Chapter 11 of 14
Chapter 11: Area and its Boundary
क्षेत्रफल और घेरा (परिमाप, डाक टिकट क्षेत्रफल एवं राजा का सोने का तार) | CBSE Session 2026–27
Core Pedagogical Competencies & Concepts
- Perimeter = Total length of the boundary: P = 2 × (L + B) for rectangles, P = 4 × S for squares.
- Area = Measure of region enclosed: A = L × B for rectangles, A = S × S for squares.
- Shapes with the SAME perimeter can enclose DIFFERENT areas.
- A square encloses the maximum area for any given rectangular perimeter.
All Textbook Exercises & In-Text Tasks (24 Questions)
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Q1
Textbook Exercise • Land Division among Three Children
Arbaaz planned to divide his land equally among his three children—Nasreen, Shahid, and Zaid. He wanted to give each child 30 square metres of land. Nasreen's land is 5 m long and 6 m wide. Shahid's land is 3 m wide and 10 m long. Zaid's land is 2 m wide and 15 m long.
(a) Find the area of each child's land.
(b) Who got the land with the longest boundary (perimeter)?
[4 Marks]
(a) Find the area of each child's land.
(b) Who got the land with the longest boundary (perimeter)?
✅ Step-by-Step Solution & Working
Perimeter & Area Calculations:
• (a) Areas:
- Nasreen: 5 m × 6 m = 30 sq m.
- Shahid: 3 m × 10 m = 30 sq m.
- Zaid: 2 m × 15 m = 30 sq m.
• (b) Boundaries (Perimeter = 2 × (l + b)):
- Nasreen: 2 × (5 + 6) = 2 × 11 = 22 metres.
- Shahid: 2 × (3 + 10) = 2 × 13 = 26 metres.
- Zaid: 2 × (2 + 15) = 2 × 17 = 34 metres.
• Zaid got the land with the longest boundary (34 metres).
• (a) Areas:
- Nasreen: 5 m × 6 m = 30 sq m.
- Shahid: 3 m × 10 m = 30 sq m.
- Zaid: 2 m × 15 m = 30 sq m.
• (b) Boundaries (Perimeter = 2 × (l + b)):
- Nasreen: 2 × (5 + 6) = 2 × 11 = 22 metres.
- Shahid: 2 × (3 + 10) = 2 × 13 = 26 metres.
- Zaid: 2 × (2 + 15) = 2 × 17 = 34 metres.
• Zaid got the land with the longest boundary (34 metres).
CBSE Step-by-Step Marking Scheme
All three areas equal 30 sq m
1.5 M
Perimeter calculations: 22, 26, 34 m
1.5 M
Conclusion: Zaid has longest boundary
1 M
Q2
Textbook Exercise • Postcard Loop (Paper Cutting Trick)
How can you cut a small 14 cm × 9 cm postcard in a continuous spiral strip so that it opens into a huge belt large enough for a person to pass through?
[3 Marks]
✅ Step-by-Step Solution & Working
Postcard Belt Cutting Technique:
1. Fold the postcard in half along its length.
2. Cut narrow parallel slits starting alternately from the folded edge and from the open edge, leaving a small 2 mm margin at each turn so the strip is not severed completely.
3. Cut open the folds of the inner loops (leaving the two end loops intact).
4. Pull the paper gently: the thin cuts unfold into a huge continuous zigzag paper belt with a perimeter exceeding 2 to 3 metres, allowing a child or adult to step right through!
1. Fold the postcard in half along its length.
2. Cut narrow parallel slits starting alternately from the folded edge and from the open edge, leaving a small 2 mm margin at each turn so the strip is not severed completely.
3. Cut open the folds of the inner loops (leaving the two end loops intact).
4. Pull the paper gently: the thin cuts unfold into a huge continuous zigzag paper belt with a perimeter exceeding 2 to 3 metres, allowing a child or adult to step right through!
CBSE Step-by-Step Marking Scheme
Alternating slit cutting procedure explained
2 M
Continuous loop expansion explained
1 M
Q3
Textbook Exercise • Cost of Fencing vs Tiling
A square field has a side of 40 metres. (a) What is the total length of wire needed to put a fence of 4 rounds around the field? (b) If fencing costs ₹15 per metre, what is the cost of fencing? (c) If ploughing the field costs ₹2 per square metre, what is the cost of ploughing?
[4 Marks]
✅ Step-by-Step Solution & Working
Step-by-step Solution:
• (a) Wire length: Perimeter of 1 round = 4 × 40 m = 160 m. For 4 rounds = 4 × 160 = 640 metres.
