Chapter 3 of 14
Chapter 3: How Many Squares?
कितने वर्ग? (क्षेत्रफल, परिमाप, डाक टिकट एवं ग्रिड पहेलियाँ) | CBSE Session 2026–27
Core Pedagogical Competencies & Concepts
- Measuring area by counting 1 cm grid squares: full squares, more than half squares as 1, exactly half squares as 1/2, less than half ignored.
- Comparing areas of postal stamps, footprints, and coin impressions.
- Relationship between area and perimeter: shapes with the same area can have different perimeters.
- Finding the area of right triangles as half of the surrounding rectangle: Area of Triangle = 1/2 × Base × Height.
- Given: $\text{Perimeter} = 24\text{ cm}$, $\text{Width} (w) = 4\text{ cm}$.
All Textbook Exercises & In-Text Tasks (24 Questions)
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Q1
Textbook Exercise • Postage Stamp Area & Perimeter
Look at postage stamps on a 1 cm square grid paper:
• Stamp A covers 18 squares.
• Stamp B covers 8 squares.
• Stamp C covers 6 squares.
• Stamp D covers 12 squares.
(a) What is the area of Stamp A? (b) Which stamp has the smallest area? (c) What is the difference between the area of the biggest and smallest stamp?
[3 Marks]
• Stamp A covers 18 squares.
• Stamp B covers 8 squares.
• Stamp C covers 6 squares.
• Stamp D covers 12 squares.
(a) What is the area of Stamp A? (b) Which stamp has the smallest area? (c) What is the difference between the area of the biggest and smallest stamp?
✅ Step-by-Step Solution & Working
Stamp Area Analysis:
• (a) Area of Stamp A: Since each square has an area of 1 sq cm, Stamp A = 18 × 1 = 18 sq cm.
• (b) Smallest stamp: Stamp C covers only 6 squares, so Stamp C has the smallest area (6 sq cm).
• (c) Difference: Area of Stamp A - Area of Stamp C = 18 - 6 = 12 sq cm.
• (a) Area of Stamp A: Since each square has an area of 1 sq cm, Stamp A = 18 × 1 = 18 sq cm.
• (b) Smallest stamp: Stamp C covers only 6 squares, so Stamp C has the smallest area (6 sq cm).
• (c) Difference: Area of Stamp A - Area of Stamp C = 18 - 6 = 12 sq cm.
CBSE Step-by-Step Marking Scheme
Part (a): 18 sq cm
1 M
Part (b): Stamp C (6 sq cm)
1 M
Part (c): 12 sq cm difference
1 M
Q2
Textbook Exercise • Perimeter Calculation
Stamp D has a length of 4 cm and a width of 3 cm. Stamp B has a length of 4 cm and a width of 2 cm. (a) Find the perimeter of Stamp D. (b) Find the perimeter of Stamp B.
[2 Marks]
✅ Step-by-Step Solution & Working
Perimeter Formula: Perimeter = 2 × (Length + Width)
• (a) Perimeter of Stamp D: 2 × (4 + 3) = 2 × 7 = 14 cm.
• (b) Perimeter of Stamp B: 2 × (4 + 2) = 2 × 6 = 12 cm.
• (a) Perimeter of Stamp D: 2 × (4 + 3) = 2 × 7 = 14 cm.
• (b) Perimeter of Stamp B: 2 × (4 + 2) = 2 × 6 = 12 cm.
CBSE Step-by-Step Marking Scheme
Part (a): 14 cm
1 M
Part (b): 12 cm
1 M
Q3
Textbook Exercise • Footprints & Estimating Area
How can you estimate the area of an irregular footprint (like a human footprint or dog paw) drawn on a 1 cm grid sheet? State the counting rules.
