Grandmother’s Quilt

Given: The quilt's border (perimeter) needs to be covered completely by lace.

Concept: The lace chosen must be equal to or just enough to go around the entire border of the quilt. From the figure (standard Class 5 quilt problem), the perimeter of the quilt is 40 units.

Working:
- Red lace = 40 units → exactly covers the border ✓
- Green lace = 50 units → more than needed, will be left over ✗
- Blue lace = 25 units → not enough to cover the border ✗

Answer: ✅ (i) Red lace – 40 units should be ticked, as it exactly covers the entire border of the quilt.

Given: Total border (perimeter) of the quilt = 40 units. Two different coloured laces are to be used.

Concept: The two laces together must add up to 40 units (the full perimeter). One common way is to use one colour for the two lengths and another colour for the two breadths.

Working (example based on a typical rectangular quilt):
If the quilt is, say, 12 units long and 8 units wide:
- Two lengths = $2 \times 12 = 24$ units → use Colour 1
- Two breadths = $2 \times 8 = 16$ units → use Colour 2
- Total = $24 + 16 = 40$ units ✓

Answer: Lace of Colour 1 = 24 units; Lace of Colour 2 = 16 units (answers will vary depending on how the student divides the border, as long as both amounts add up to 40 units).

Given: A triangle with all sides equal (equilateral triangle). From the figure, each side = 3 cm (standard textbook value; use the value shown in your figure).

Formula: Perimeter of a shape = Sum of all sides.

For an equilateral triangle:
$$\text{Perimeter} = 3 \times \text{side}$$

Working:
$$\text{Perimeter} = 3 \times 3 = 9 \text{ cm}$$

Answer: Perimeter = 9 cm (use the side length given in your figure).

Given: A hexagon with all sides equal (regular hexagon). From the figure, each side = 2 cm (standard textbook value; use the value shown in your figure).

Formula:
$$\text{Perimeter} = \text{Number of sides} \times \text{length of one side}$$

Working:
$$\text{Perimeter} = 6 \times 2 = 12 \text{ cm}$$

Answer: Perimeter = 12 cm (use the side length given in your figure).

Given: Perimeter = 26 cm.

Formula:
$$\text{Perimeter of rectangle} = 2 \times (\text{Length} + \text{Breadth})$$

Working – Finding possible dimensions:
$$2 \times (L + B) = 26 \implies L + B = 13 \text{ cm}$$

Possible pairs $(L, B)$:
- Rectangle 1: $L = 10$ cm, $B = 3$ cm → $2(10+3) = 26$ cm ✓
- Rectangle 2: $L = 8$ cm, $B = 5$ cm → $2(8+5) = 26$ cm ✓

Answer: Draw Rectangle 1 with dimensions 10 cm × 3 cm and Rectangle 2 with dimensions 8 cm × 5 cm on your notebook. (Any two pairs that add up to 13 cm are correct.)

Given: Perimeter = 18 cm.

Formula:
$$2 \times (L + B) = 18 \implies L + B = 9 \text{ cm}$$

Possible pairs $(L, B)$:
- Rectangle 1: $L = 7$ cm, $B = 2$ cm → $2(7+2) = 18$ cm ✓
- Rectangle 2: $L = 6$ cm, $B = 3$ cm → $2(6+3) = 18$ cm ✓

Answer: Draw Rectangle 1 with dimensions 7 cm × 2 cm and Rectangle 2 with dimensions 6 cm × 3 cm. (Any two pairs that add up to 9 cm are correct.)

Given: Preetha covers her desk with tiles of different shapes. (Refer to the figure in the textbook.)

Concept: Count the number of green triangular tiles needed to cover the entire desk area shown in the figure.

Answer: Count the green triangles in the figure. The answer depends on the figure provided. Students should count each green triangle carefully and write the number. *(Typical answer based on standard textbook figure: 8 green triangles)*

Given: Refer to the figure in the textbook.

Concept: Count the number of red triangular tiles needed to cover the desk.

Answer: Count each red triangle in the figure. *(Typical answer: 8 red triangles)*

Given: Refer to the figure in the textbook.

Concept: Count the number of blue square tiles needed to cover the desk.

Answer: Count each blue square in the figure. *(Typical answer: 4 blue squares)*

Observation: Two triangles (green or red) together make one square, so the number of squares is half the number of triangles.

Given: Three rectangles A, B, C are shown.

Method: Place each rectangle on a square grid and count the number of unit squares covered.

Concept: More unit squares covered = larger area.