• (b) Cost of fencing: 640 m × ₹15 = ₹9,600.
• (c) Cost of ploughing: Area = 40 m × 40 m = 1,600 sq m. Cost = 1,600 × ₹2 = ₹3,200.
• (a) Wire length: Perimeter of 1 round = 4 × 40 m = 160 m. For 4 rounds = 4 × 160 = 640 metres.
• (b) Cost of fencing: 640 m × ₹15 = ₹9,600.
• (c) Cost of ploughing: Area = 40 m × 40 m = 1,600 sq m. Cost = 1,600 × ₹2 = ₹3,200.
CBSE Step-by-Step Marking Scheme
Wire length: 640 m
1.5 M
Fencing cost: ₹9,600
1 M
Ploughing cost: ₹3,200
1.5 M
Q4
Textbook Exercise • Saree Border Length
A saree is 5.5 metres long and 1.2 metres wide. A decorative zari border is to be stitched along all 4 outer edges of the saree. What is the total length of border required in metres?
[2 Marks]
✅ Step-by-Step Solution & Working
Perimeter of Saree:
• Perimeter = 2 × (Length + Width) = 2 × (5.5 + 1.2) = 2 × 6.7 = 13.4 metres.
• Perimeter = 2 × (Length + Width) = 2 × (5.5 + 1.2) = 2 × 6.7 = 13.4 metres.
CBSE Step-by-Step Marking Scheme
Formula 2(l + w)
1 M
Answer: 13.4 metres
1 M
Q5
Textbook Exercise • Comparing Rectangles with Equal Perimeter
Two rectangular vegetable plots both have a perimeter of 36 metres: Plot A is 10 m long and 8 m wide. Plot B is 12 m long and 6 m wide. (a) Verify that both plots have the same perimeter. (b) Find the area of Plot A and Plot B. (c) Which plot gives more area to grow vegetables?
[4 Marks]
✅ Step-by-Step Solution & Working
Comparison:
• (a) Perimeters:
- Plot A: 2 × (10 + 8) = 2 × 18 = 36 m.
- Plot B: 2 × (12 + 6) = 2 × 18 = 36 m. (Verified!).
• (b) Areas:
- Area of Plot A = 10 × 8 = 80 sq m.
- Area of Plot B = 12 × 6 = 72 sq m.
• (c) Conclusion: Plot A gives 80 - 72 = 8 sq m more area (shapes closer to a square maximize area!).
• (a) Perimeters:
- Plot A: 2 × (10 + 8) = 2 × 18 = 36 m.
- Plot B: 2 × (12 + 6) = 2 × 18 = 36 m. (Verified!).
• (b) Areas:
- Area of Plot A = 10 × 8 = 80 sq m.
- Area of Plot B = 12 × 6 = 72 sq m.
• (c) Conclusion: Plot A gives 80 - 72 = 8 sq m more area (shapes closer to a square maximize area!).
CBSE Step-by-Step Marking Scheme
Verification of perimeters
1 M
Areas: 80 sq m and 72 sq m
2 M
Plot A gives more area
1 M
Q6
Textbook Exercise • Carpet Fitting in a Hall
A hall floor is 15 metres long and 8 metres wide. A square carpet of side 6 metres is laid on the floor. Find the area of the floor that is left bare (not carpeted).
[3 Marks]
✅ Step-by-Step Solution & Working
Area Difference:
1. Area of hall floor = 15 m × 8 m = 120 sq metres.
2. Area of square carpet = 6 m × 6 m = 36 sq metres.
3. Bare floor area: 120 - 36 = 84 sq metres.
1. Area of hall floor = 15 m × 8 m = 120 sq metres.
2. Area of square carpet = 6 m × 6 m = 36 sq metres.
3. Bare floor area: 120 - 36 = 84 sq metres.
CBSE Step-by-Step Marking Scheme
Hall area: 120 sq m
1 M
Carpet area: 36 sq m
1 M
Bare area: 84 sq m
1 M
Q7
Textbook Exercise • Thread Loop Shape Shifting
A loop of string of length 20 cm is placed on a table: (a) If it forms a square, what is its area? (b) If it forms a rectangle of width 4 cm, what is its area? (c) Which shape has greater area?
[3 Marks]
✅ Step-by-Step Solution & Working
String Loop (Perimeter = 20 cm):
• (a) Square: Side = 20 ÷ 4 = 5 cm. Area = 5 × 5 = 25 sq cm.