[3 Marks]
✅ Step-by-Step Solution & Working
Grid Counting Rules for Irregular Shapes:
1. Count every square that is fully covered as 1 whole sq cm.
2. Count every square that is more than half covered as 1 whole sq cm.
3. If a square is exactly half covered, count it as 1/2 sq cm (two half squares make 1 sq cm).
4. Ignore squares that are less than half covered.
5. Add all counted squares together to find the estimated total area.
1. Count every square that is fully covered as 1 whole sq cm.
2. Count every square that is more than half covered as 1 whole sq cm.
3. If a square is exactly half covered, count it as 1/2 sq cm (two half squares make 1 sq cm).
4. Ignore squares that are less than half covered.
5. Add all counted squares together to find the estimated total area.
CBSE Step-by-Step Marking Scheme
Full & >half square rule
1 M
Half square rule
1 M
Q4
Textbook Exercise • Triangle in a Rectangle Area
A rectangle drawn on grid paper is 5 cm long and 4 cm wide. A diagonal line is drawn dividing it into two equal triangles. (a) What is the area of the rectangle? (b) What is the area of each triangle?
[3 Marks]
✅ Step-by-Step Solution & Working
Area Calculations:
• (a) Area of rectangle: Length × Width = 5 cm × 4 cm = 20 sq cm.
• (b) Area of each triangle: The diagonal divides the rectangle into two equal halves. Area of triangle = 1/2 × Area of rectangle = 20 ÷ 2 = 10 sq cm.
• (a) Area of rectangle: Length × Width = 5 cm × 4 cm = 20 sq cm.
• (b) Area of each triangle: The diagonal divides the rectangle into two equal halves. Area of triangle = 1/2 × Area of rectangle = 20 ÷ 2 = 10 sq cm.
CBSE Step-by-Step Marking Scheme
Part (a): 20 sq cm
1.5 M
Part (b): 10 sq cm
1.5 M
Q5
Textbook Exercise • 12 Squares Puzzle (Pentominoes)
Arrange 12 squares of side 1 cm to form rectangles of different dimensions. (a) List all possible rectangles that can be formed using all 12 squares. (b) Which rectangle has the longest perimeter? (c) Which rectangle has the shortest perimeter?
[4 Marks]
✅ Step-by-Step Solution & Working
Possible Rectangles with Area 12 sq cm:
• 1 × 12 rectangle: Perimeter = 2 × (1 + 12) = 2 × 13 = 26 cm.
• 2 × 6 rectangle: Perimeter = 2 × (2 + 6) = 2 × 8 = 16 cm.
• 3 × 4 rectangle: Perimeter = 2 × (3 + 4) = 2 × 7 = 14 cm.
• (b) Longest perimeter: The 1 × 12 rectangle with perimeter 26 cm.
• (c) Shortest perimeter: The 3 × 4 rectangle with perimeter 14 cm.
• 1 × 12 rectangle: Perimeter = 2 × (1 + 12) = 2 × 13 = 26 cm.
• 2 × 6 rectangle: Perimeter = 2 × (2 + 6) = 2 × 8 = 16 cm.
• 3 × 4 rectangle: Perimeter = 2 × (3 + 4) = 2 × 7 = 14 cm.
• (b) Longest perimeter: The 1 × 12 rectangle with perimeter 26 cm.
• (c) Shortest perimeter: The 3 × 4 rectangle with perimeter 14 cm.
CBSE Step-by-Step Marking Scheme
All 3 rectangle dimensions listed
1.5 M
Longest perimeter: 26 cm
1.5 M
Shortest perimeter: 14 cm
1 M
Q6
Textbook Exercise • Same Area Different Perimeter
Can two shapes have the same area but different perimeters? Explain with an example using 16 square tiles of side 1 cm.
[3 Marks]
✅ Step-by-Step Solution & Working
Explanation & Example:
Yes, two shapes can have identical areas but different perimeters.
• Example with 16 square tiles (Area = 16 sq cm in both cases):
- Shape 1 (4 cm × 4 cm square): Area = 16 sq cm. Perimeter = 4 × 4 = 16 cm.
- Shape 2 (1 cm × 16 cm ribbon): Area = 16 sq cm. Perimeter = 2 × (1 + 16) = 2 × 17 = 34 cm.
Both shapes have an area of 16 sq cm, but their perimeters are 16 cm and 34 cm.
Yes, two shapes can have identical areas but different perimeters.
• Example with 16 square tiles (Area = 16 sq cm in both cases):
- Shape 1 (4 cm × 4 cm square): Area = 16 sq cm. Perimeter = 4 × 4 = 16 cm.
- Shape 2 (1 cm × 16 cm ribbon): Area = 16 sq cm. Perimeter = 2 × (1 + 16) = 2 × 17 = 34 cm.
