Chapter 11 of 15
Chapter 11: Grandmother's Quilt
दादी की रजाई (क्षेत्रफल, परिमाप एवं वर्ग ग्रिड) | CBSE Session 2026–27
Core Pedagogical Competencies & Concepts
- Difference between Perimeter (length of outer boundary) and Area (measure of flat surface enclosed).
- Formulas: Perimeter of Rectangle = 2 × (Length + Breadth); Area of Rectangle = Length × Breadth.
- Formulas for Squares: Perimeter = 4 × Side; Area = Side × Side.
- Counting 1 cm grid squares to estimate areas of irregular shapes and quilt patches.
- Definition: Tessellation is the tiling of a flat two-dimensional plane using one or more geometric shapes with zero gaps and zero overlaps.
All Textbook Exercises & In-Text Tasks (24 Questions)
100% Complete Ingestion
Q1
Patchwork Quilt Dimensions • Perimeter & Area
Grandmother stitches a rectangular patchwork quilt measuring 240 cm in length and 150 cm in width.
(a) Calculate the total perimeter of the quilt to find how much decorative ribbon border she needs.
(b) Calculate the total area of the quilt in square centimetres and square metres.
[3 Marks]
(a) Calculate the total perimeter of the quilt to find how much decorative ribbon border she needs.
(b) Calculate the total area of the quilt in square centimetres and square metres.
✅ Step-by-Step Solution & Working
Perimeter & Area Formulas:
• (a) Perimeter: $P = 2 \times (L + B) = 2 \times (240 + 150) = 2 \times 390 = \mathbf{780\text{ cm}}$ (or 7.8 metres of ribbon).
• (b) Area:
- In cm²: $A = L \times B = 240 \times 150 = \mathbf{36,000\text{ sq cm}}$.
- In m²: $36,000 ÷ 10,000 = \mathbf{3.6\text{ sq metres}}$ ($1\text{ m}^2 = 10,000\text{ cm}^2$).
• (a) Perimeter: $P = 2 \times (L + B) = 2 \times (240 + 150) = 2 \times 390 = \mathbf{780\text{ cm}}$ (or 7.8 metres of ribbon).
• (b) Area:
- In cm²: $A = L \times B = 240 \times 150 = \mathbf{36,000\text{ sq cm}}$.
- In m²: $36,000 ÷ 10,000 = \mathbf{3.6\text{ sq metres}}$ ($1\text{ m}^2 = 10,000\text{ cm}^2$).
CBSE Step-by-Step Marking Scheme
Perimeter = 780 cm (7.8 m)
1.5 M
Area = 36,000 cm² = 3.6 m²
1.5 M
Q2
Square vs Rectangular Patches • Comparison
Grandmother compares two cloth patches:
• Patch 1: Square with side 20 cm.
• Patch 2: Rectangle measuring 25 cm by 16 cm.
(a) Find the perimeter of both patches.
(b) Find the area of both patches.
(c) What do you observe about their areas and perimeters?
[3 Marks]
• Patch 1: Square with side 20 cm.
• Patch 2: Rectangle measuring 25 cm by 16 cm.
(a) Find the perimeter of both patches.
(b) Find the area of both patches.
(c) What do you observe about their areas and perimeters?
✅ Step-by-Step Solution & Working
Calculations:
• (a) Perimeters:
- Square: $4 \times 20 = \mathbf{80\text{ cm}}$.
- Rectangle: $2 \times (25 + 16) = 2 \times 41 = \mathbf{82\text{ cm}}$.
• (b) Areas:
- Square: $20 \times 20 = \mathbf{400\text{ sq cm}}$.
- Rectangle: $25 \times 16 = \mathbf{400\text{ sq cm}}$.
• (c) Observation: Both patches have the exact same area (400 cm²), but the rectangular patch has a larger perimeter (82 cm > 80 cm). For a given area, a square always has the minimum perimeter.
• (a) Perimeters:
- Square: $4 \times 20 = \mathbf{80\text{ cm}}$.
- Rectangle: $2 \times (25 + 16) = 2 \times 41 = \mathbf{82\text{ cm}}$.
• (b) Areas:
- Square: $20 \times 20 = \mathbf{400\text{ sq cm}}$.
- Rectangle: $25 \times 16 = \mathbf{400\text{ sq cm}}$.
• (c) Observation: Both patches have the exact same area (400 cm²), but the rectangular patch has a larger perimeter (82 cm > 80 cm). For a given area, a square always has the minimum perimeter.
CBSE Step-by-Step Marking Scheme
Perimeters: 80 cm and 82 cm
1.0 M
Areas: 400 cm² each
1.0 M
Valid geometric observation
1.0 M
Q3
Tiling Quilt with Small Unit Squares
The quilt measures 200 cm by 120 cm. Grandmother wants to cover the entire quilt using square fabric patches of side 10 cm.