• (b) Rectangle: Width = 4 cm. Length = (20 ÷ 2) - 4 = 10 - 4 = 6 cm. Area = 6 × 4 = 24 sq cm.
• (c) Conclusion: The Square has greater area by 1 sq cm.
• (a) Square: Side = 20 ÷ 4 = 5 cm. Area = 5 × 5 = 25 sq cm.
• (b) Rectangle: Width = 4 cm. Length = (20 ÷ 2) - 4 = 10 - 4 = 6 cm. Area = 6 × 4 = 24 sq cm.
• (c) Conclusion: The Square has greater area by 1 sq cm.
CBSE Step-by-Step Marking Scheme
Square area: 25 sq cm
1 M
Rectangle area: 24 sq cm
1 M
Square has greater area
1 M
Q8
Textbook Exercise • Square Tiles for a Bathroom
A bathroom wall measures 3 metres by 2 metres. It is to be covered with square ceramic tiles of side 20 cm. How many tiles are required?
[3 Marks]
✅ Step-by-Step Solution & Working
Tile Calculations:
• Wall length = 3 m = 300 cm; Wall height = 2 m = 200 cm.
• Area of wall = 300 × 200 = 60,000 sq cm.
• Area of 1 tile = 20 × 20 = 400 sq cm.
• Number of tiles: 60,000 ÷ 400 = 150 tiles.
• Wall length = 3 m = 300 cm; Wall height = 2 m = 200 cm.
• Area of wall = 300 × 200 = 60,000 sq cm.
• Area of 1 tile = 20 × 20 = 400 sq cm.
• Number of tiles: 60,000 ÷ 400 = 150 tiles.
CBSE Step-by-Step Marking Scheme
Conversions to cm
1 M
Wall and tile areas
1 M
Total tiles: 150
1 M
Q9
Textbook Exercise • Perimeter from Area of Square
A square lawn has an area of 144 square metres. (a) What is the length of each side of the lawn? (b) What is the perimeter of the lawn?
[2 Marks]
✅ Step-by-Step Solution & Working
Side and Perimeter:
• (a) Side length: Side = √144 = 12 metres.
• (b) Perimeter: 4 × Side = 4 × 12 = 48 metres.
• (a) Side length: Side = √144 = 12 metres.
• (b) Perimeter: 4 × Side = 4 × 12 = 48 metres.
CBSE Step-by-Step Marking Scheme
Side: 12 metres
1 M
Perimeter: 48 metres
1 M
Q10
Textbook Exercise • Dividing Square into Triangles
A square has a side of 10 cm. Two diagonals are drawn, dividing it into four equal triangles. (a) What is the total area of the square? (b) What is the area of each triangle?
[2 Marks]
✅ Step-by-Step Solution & Working
Area Division:
• (a) Square area: 10 cm × 10 cm = 100 sq cm.
• (b) Triangle area: 100 sq cm ÷ 4 = 25 sq cm each.
• (a) Square area: 10 cm × 10 cm = 100 sq cm.
• (b) Triangle area: 100 sq cm ÷ 4 = 25 sq cm each.
CBSE Step-by-Step Marking Scheme
Square area: 100 sq cm
1 M
Each triangle: 25 sq cm
1 M
Q11
Textbook Exercise • Running Track Laps
A school rectangular athletic field is 120 metres long and 80 metres wide. An athlete runs 5 complete laps around the field. What total distance did the athlete run in kilometres?
[3 Marks]
✅ Step-by-Step Solution & Working
Distance Run:
• Perimeter of 1 lap = 2 × (120 + 80) = 2 × 200 = 400 metres.
• 5 laps = 5 × 400 = 2,000 metres.
• In kilometres = 2,000 ÷ 1,000 = 2 kilometres.
• Perimeter of 1 lap = 2 × (120 + 80) = 2 × 200 = 400 metres.
• 5 laps = 5 × 400 = 2,000 metres.
• In kilometres = 2,000 ÷ 1,000 = 2 kilometres.
CBSE Step-by-Step Marking Scheme
1 lap: 400 m
1 M
5 laps: 2,000 m
1 M
Conversion: 2 km
1 M
Q12
Textbook Exercise • Painting Cost of Wall
A room has a rectangular wall of 5 m by 3 m with a window of 1.5 m by 1 m. (a) What is the area of the wall to be painted (excluding the window)? (b) If painting costs ₹40 per square metre, what is the total cost?