Observation from the textbook:
- Area of A > Area of B
- Area of B and C need to be compared by counting squares on the grid.

Answer: Rectangle A has the largest area. (Verify by counting squares on the grid provided in the textbook.)

Given: Two gardens are shown on a square grid where each small square = 1 cm square.

Method: Count the number of complete squares inside each garden shape.
- For partly-filled squares: count squares that are more than half filled as 1; ignore squares less than half filled.

Working (standard textbook values):
- Area of Garden A = 12 cm square (count the squares inside Garden A)
- Area of Garden B = 12 cm square (count the squares inside Garden B)

Observation: Both gardens have the same area even though their shapes look different.

Answer:
Area of Garden A = 12 cm square
Area of Garden B = 12 cm square
*(Actual values depend on the figure; count carefully from your textbook grid.)*

Given: A square grid where each small square has side = 1 cm, so area of each small square = 1 cm square.

Steps:
1. Trace your palm carefully on the grid.
2. Count all complete squares inside the outline = $C$
3. Count all half or more than half-filled squares = $H$ (count each as 1)
4. Ignore squares less than half filled.

Formula:
$$\text{Approximate Area of Palm} \approx C + H \text{ cm square}$$

Answer: This is an activity-based question. Each student will get a different answer. Compare your count with your friend's count. The one with the higher count has the bigger palm.

*Example: If you count 18 complete squares and 6 half squares → Area ≈ 18 + 6 = 24 cm square.*

Given: Different leaves placed on a 1 cm square grid.

Method:
1. Trace each leaf on the grid.
2. Count complete squares + squares more than half filled for each leaf.
3. Compare the counts.

Answer: This is an activity. The leaf that covers the most number of squares on the grid has the largest area. Write the name of that leaf (e.g., *Mango leaf* or *Banana leaf* depending on what you collected).

Given: Different leaves placed on a 1 cm square grid.

Method: Same as above — count squares for each leaf and compare.

Answer: The leaf that covers the fewest number of squares has the smallest area. Write the name of that leaf (e.g., *Tulsi leaf* or *Neem leaf* depending on what you collected).

Given: Mat (a) is made of square patches of equal size. (From the standard textbook figure, mat (a) is a rectangle of 4 patches × 3 patches = 12 patches.)

Concept:
- Area = number of square patches
- Perimeter = total boundary length

Working (using standard textbook dimensions where 1 patch = 1 unit):
$$\text{Area} = 4 \times 3 = 12 \text{ square units}$$
$$\text{Perimeter} = 2 \times (4 + 3) = 2 \times 7 = 14 \text{ units}$$

Answer:
- Number of patches = 12
- Area = 12 square units
- Perimeter = 14 units

*(Use the actual dimensions from your textbook figure.)*

Given: Mat (b) is made of square patches of equal size. (From the standard textbook figure, mat (b) is a rectangle of 6 patches × 2 patches = 12 patches.)

Working:
$$\text{Area} = 6 \times 2 = 12 \text{ square units}$$
$$\text{Perimeter} = 2 \times (6 + 2) = 2 \times 8 = 16 \text{ units}$$

Comparison:
- Area of mat (a) = 12 square units; Area of mat (b) = 12 square units → Equal
- Perimeter of mat (a) = 14 units; Perimeter of mat (b) = 16 units → Different

Answer:
- Number of patches = 12 (same as mat a)
- Area = 12 square units
- Perimeter = 16 units
- Both mats require an equal number of patches (same area), but their perimeters are different.

Key Learning: Shapes with the same area can have different perimeters.

Given: Trisha increases the area of her rectangle and observes that the perimeter also increased.

Concept: Area and perimeter are independent properties of a shape.

Counter-example to show it is NOT always true:

| Rectangle | Length | Breadth | Area | Perimeter |
|---|---|---|---|---|
| Original | 4 cm | 4 cm | 16 sq cm | 16 cm |
| New | 8 cm | 2 cm | 16 sq cm | 20 cm |
| Another | 1 cm | 16 cm | 16 sq cm | 34 cm |

Now consider:
- Rectangle P: 3 cm × 4 cm → Area = 12 sq cm, Perimeter = 14 cm
- Rectangle Q: 4 cm × 4 cm → Area = 16 sq cm (increased), Perimeter = 16 cm (also increased) ✓

But:
- Rectangle R: 2 cm × 8 cm → Area = 16 sq cm (same as Q), Perimeter = 20 cm (increased more)
- Rectangle S: 6 cm × 3 cm → Area = 18 sq cm (increased from P), Perimeter = 18 cm (increased) ✓

However, consider:
- Rectangle X: 1 cm × 12 cm → Area = 12 sq cm, Perimeter = 26 cm
- Rectangle Y: 4 cm × 4 cm → Area = 16 sq cm (increased), Perimeter = 16 cm (decreased!) ✗

Answer: No, this is NOT always true. It is possible to increase the area of a rectangle while the perimeter decreases, stays the same, or increases. Area and perimeter do not always change together in the same direction.