Both shapes have an area of 16 sq cm, but their perimeters are 16 cm and 34 cm.
CBSE Step-by-Step Marking Scheme
Affirmative 'Yes' stated
1 M
Square calculation (Area 16, Perim 16)
1 M
Rectangle calculation (Area 16, Perim 34)
1 M
Q7
Textbook Exercise • Postcard Coverage
A standard postcard measures 14 cm in length and 9 cm in width. (a) What is the area of the postcard? (b) How many postage stamps of size 3 cm by 2 cm can fit completely on the postcard without overlapping?
[3 Marks]
✅ Step-by-Step Solution & Working
Calculations:
• (a) Area of postcard: 14 cm × 9 cm = 126 sq cm.
• (b) Stamps fitting: Area of one stamp = 3 cm × 2 cm = 6 sq cm.
- Along 14 cm length: 14 ÷ 3 = 4 stamps (with 2 cm left over).
- Along 9 cm width: 9 ÷ 2 = 4 stamps (with 1 cm left over).
- Total stamps fitted = 4 × 4 = 16 stamps.
(By pure area quotient: 126 ÷ 6 = 21 stamps, but physical grid fitting allows 16 to 18 stamps).
• (a) Area of postcard: 14 cm × 9 cm = 126 sq cm.
• (b) Stamps fitting: Area of one stamp = 3 cm × 2 cm = 6 sq cm.
- Along 14 cm length: 14 ÷ 3 = 4 stamps (with 2 cm left over).
- Along 9 cm width: 9 ÷ 2 = 4 stamps (with 1 cm left over).
- Total stamps fitted = 4 × 4 = 16 stamps.
(By pure area quotient: 126 ÷ 6 = 21 stamps, but physical grid fitting allows 16 to 18 stamps).
CBSE Step-by-Step Marking Scheme
Area of postcard: 126 sq cm
1.5 M
Stamp fitting calculation
1.5 M
Q8
Textbook Exercise • Floor Tile Pattern
A square floor of side 6 metres needs to be paved with square tiles of side 30 cm. (a) Convert 6 m to cm. (b) Find the area of the floor in sq cm. (c) How many tiles are needed to pave the entire floor?
[3 Marks]
✅ Step-by-Step Solution & Working
Tiling Calculations:
• (a) Conversion: 6 m = 6 × 100 = 600 cm.
• (b) Area of floor: 600 cm × 600 cm = 360,000 sq cm.
• (c) Number of tiles: Area of one tile = 30 cm × 30 cm = 900 sq cm.
Number of tiles = 360,000 ÷ 900 = 400 tiles.
• (a) Conversion: 6 m = 6 × 100 = 600 cm.
• (b) Area of floor: 600 cm × 600 cm = 360,000 sq cm.
• (c) Number of tiles: Area of one tile = 30 cm × 30 cm = 900 sq cm.
Number of tiles = 360,000 ÷ 900 = 400 tiles.
CBSE Step-by-Step Marking Scheme
Conversion to 600 cm
1 M
Area of floor: 360,000 sq cm
1 M
Tiles needed: 400 tiles
1 M
Q9
Textbook Exercise • Area of Composite Grid Figure
A figure drawn on a 1 cm grid paper consists of 14 full squares, 8 half squares, and 4 squares that are more than half covered. What is the total area of the figure?
[2 Marks]
✅ Step-by-Step Solution & Working
Grid Area Calculation:
• 14 full squares = 14 × 1 = 14 sq cm.
• 8 half squares = 8 × 0.5 = 4 sq cm.
• 4 more-than-half squares = 4 × 1 = 4 sq cm.
• Total Area: 14 + 4 + 4 = 22 sq cm.
• 14 full squares = 14 × 1 = 14 sq cm.
• 8 half squares = 8 × 0.5 = 4 sq cm.
• 4 more-than-half squares = 4 × 1 = 4 sq cm.
• Total Area: 14 + 4 + 4 = 22 sq cm.
CBSE Step-by-Step Marking Scheme
Stepwise square values
1 M
Total area: 22 sq cm
1 M
Q10
Textbook Exercise • Perimeter Comparison of Shapes
A wire of length 36 cm is bent into: (a) A square (b) An equilateral triangle (c) A regular hexagon. What will be the length of each side in each shape?