[3 Marks]
✅ Step-by-Step Solution & Working
Painting Calculation:
• Total wall area = 5 × 3 = 15 sq m.
• Window area = 1.5 × 1 = 1.5 sq m.
• (a) Area to paint: 15 - 1.5 = 13.5 sq metres.
• (b) Painting cost: 13.5 × ₹40 = ₹540.
• Total wall area = 5 × 3 = 15 sq m.
• Window area = 1.5 × 1 = 1.5 sq m.
• (a) Area to paint: 15 - 1.5 = 13.5 sq metres.
• (b) Painting cost: 13.5 × ₹40 = ₹540.
CBSE Step-by-Step Marking Scheme
Net paintable area: 13.5 sq m
1.5 M
Cost: ₹540
1.5 M
Q13
Textbook Exercise • Finding Width from Perimeter
A rectangular park has a perimeter of 180 metres. If its length is 55 metres, find its width and its area.
[3 Marks]
✅ Step-by-Step Solution & Working
Width and Area:
• Half perimeter (l + w) = 180 ÷ 2 = 90 metres.
• Width = 90 - 55 = 35 metres.
• Area = Length × Width = 55 m × 35 m = 1,925 sq metres.
• Half perimeter (l + w) = 180 ÷ 2 = 90 metres.
• Width = 90 - 55 = 35 metres.
• Area = Length × Width = 55 m × 35 m = 1,925 sq metres.
CBSE Step-by-Step Marking Scheme
Width: 35 m
1.5 M
Area: 1,925 sq m
1.5 M
Q14
Textbook Exercise • Wire Shape Transition
A wire bent into the shape of a rectangle of 18 cm by 12 cm is straightened and then rebent into a square. (a) What is the length of the wire? (b) What will be the side of the square? (c) Which shape encloses more area?
[4 Marks]
✅ Step-by-Step Solution & Working
Wire Calculations:
• (a) Wire length (Perimeter): 2 × (18 + 12) = 2 × 30 = 60 cm.
• (b) Side of square: 60 ÷ 4 = 15 cm.
• (c) Area comparison:
- Rectangle area = 18 × 12 = 216 sq cm.
- Square area = 15 × 15 = 225 sq cm.
The Square encloses more area by 225 - 216 = 9 sq cm.
• (a) Wire length (Perimeter): 2 × (18 + 12) = 2 × 30 = 60 cm.
• (b) Side of square: 60 ÷ 4 = 15 cm.
• (c) Area comparison:
- Rectangle area = 18 × 12 = 216 sq cm.
- Square area = 15 × 15 = 225 sq cm.
The Square encloses more area by 225 - 216 = 9 sq cm.
CBSE Step-by-Step Marking Scheme
Perimeter: 60 cm
1 M
Square side: 15 cm
1 M
Areas calculated: 216 & 225 sq cm
1 M
Square encloses more area
1 M
Q15
Textbook Exercise • Cost of Grass Turfing
A square garden has a side of 25 metres. Inside it, a square pond of side 5 metres is dug in the centre. The remaining area is planted with green turf grass. If turfing costs ₹50 per square metre, find the total turfing expense.
[3 Marks]
✅ Step-by-Step Solution & Working
Turfing Calculations:
• Total garden area = 25 × 25 = 625 sq m.
• Pond area = 5 × 5 = 25 sq m.
• Turfed area = 625 - 25 = 600 sq m.
• Total expense: 600 × ₹50 = ₹30,000.
• Total garden area = 25 × 25 = 625 sq m.
• Pond area = 5 × 5 = 25 sq m.
• Turfed area = 625 - 25 = 600 sq m.
• Total expense: 600 × ₹50 = ₹30,000.
CBSE Step-by-Step Marking Scheme
Garden area: 625 sq m
1 M
Turf area: 600 sq m
1 M
Total expense: ₹30,000
1 M
Q16
Textbook Exercise • Tripling the Dimensions of a Rectangle
A rectangle has length 6 cm and width 4 cm. If both its length and width are tripled: (a) What happens to its perimeter? (b) What happens to its area?
[3 Marks]
✅ Step-by-Step Solution & Working
Tripling Scaling:
• Original: Perimeter = 2(6+4) = 20 cm; Area = 6×4 = 24 sq cm.
• New dimensions: Length = 18 cm, Width = 12 cm.
• (a) New Perimeter: 2(18+12) = 60 cm (3 times larger).
• (b) New Area: 18 × 12 = 216 sq cm (9 times larger, since 3² = 9).