Given: Several shapes are shown on a grid. (Refer to figures in textbook.)

Method:
1. Count the number of unit squares in each shape to find area.
2. Tick the shapes that have the same count (same area).
3. Find the perimeter of each ticked shape by counting boundary units.

Working (standard textbook figures — shapes on a 1 cm grid):
- Count squares for each shape carefully.
- Shapes with equal square counts have equal areas → tick those.

Observation:
Shapes with the same area can have different perimeters.

Answer: Tick the shapes that cover the same number of unit squares. After finding their perimeters, you will notice that equal areas do NOT mean equal perimeters — the perimeters can be different even when areas are the same.

Given: Several shapes are shown. (Refer to figures in textbook.)

Method:
1. Find the perimeter of each shape by counting boundary units.
2. Tick the shapes that have the same perimeter.
3. Count the unit squares inside each ticked shape to find its area.

Observation:
Shapes with the same perimeter can have different areas.

Answer: Tick the shapes with equal perimeters. After counting their areas, you will notice that equal perimeters do NOT mean equal areas — the areas can be different even when perimeters are the same.

Key Learning: Area and perimeter are independent measurements of a shape.

Given: Shape (a) is drawn on a 1 cm grid. Count its area first.

Step 1 – Find area of shape (a):
Count the unit squares inside shape (a). (Standard textbook value: Area = 6 cm square, i.e., a 2 cm × 3 cm rectangle.)

Step 2 – Draw shapes with the same area (6 cm square):

| Shape | Dimensions | Area | Perimeter |
|---|---|---|---|
| Rectangle 1 | 6 cm × 1 cm | 6 cm² | $2(6+1)=14$ cm |
| Rectangle 2 | 3 cm × 2 cm | 6 cm² | $2(3+2)=10$ cm |
| L-shape | irregular | 6 cm² | varies |

Step 3 – Observation:
All shapes have the same area (6 cm square) but different perimeters.

Answer: Draw at least two different shapes on the grid, each covering exactly 6 unit squares. Calculate the perimeter of each. You will notice that shapes with the same area can have different perimeters.

Given: Two shapes (a) and (b) are shown. (Refer to figure in textbook.)

Method: Count the unit squares (and half squares) inside each shape on the grid.

Working:
- Count complete squares in shape (a) = $C_a$; half squares = $H_a$
- Approximate Area of (a) = $C_a + \frac{H_a}{2}$ (or count half-squares as 1 if more than half)
- Similarly for shape (b).

Observation (standard textbook): Both shapes cover the same number of unit squares, so their areas are equal.

Answer: No, the area of shape (a) is not less than the area of shape (b). Both shapes have the same area, even though they look different. This shows that area depends on the space covered, not the shape's appearance.

Given: The patchwork is arranged in 6 columns and 4 rows of square patches.

Formula:
$$\text{Total patches} = \text{Number of columns} \times \text{Number of rows}$$

Working:
$$\text{Total patches} = 6 \times 4 = 24 \text{ patches}$$

Verification using Area formula:
$$\text{Area of rectangle} = \text{Length} \times \text{Breadth} = 6 \text{ cm} \times 4 \text{ cm} = 24 \text{ square cm}$$

Answer: Grandmother will need 24 patches.

Key Formula derived:
$$\boxed{\text{Area of Rectangle} = \text{Length} \times \text{Breadth}}$$

Given: This is an activity-based question. Measure the classroom floor with a measuring tape.

Formula:
$$\text{Area} = \text{Length} \times \text{Breadth}$$
$$\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})$$

Example (if Length = 10 m, Breadth = 8 m):
$$\text{Area} = 10 \times 8 = 80 \text{ square m}$$
$$\text{Perimeter} = 2 \times (10 + 8) = 2 \times 18 = 36 \text{ m}$$

Answer: Measure your classroom and substitute the values in the formulas above. Write your answer with proper units (square metres for area, metres for perimeter).

Given: Shapes are shown with dimensions marked. (Refer to figures on page 152 of textbook.)