[3 Marks]
✅ Step-by-Step Solution & Working
Side Length Calculations (Side = Perimeter ÷ Number of sides):
• (a) Square (4 sides): 36 ÷ 4 = 9 cm.
• (b) Equilateral triangle (3 sides): 36 ÷ 3 = 12 cm.
• (c) Regular hexagon (6 sides): 36 ÷ 6 = 6 cm.
• (a) Square (4 sides): 36 ÷ 4 = 9 cm.
• (b) Equilateral triangle (3 sides): 36 ÷ 3 = 12 cm.
• (c) Regular hexagon (6 sides): 36 ÷ 6 = 6 cm.
CBSE Step-by-Step Marking Scheme
Square side: 9 cm
1 M
Triangle side: 12 cm
1 M
Hexagon side: 6 cm
1 M
Q11
Textbook Exercise • Dividing Rectangles into Equal Shapes
A 10 cm × 6 cm rectangle has an area of 60 sq cm. Show how it can be divided into: (a) 3 equal rectangles (b) 2 equal triangles. State the area of each resulting piece.
[3 Marks]
✅ Step-by-Step Solution & Working
Divisions:
• (a) 3 equal rectangles: Cut along the 6 cm width into three strips of 2 cm width each (10 cm × 2 cm). Area of each strip = 60 ÷ 3 = 20 sq cm.
• (b) 2 equal triangles: Cut diagonally from one corner to the opposite corner. Area of each triangle = 60 ÷ 2 = 30 sq cm.
• (a) 3 equal rectangles: Cut along the 6 cm width into three strips of 2 cm width each (10 cm × 2 cm). Area of each strip = 60 ÷ 3 = 20 sq cm.
• (b) 2 equal triangles: Cut diagonally from one corner to the opposite corner. Area of each triangle = 60 ÷ 2 = 30 sq cm.
CBSE Step-by-Step Marking Scheme
Part (a): 3 strips of 20 sq cm each
1.5 M
Part (b): 2 triangles of 30 sq cm each
1.5 M
Q12
Textbook Exercise • Comparing Footprint Sizes
Rohan's footprint covers 18 squares on a grid sheet, while his baby sister's footprint covers 7 squares. (a) By how many square centimetres is Rohan's footprint larger? (b) If his father's footprint is double Rohan's, what is its area?
[2 Marks]
✅ Step-by-Step Solution & Working
Footprint Calculations:
• (a) Difference: 18 sq cm - 7 sq cm = 11 sq cm larger.
• (b) Father's footprint area: 2 × 18 sq cm = 36 sq cm.
• (a) Difference: 18 sq cm - 7 sq cm = 11 sq cm larger.
• (b) Father's footprint area: 2 × 18 sq cm = 36 sq cm.
CBSE Step-by-Step Marking Scheme
Part (a): 11 sq cm difference
1 M
Part (b): 36 sq cm area
1 M
Q13
Textbook Exercise • Perimeter with Missing Side
A five-sided garden polygon has a total perimeter of 45 metres. Four of its sides measure 8 m, 11 m, 9 m, and 10 m. What is the length of the fifth side?
[2 Marks]
✅ Step-by-Step Solution & Working
Missing Side Calculation:
• Sum of given four sides = 8 + 11 + 9 + 10 = 38 m.
• Total perimeter = 45 m.
• Length of fifth side = 45 - 38 = 7 metres.
• Sum of given four sides = 8 + 11 + 9 + 10 = 38 m.
• Total perimeter = 45 m.
• Length of fifth side = 45 - 38 = 7 metres.
CBSE Step-by-Step Marking Scheme
Sum of 4 sides: 38 m
1 M
Fifth side: 7 m
1 M
Q14
Textbook Exercise • Square vs Rectangle Area
A square has a side of 8 cm. A rectangle has a length of 10 cm and a width of 6 cm. (a) Which figure has a greater area, and by how much? (b) Which figure has a greater perimeter, and by how much?
[4 Marks]
✅ Step-by-Step Solution & Working
Comparison:
• (a) Area:
- Area of Square = 8 × 8 = 64 sq cm.
- Area of Rectangle = 10 × 6 = 60 sq cm.
- The Square has greater area by 64 - 60 = 4 sq cm.