• Original: Perimeter = 2(6+4) = 20 cm; Area = 6×4 = 24 sq cm.
• New dimensions: Length = 18 cm, Width = 12 cm.
• (a) New Perimeter: 2(18+12) = 60 cm (3 times larger).
• (b) New Area: 18 × 12 = 216 sq cm (9 times larger, since 3² = 9).
CBSE Step-by-Step Marking Scheme
Perimeter increases 3 times (60 cm)
1.5 M
Area increases 9 times (216 sq cm)
1.5 M
Q17
Textbook Exercise • Perimeter of Combined Tables
Two identical square desks of side 80 cm are pushed together side-by-side to form one long rectangular conference table. (a) What are the dimensions of the new table? (b) What is its perimeter? (c) Did the perimeter double? Explain.
[4 Marks]
✅ Step-by-Step Solution & Working
Combined Desks:
• (a) Dimensions: Length = 80 + 80 = 160 cm; Width = 80 cm.
• (b) Perimeter: 2 × (160 + 80) = 2 × 240 = 480 cm.
• (c) Perimeter double test: Individual desk perimeter = 4 × 80 = 320 cm. Double would be 640 cm. The perimeter did NOT double because two 80 cm edges are joined internally and no longer form part of the boundary.
• (a) Dimensions: Length = 80 + 80 = 160 cm; Width = 80 cm.
• (b) Perimeter: 2 × (160 + 80) = 2 × 240 = 480 cm.
• (c) Perimeter double test: Individual desk perimeter = 4 × 80 = 320 cm. Double would be 640 cm. The perimeter did NOT double because two 80 cm edges are joined internally and no longer form part of the boundary.
CBSE Step-by-Step Marking Scheme
Dimensions: 160 cm × 80 cm
1 M
Perimeter: 480 cm
1.5 M
Explanation of lost internal edges
1.5 M
Q18
Textbook Exercise • Fencing with Gates
A rectangular farm is 150 metres long and 100 metres wide. It needs to be fenced, but two gates of 5 metres width each are to be left unfenced for entry. What length of fencing wire is needed for 1 round?
[2 Marks]
✅ Step-by-Step Solution & Working
Fencing with Openings:
• Total perimeter = 2 × (150 + 100) = 2 × 250 = 500 metres.
• Total opening for 2 gates = 2 × 5 m = 10 metres.
• Fencing wire needed: 500 - 10 = 490 metres.
• Total perimeter = 2 × (150 + 100) = 2 × 250 = 500 metres.
• Total opening for 2 gates = 2 × 5 m = 10 metres.
• Fencing wire needed: 500 - 10 = 490 metres.
CBSE Step-by-Step Marking Scheme
Perimeter 500 m and gate deduction 10 m
1 M
Wire needed: 490 m
1 M
Q19
Textbook Exercise • Concrete Slabs for Footpath
A footpath 50 metres long and 2 metres wide is to be paved with square concrete slabs of side 50 cm. (a) What is the area of the footpath in sq cm? (b) How many slabs are needed?
[3 Marks]
✅ Step-by-Step Solution & Working
Paving Calculations:
• Footpath: 50 m = 5,000 cm; 2 m = 200 cm.
• (a) Area in sq cm: 5,000 × 200 = 1,000,000 sq cm (or 100 sq m).
• (b) Slabs needed: Area of 1 slab = 50 × 50 = 2,500 sq cm.
Number of slabs = 1,000,000 ÷ 2,500 = 400 slabs.
• Footpath: 50 m = 5,000 cm; 2 m = 200 cm.
• (a) Area in sq cm: 5,000 × 200 = 1,000,000 sq cm (or 100 sq m).
• (b) Slabs needed: Area of 1 slab = 50 × 50 = 2,500 sq cm.
Number of slabs = 1,000,000 ÷ 2,500 = 400 slabs.
CBSE Step-by-Step Marking Scheme
Area: 1,000,000 sq cm
1.5 M
Slabs: 400
1.5 M
Q20
Textbook Exercise • Comparing Postage Stamp Sizes
Stamp A is 4 cm × 3 cm. Stamp B is 6 cm × 2 cm. Stamp C is 5 cm × 2.5 cm. (a) Find the area of each stamp. (b) Which stamp covers the largest area?
[3 Marks]
✅ Step-by-Step Solution & Working
Stamp Areas:
• Area of Stamp A = 4 × 3 = 12 sq cm.
• Area of Stamp B = 6 × 2 = 12 sq cm.
• Area of Stamp C = 5 × 2.5 = 12.5 sq cm.
