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.)*