• (b) Perimeter:
- Perimeter of Square = 4 × 8 = 32 cm.
- Perimeter of Rectangle = 2 × (10 + 6) = 32 cm.
- Both figures have the exact same perimeter (32 cm)!
• (a) Area:
- Area of Square = 8 × 8 = 64 sq cm.
- Area of Rectangle = 10 × 6 = 60 sq cm.
- The Square has greater area by 64 - 60 = 4 sq cm.
• (b) Perimeter:
- Perimeter of Square = 4 × 8 = 32 cm.
- Perimeter of Rectangle = 2 × (10 + 6) = 32 cm.
- Both figures have the exact same perimeter (32 cm)!
CBSE Step-by-Step Marking Scheme
Square area 64, rectangle 60, diff 4 sq cm
2 M
Perimeters both 32 cm comparison
2 M
Q15
Textbook Exercise • Chequered Chessboard Problem
A standard chessboard has 8 rows and 8 columns of alternating black and white squares. If each small square has a side of 3 cm: (a) What is the side length of the entire board? (b) What is the total area of the chessboard? (c) What is the total area of all the black squares?
[3 Marks]
✅ Step-by-Step Solution & Working
Chessboard Calculations:
• (a) Side length: 8 squares × 3 cm = 24 cm.
• (b) Total area: 24 cm × 24 cm = 576 sq cm.
• (c) Black squares area: Exactly half of the 64 squares are black (32 squares). Area = 576 ÷ 2 = 288 sq cm.
• (a) Side length: 8 squares × 3 cm = 24 cm.
• (b) Total area: 24 cm × 24 cm = 576 sq cm.
• (c) Black squares area: Exactly half of the 64 squares are black (32 squares). Area = 576 ÷ 2 = 288 sq cm.
CBSE Step-by-Step Marking Scheme
Side length: 24 cm
1 M
Total area: 576 sq cm
1 M
Black squares area: 288 sq cm
1 M
Q16
Textbook Exercise • Pentomino Puzzle (T-Shape)
A 'pentomino' is a shape made by joining 5 squares edge-to-edge. (a) How many square centimetres is the area of any pentomino made with 1 cm squares? (b) If a 'T-shape' pentomino has 3 squares across the top and 2 squares going down the middle stem, calculate its perimeter.
[3 Marks]
✅ Step-by-Step Solution & Working
Pentomino Analysis:
• (a) Area: Since it always consists of 5 squares of 1 sq cm, its area is always 5 sq cm.
• (b) Perimeter of T-pentomino:
Counting outer edge segments: Top edge = 3 cm; left & right ends of crossbar = 1 cm each; under-crossbar edges = 1 cm each; stem sides = 2 cm each; bottom edge = 1 cm.
Total perimeter = 3 + 1 + 1 + 1 + 2 + 1 + 2 + 1 = 12 cm.
• (a) Area: Since it always consists of 5 squares of 1 sq cm, its area is always 5 sq cm.
• (b) Perimeter of T-pentomino:
Counting outer edge segments: Top edge = 3 cm; left & right ends of crossbar = 1 cm each; under-crossbar edges = 1 cm each; stem sides = 2 cm each; bottom edge = 1 cm.
Total perimeter = 3 + 1 + 1 + 1 + 2 + 1 + 2 + 1 = 12 cm.
CBSE Step-by-Step Marking Scheme
Area: 5 sq cm
1 M
Perimeter counting: 12 cm
2 M
Q17
Textbook Exercise • Cost of Fencing a Garden
A rectangular park is 25 metres long and 15 metres wide. Fencing wire costs ₹20 per metre. What is the total cost of fencing the park with 3 rounds of wire?
[3 Marks]
✅ Step-by-Step Solution & Working
Fencing Cost Calculation:
1. Perimeter of park = 2 × (25 + 15) = 2 × 40 = 80 metres.
2. Total wire needed for 3 rounds = 3 × 80 = 240 metres.
3. Total cost = 240 m × ₹20 = ₹4,800.
1. Perimeter of park = 2 × (25 + 15) = 2 × 40 = 80 metres.
2. Total wire needed for 3 rounds = 3 × 80 = 240 metres.
3. Total cost = 240 m × ₹20 = ₹4,800.
CBSE Step-by-Step Marking Scheme
Perimeter: 80 m
1 M
3 rounds length: 240 m
1 M
Total cost: ₹4,800
1 M
Q18
Textbook Exercise • Doubling the Side of a Square
A square has a side of 4 cm. If its side is doubled to 8 cm: (a) What happens to its perimeter? (b) What happens to its area?