(a) How many patches fit along the length?
(b) How many patches fit along the width?
(c) How many total fabric patches are required?
[3 Marks]
(a) How many patches fit along the length?
(b) How many patches fit along the width?
(c) How many total fabric patches are required?
✅ Step-by-Step Solution & Working
Grid Tiling:
• (a) Along Length: 200 ÷ 10 = 20 patches.
• (b) Along Width: 120 ÷ 10 = 12 patches.
• (c) Total Patches: 20 × 12 = 240 patches.
(Check: Total area = 24,000 cm². Patch area = 100 cm². 24,000 ÷ 100 = 240).
• (a) Along Length: 200 ÷ 10 = 20 patches.
• (b) Along Width: 120 ÷ 10 = 12 patches.
• (c) Total Patches: 20 × 12 = 240 patches.
(Check: Total area = 24,000 cm². Patch area = 100 cm². 24,000 ÷ 100 = 240).
CBSE Step-by-Step Marking Scheme
20 along length, 12 along width
1.5 M
Total = 240 patches
1.5 M
Q4
Park Fencing & Turfing Cost
A rectangular community park is 45 m long and 30 m wide.
(a) Find the length of wire required to fence it with 3 rounds of barbed wire.
(b) If turfing the lawn costs ₹20 per square metre, calculate the total cost of planting grass across the entire park.
[3 Marks]
(a) Find the length of wire required to fence it with 3 rounds of barbed wire.
(b) If turfing the lawn costs ₹20 per square metre, calculate the total cost of planting grass across the entire park.
✅ Step-by-Step Solution & Working
Calculations:
• (a) Barbed Wire (Perimeter):
- Perimeter = 2 × (45 + 30) = 2 × 75 = 150 metres.
- 3 rounds of wire = 3 × 150 = 450 metres of wire.
• (b) Turfing Cost (Area):
- Area = 45 m × 30 m = 1,350 sq m.
- Cost = 1,350 × ₹20 = ₹27,000.
• (a) Barbed Wire (Perimeter):
- Perimeter = 2 × (45 + 30) = 2 × 75 = 150 metres.
- 3 rounds of wire = 3 × 150 = 450 metres of wire.
• (b) Turfing Cost (Area):
- Area = 45 m × 30 m = 1,350 sq m.
- Cost = 1,350 × ₹20 = ₹27,000.
CBSE Step-by-Step Marking Scheme
Perimeter = 150 m; 3 rounds = 450 m
1.5 M
Area = 1,350 m²; Cost = ₹27,000
1.5 M
Q5
Square Field • Missing Side from Perimeter
The perimeter of a square vegetable garden is 64 metres.
(a) What is the length of each side of the garden?
(b) What is the total area of the garden in square metres?
[3 Marks]
(a) What is the length of each side of the garden?
(b) What is the total area of the garden in square metres?
✅ Step-by-Step Solution & Working
Working:
• (a) Side Length: Perimeter of square = $4 \times \text{Side}$.
- Side = Perimeter ÷ 4 = 64 ÷ 4 = 16 metres.
• (b) Area: Area = Side × Side = 16 × 16 = 256 square metres.
• (a) Side Length: Perimeter of square = $4 \times \text{Side}$.
- Side = Perimeter ÷ 4 = 64 ÷ 4 = 16 metres.
• (b) Area: Area = Side × Side = 16 × 16 = 256 square metres.
CBSE Step-by-Step Marking Scheme
Side = 16 m
1.5 M
Area = 256 sq m
1.5 M
Q6
Missing Dimension from Area
A rectangular quilt has an area of 4,800 sq cm and a width of 60 cm.
(a) Find the length of the quilt.
(b) Find its perimeter.
[3 Marks]
(a) Find the length of the quilt.
(b) Find its perimeter.
✅ Step-by-Step Solution & Working
Calculations:
• (a) Length: Area = Length × Width → Length = Area ÷ Width.
- Length = 4,800 ÷ 60 = 80 cm.
• (b) Perimeter: $P = 2 \times (L + B) = 2 \times (80 + 60) = 2 \times 140 = \mathbf{280\text{ cm}}$.
• (a) Length: Area = Length × Width → Length = Area ÷ Width.
- Length = 4,800 ÷ 60 = 80 cm.
• (b) Perimeter: $P = 2 \times (L + B) = 2 \times (80 + 60) = 2 \times 140 = \mathbf{280\text{ cm}}$.
CBSE Step-by-Step Marking Scheme
Length = 80 cm
1.5 M
Perimeter = 280 cm
1.5 M
Q7
L-Shaped Composite Figure Area & Perimeter
An L-shaped cloth piece has horizontal base 12 cm, left vertical height 10 cm, top horizontal width 4 cm, interior vertical drop 6 cm, interior horizontal shelf 8 cm, and right vertical edge 4 cm.
