossible.
4. Counting overlapping tetromino placements — Each cell must be covered by exactly one tetromino. Overlapping placements are invalid.
5. Forgetting that the sum of each row in Pascal's triangle follows 2^(n−1) — This is a consistent pattern. Row 1 sums to 1, Row 2 to 2, Row 3 to 4, and so on, always doubling.
6. Mixing up rotational symmetry order with lines of symmetry — Some figures have different counts. A parallelogram (non-rectangle) has rotational symmetry of order 2 but zero lines of symmetry.
7. Not checking all possible routes before declaring the shortest path — You must verify every route's total distance. Declaring the shortest route without enumerating all possibilities can lead to wrong answers.
Exam Checklist
- [ ] Tetromino shapes: can you name and draw all 5 free tetrominoes?
- [ ] Tiling problems: verify that the grid area is a multiple of 4 before attempting to tile with tetrominoes
- [ ] Finding shortest routes in networks: list all routes, calculate each total, then pick the minimum
- [ ] Lines of symmetry for regular polygons: n-sided regular polygon has n lines of symmetry
- [ ] Order of rotational symmetry: equilateral triangle = 3, square = 4, regular hexagon = 6
- [ ] Pascal's triangle: each entry = sum of two entries above; row sums = powers of 2
- [ ] Number patterns: arithmetic sequences, geometric sequences, Fibonacci sequence, perfect squares, triangular numbers
Internal Links
- [Term 1 Number System Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions/term-1-number-system-exercise-solutions)
- [Term 1 Geometry Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions/term-1-geometry-exercise-solutions)
- [All Class 7 Maths Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions)
- [Class 7 Maths Subject Guide](/school/tamil-nadu-state-board/class/7/maths)