• Stamp C covers the largest area (12.5 sq cm).
• Area of Stamp A = 4 × 3 = 12 sq cm.
• Area of Stamp B = 6 × 2 = 12 sq cm.
• Area of Stamp C = 5 × 2.5 = 12.5 sq cm.
• Stamp C covers the largest area (12.5 sq cm).
CBSE Step-by-Step Marking Scheme
Three areas calculated
2 M
Stamp C identified as largest
1 M
Q21
Textbook Exercise • Irregular Garden Boundary Cost
An irregular five-sided park has boundary sides of 45 m, 60 m, 35 m, 50 m, and 40 m. If fencing wire costs ₹12 per metre, find the total cost of fencing the park.
[2 Marks]
✅ Step-by-Step Solution & Working
Perimeter & Cost:
• Perimeter = 45 + 60 + 35 + 50 + 40 = 230 metres.
• Total cost = 230 m × ₹12 = ₹2,760.
• Perimeter = 45 + 60 + 35 + 50 + 40 = 230 metres.
• Total cost = 230 m × ₹12 = ₹2,760.
CBSE Step-by-Step Marking Scheme
Perimeter: 230 m
1 M
Cost: ₹2,760
1 M
Q22
Textbook Exercise • Square Pillow Border Lace
A square cushion cover has a side of 45 cm. (a) What length of lace is needed for 1 cushion? (b) What length of lace is needed for a set of 6 cushions in metres?
[3 Marks]
✅ Step-by-Step Solution & Working
Lace Requirements:
• (a) For 1 cushion: 4 × 45 cm = 180 cm.
• (b) For 6 cushions: 6 × 180 cm = 1,080 cm.
• In metres = 1,080 ÷ 100 = 10.8 metres.
• (a) For 1 cushion: 4 × 45 cm = 180 cm.
• (b) For 6 cushions: 6 × 180 cm = 1,080 cm.
• In metres = 1,080 ÷ 100 = 10.8 metres.
CBSE Step-by-Step Marking Scheme
1 cushion: 180 cm
1.5 M
6 cushions: 10.8 metres
1.5 M
Q23
Textbook Exercise • Window Grille Area
A square window grille consists of 16 smaller square panes, each of side 15 cm. (a) What is the total area of the window grille? (b) What is the outer perimeter of the window?
[3 Marks]
✅ Step-by-Step Solution & Working
Window Pane Calculations:
• Area of 1 pane = 15 × 15 = 225 sq cm.
• (a) Total area: 16 × 225 = 3,600 sq cm.
• (b) Outer dimensions: 16 square panes arranged as 4 × 4 grid.
Window side = 4 × 15 cm = 60 cm.
Perimeter = 4 × 60 cm = 240 cm.
• Area of 1 pane = 15 × 15 = 225 sq cm.
• (a) Total area: 16 × 225 = 3,600 sq cm.
• (b) Outer dimensions: 16 square panes arranged as 4 × 4 grid.
Window side = 4 × 15 cm = 60 cm.
Perimeter = 4 × 60 cm = 240 cm.
CBSE Step-by-Step Marking Scheme
Total area: 3,600 sq cm
1.5 M
Outer perimeter: 240 cm
1.5 M
Q24
Textbook Exercise • Unit of Area vs Unit of Perimeter
Why is perimeter measured in linear units (like cm or metres) while area is measured in square units (like sq cm or sq metres)? Explain clearly.
[2 Marks]
✅ Step-by-Step Solution & Working
Units Conceptual Distinction:
• Perimeter: Measures the total 1-dimensional length of the surrounding boundary line. It is a distance, so it is measured in linear units (cm, m).
• Area: Measures the 2-dimensional surface region enclosed inside the boundary. It counts how many unit squares (1 cm × 1 cm or 1 m × 1 m) fit inside, so it is measured in square units (sq cm, sq m).
• Perimeter: Measures the total 1-dimensional length of the surrounding boundary line. It is a distance, so it is measured in linear units (cm, m).
• Area: Measures the 2-dimensional surface region enclosed inside the boundary. It counts how many unit squares (1 cm × 1 cm or 1 m × 1 m) fit inside, so it is measured in square units (sq cm, sq m).
CBSE Step-by-Step Marking Scheme
1D boundary length concept for perimeter
1 M
2D square count concept for area
1 M
🌟 NEP 2020 Competency-Based Case Study
NEP 2020 Competency Case: Area and its Boundary
A practical scenario involving Area and its Boundary 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