[3 Marks]
✅ Step-by-Step Solution & Working
Effect of Doubling Side:
• (a) Perimeter:
- Original: 4 × 4 = 16 cm.
- New: 4 × 8 = 32 cm.
- The perimeter doubles (increases by 2 times).
• (b) Area:
- Original: 4 × 4 = 16 sq cm.
- New: 8 × 8 = 64 sq cm.
- The area becomes 4 times larger (2² = 4).
• (a) Perimeter:
- Original: 4 × 4 = 16 cm.
- New: 4 × 8 = 32 cm.
- The perimeter doubles (increases by 2 times).
• (b) Area:
- Original: 4 × 4 = 16 sq cm.
- New: 8 × 8 = 64 sq cm.
- The area becomes 4 times larger (2² = 4).
CBSE Step-by-Step Marking Scheme
Perimeter doubles (16 cm to 32 cm)
1.5 M
Area quadruples (16 to 64 sq cm)
1.5 M
Q19
Textbook Exercise • L-Shaped Room Area
An L-shaped room can be divided into two rectangles: Rectangle 1 is 4 m × 3 m, and Rectangle 2 is 2 m × 2 m. (a) What is the total area of the room? (b) If carpet costs ₹150 per square metre, what is the cost of carpeting the room?
[3 Marks]
✅ Step-by-Step Solution & Working
Area & Carpet Cost:
• (a) Total area:
- Area of Rectangle 1 = 4 × 3 = 12 sq m.
- Area of Rectangle 2 = 2 × 2 = 4 sq m.
- Total area = 12 + 4 = 16 sq metres.
• (b) Cost of carpeting: 16 sq m × ₹150 = ₹2,400.
• (a) Total area:
- Area of Rectangle 1 = 4 × 3 = 12 sq m.
- Area of Rectangle 2 = 2 × 2 = 4 sq m.
- Total area = 12 + 4 = 16 sq metres.
• (b) Cost of carpeting: 16 sq m × ₹150 = ₹2,400.
CBSE Step-by-Step Marking Scheme
Total area: 16 sq m
1.5 M
Carpet cost: ₹2,400
1.5 M
Q20
Textbook Exercise • Border Ribbon on Table Cloth
A square table cloth has an area of 49 sq metres. (a) What is the length of one side of the table cloth? (b) How many metres of lace border will be needed to sew all along its edges?
[2 Marks]
✅ Step-by-Step Solution & Working
Calculations:
• (a) Side length: Since Area = Side × Side = 49, Side = √49 = 7 metres.
• (b) Lace border needed: Perimeter = 4 × Side = 4 × 7 = 28 metres of lace.
• (a) Side length: Since Area = Side × Side = 49, Side = √49 = 7 metres.
• (b) Lace border needed: Perimeter = 4 × Side = 4 × 7 = 28 metres of lace.
CBSE Step-by-Step Marking Scheme
Side: 7 m
1 M
Perimeter/Lace: 28 m
1 M
Q21
Textbook Exercise • Perimeter of Irregular Grid Figures
On a 1 cm grid paper, draw a staircase shape going 3 units right, 1 unit up, 2 units right, 1 unit up, 1 unit left, 2 units down, 4 units left, and 1 unit down to close. Count the total unit segments along the boundary to find its perimeter.
[2 Marks]
✅ Step-by-Step Solution & Working
Perimeter Determination:
Tracing each segment along the outer contour: 3 + 1 + 2 + 1 + 1 + 2 + 4 + 1 = 15 units (or 14-16 cm depending on closed vertex coordinates). Tracing a closed staircase with baseline 5 cm and height 2 cm gives a boundary of 14 cm.
Tracing each segment along the outer contour: 3 + 1 + 2 + 1 + 1 + 2 + 4 + 1 = 15 units (or 14-16 cm depending on closed vertex coordinates). Tracing a closed staircase with baseline 5 cm and height 2 cm gives a boundary of 14 cm.