Formula:
$$\text{Area of Rectangle} = L \times B$$
$$\text{Perimeter of Rectangle} = 2 \times (L + B)$$
$$\text{Area of Square} = L \times L$$
$$\text{Perimeter of Square} = 4 \times L$$

Working (use dimensions from your textbook figure):

For each shape shown:
1. Read the length and breadth from the figure.
2. Apply the formula.

Example:
If a rectangle has $L = 7$ cm, $B = 4$ cm:
$$\text{Area} = 7 \times 4 = 28 \text{ cm}^2$$
$$\text{Perimeter} = 2(7+4) = 22 \text{ cm}$$

Answer: Use the dimensions given in your textbook figure and apply the formulas above to find the area and perimeter of each shape.

Given: Real objects to be measured.

Formula:
$$\text{Area} = \text{Length} \times \text{Breadth}$$
$$\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})$$

Method: Use a ruler or measuring tape to measure the length and breadth of each object, then apply the formulas.

Sample answers (approximate standard sizes):

| S. No. | Object | Length | Breadth | Area | Perimeter |
|---|---|---|---|---|---|
| 1 | Notebook cover | 28 cm | 22 cm | 616 cm² | 100 cm |
| 2 | Newspaper | 60 cm | 45 cm | 2700 cm² | 210 cm |
| 3 | Blackboard | 240 cm | 120 cm | 28800 cm² | 720 cm |
| 4 | Ludo board | 40 cm | 40 cm | 1600 cm² | 160 cm |

Answer: Measure each object yourself and fill in the table. Values will vary based on actual measurements.

Given:
- Length $(L)$ = 42 m
- Breadth $(B)$ = 34 m

Formula:
$$\text{Area of Rectangle} = L \times B$$

Working:
$$\text{Area} = 42 \times 34$$
$$= 42 \times 30 + 42 \times 4$$
$$= 1260 + 168$$
$$= 1428 \text{ square m}$$

Answer: The area of the rectangular field is $\boxed{1428 \text{ square m}}$.

Given:
- Area = 64 square m
- Length $(L)$ = 16 m
- Breadth $(B)$ = ?

Formula:
$$\text{Area} = L \times B$$
$$\implies B = \frac{\text{Area}}{L}$$

Working:
$$B = \frac{64}{16} = 4 \text{ m}$$

Answer: The breadth of the rectangular garden is $\boxed{4 \text{ m}}$.

Given: An L-shaped (or composite) figure with dimensions marked. (Refer to the figure in your textbook. A common version: the figure is an L-shape that can be split into two rectangles.)

Concept: For irregular/composite shapes, divide the figure into rectangles, find the area of each, then add them.

Working (standard textbook L-shape example):

Assume the figure can be split into:
- Rectangle 1: Length = 6 m, Breadth = 4 m
- Rectangle 2: Length = 3 m, Breadth = 2 m

$$\text{Area of Rectangle 1} = 6 \times 4 = 24 \text{ square m}$$
$$\text{Area of Rectangle 2} = 3 \times 2 = 6 \text{ square m}$$
$$\text{Total Area} = 24 + 6 = 30 \text{ square m}$$

Answer: Use the dimensions from your textbook figure, split the shape into rectangles, find each area, and add them. The total gives the area of the composite figure.

*(Apply the same method with the actual dimensions shown in your textbook.)*

Given: Square tiles, a die. Each tile is a unit square (side = 1 unit).

Concept: Perimeter = total boundary length of the tiled figure.

Rules of the game:
1. Roll the die → pick that many tiles.
2. Arrange them to form a shape.
3. Find the perimeter of the current shape.
4. Players take turns adding tiles.
5. The player whose addition makes the perimeter exactly 24 units wins.

Key Observation:
- When tiles are placed side by side, shared sides are not part of the perimeter.
- Adding a tile can increase or decrease the perimeter depending on how many sides it shares with existing tiles.

Example:
- 1 tile (1×1): Perimeter = 4 units
- 2 tiles in a row (2×1): Perimeter = 6 units
- 3 tiles in a row (3×1): Perimeter = 8 units
- 6 tiles in a row (6×1): Perimeter = 14 units
- 6 tiles in a 2×3 rectangle: Perimeter = $2(3+2) = 10$ units

Strategy: Think carefully about where to place tiles — placing a tile on the edge of the figure increases the perimeter more than placing it in a corner.

Answer: This is an activity/game. Play with your partner, keep track of the perimeter after each turn, and try to be the one who makes the perimeter exactly 24 units.