CBSE Step-by-Step Marking Scheme
Stepwise segment addition
1 M
Total perimeter statement
1 M
Q22
Textbook Exercise • Comparing Book Cover Areas
Maths book cover is 28 cm long and 21 cm wide. Hindi book cover is 25 cm long and 20 cm wide. Find: (a) Area of Maths cover (b) Area of Hindi cover (c) Difference in their areas.
[3 Marks]
✅ Step-by-Step Solution & Working
Cover Area Calculations:
• (a) Maths cover area: 28 × 21 = 588 sq cm.
• (b) Hindi cover area: 25 × 20 = 500 sq cm.
• (c) Difference: 588 - 500 = 88 sq cm.
• (a) Maths cover area: 28 × 21 = 588 sq cm.
• (b) Hindi cover area: 25 × 20 = 500 sq cm.
• (c) Difference: 588 - 500 = 88 sq cm.
CBSE Step-by-Step Marking Scheme
Maths area: 588 sq cm
1 M
Hindi area: 500 sq cm
1 M
Difference: 88 sq cm
1 M
Q23
Textbook Exercise • Tiling without Gaps
Explain why regular hexagon tiles and square tiles can pave a floor without leaving any gaps, but regular pentagon tiles cannot.
[3 Marks]
✅ Step-by-Step Solution & Working
Tiling & Angle Sum at a Vertex:
• For tiles to meet around a point without gaps or overlaps, the sum of their corner angles must equal exactly 360°.
• Squares: Corner angle = 90°. Four squares meet: 4 × 90° = 360° (no gaps).
• Hexagons: Corner angle = 120°. Three hexagons meet: 3 × 120° = 360° (no gaps).
• Pentagons: Corner angle = 108°. Three pentagons = 324° (gap of 36°); four pentagons = 432° (overlap). Therefore, pentagons cannot tile a plane.
• For tiles to meet around a point without gaps or overlaps, the sum of their corner angles must equal exactly 360°.
• Squares: Corner angle = 90°. Four squares meet: 4 × 90° = 360° (no gaps).
• Hexagons: Corner angle = 120°. Three hexagons meet: 3 × 120° = 360° (no gaps).
• Pentagons: Corner angle = 108°. Three pentagons = 324° (gap of 36°); four pentagons = 432° (overlap). Therefore, pentagons cannot tile a plane.
CBSE Step-by-Step Marking Scheme
Vertex 360° rule explained
1 M
Square & hexagon 360° verification
1 M
Pentagon 324° failure explained
1 M
Q24
Textbook Exercise • Garden Path Area
A square flower bed has a side of 10 metres. A 1-metre wide path is built all around the outside of the flower bed. (a) What is the side length of the outer boundary including the path? (b) Find the area of the flower bed alone. (c) Find the area of the path.
[3 Marks]
✅ Step-by-Step Solution & Working
Path Area Calculations:
• (a) Outer side: Inner side = 10 m. Adding 1 m path on both sides: 10 + 1 + 1 = 12 metres.
• (b) Area of flower bed: 10 m × 10 m = 100 sq metres.
• (c) Area of the path:
- Outer total area = 12 m × 12 m = 144 sq m.
- Area of path = Outer area - Flower bed area = 144 - 100 = 44 sq metres.
• (a) Outer side: Inner side = 10 m. Adding 1 m path on both sides: 10 + 1 + 1 = 12 metres.
• (b) Area of flower bed: 10 m × 10 m = 100 sq metres.
• (c) Area of the path:
- Outer total area = 12 m × 12 m = 144 sq m.
- Area of path = Outer area - Flower bed area = 144 - 100 = 44 sq metres.
CBSE Step-by-Step Marking Scheme
Outer side: 12 m
1 M
Inner area: 100 sq m
1 M
Path area: 44 sq m
1 M
🌟 NEP 2020 Competency-Based Case Study
NEP 2020 Competency Case: How Many Squares?
A practical scenario involving How Many Squares? in everyday life and community contexts.
Analytical Task: Analyze the quantitative relationships and state the conclusion.
Evaluation & Model Answer:
Apply the step-by-step formulas and concepts mastered in this chapter.
Rubric: 1M formula + 1M working + 1M answer