(a) Find the perimeter of the figure by adding all outer sides.
(b) Find its area by splitting it into two rectangles.
[3 Marks]
(a) Find the perimeter of the figure by adding all outer sides.
(b) Find its area by splitting it into two rectangles.
✅ Step-by-Step Solution & Working
Composite Figure Geometry:
• (a) Perimeter: Sum of 6 sides = 12 + 10 + 4 + 6 + 8 + 4 = 44 cm.
• (b) Area by Splitting:
- Rectangle 1 (Left vertical column): 4 cm wide × 10 cm high = 40 cm².
- Rectangle 2 (Right base shelf): 8 cm long × 4 cm high = 32 cm².
- Total Area = 40 + 32 = 72 square cm.
• (a) Perimeter: Sum of 6 sides = 12 + 10 + 4 + 6 + 8 + 4 = 44 cm.
• (b) Area by Splitting:
- Rectangle 1 (Left vertical column): 4 cm wide × 10 cm high = 40 cm².
- Rectangle 2 (Right base shelf): 8 cm long × 4 cm high = 32 cm².
- Total Area = 40 + 32 = 72 square cm.
CBSE Step-by-Step Marking Scheme
Perimeter = 44 cm
1.5 M
Area = 40 + 32 = 72 cm²
1.5 M
Q8
Same Perimeter, Different Areas
Two rectangular frames both have a perimeter of 36 cm.
• Frame A has length 10 cm and width 8 cm.
• Frame B has length 14 cm and width 4 cm.
(a) Verify that both perimeters are 36 cm.
(b) Calculate the area of Frame A and Frame B.
(c) Which frame encloses more space and by how much?
[3 Marks]
• Frame A has length 10 cm and width 8 cm.
• Frame B has length 14 cm and width 4 cm.
(a) Verify that both perimeters are 36 cm.
(b) Calculate the area of Frame A and Frame B.
(c) Which frame encloses more space and by how much?
✅ Step-by-Step Solution & Working
Calculations:
• (a) Perimeter Verification:
- Frame A: 2 × (10 + 8) = 2 × 18 = 36 cm.
- Frame B: 2 × (14 + 4) = 2 × 18 = 36 cm. Both equal 36 cm!
• (b) Areas:
- Frame A: 10 × 8 = 80 sq cm.
- Frame B: 14 × 4 = 56 sq cm.
• (c) Comparison: Frame A encloses more area by: 80 - 56 = 24 sq cm more. (Shapes closer to a square maximize area).
• (a) Perimeter Verification:
- Frame A: 2 × (10 + 8) = 2 × 18 = 36 cm.
- Frame B: 2 × (14 + 4) = 2 × 18 = 36 cm. Both equal 36 cm!
• (b) Areas:
- Frame A: 10 × 8 = 80 sq cm.
- Frame B: 14 × 4 = 56 sq cm.
• (c) Comparison: Frame A encloses more area by: 80 - 56 = 24 sq cm more. (Shapes closer to a square maximize area).
CBSE Step-by-Step Marking Scheme
Perimeter verified = 36 cm
1.0 M
Areas: 80 cm² and 56 cm²
1.0 M
Frame A encloses 24 cm² more
1.0 M
Q9
Lace Border Cost • Currency Calculation
Grandmother buys golden lace to border 4 identical square pillow covers of side 45 cm.
(a) How many metres of lace does she need for all 4 pillow covers?
(b) If the lace costs ₹18 per metre, what is her total expenditure?
[3 Marks]
(a) How many metres of lace does she need for all 4 pillow covers?
(b) If the lace costs ₹18 per metre, what is her total expenditure?
✅ Step-by-Step Solution & Working
Calculations:
• (a) Lace Needed:
- Perimeter of 1 cover = 4 × 45 cm = 180 cm = 1.8 metres.
- 4 pillow covers = 4 × 1.8 m = 7.2 metres of lace.
• (b) Total Cost: 7.2 m × ₹18 = ₹129.60.
• (a) Lace Needed:
- Perimeter of 1 cover = 4 × 45 cm = 180 cm = 1.8 metres.
- 4 pillow covers = 4 × 1.8 m = 7.2 metres of lace.
• (b) Total Cost: 7.2 m × ₹18 = ₹129.60.
CBSE Step-by-Step Marking Scheme
Lace = 7.2 m
1.5 M
Cost = ₹129.60
1.5 M
Q10
Floor Tiling • Tile Count Problem
A bedroom floor is 5 m long and 4 m wide. The homeowner wants to tile it using square ceramic tiles of side 50 cm (0.5 m).
(a) What is the floor area in square metres?
(b) What is the area of 1 tile in square metres?
(c) How many tiles are needed to pave the floor?
[3 Marks]
(a) What is the floor area in square metres?
(b) What is the area of 1 tile in square metres?
(c) How many tiles are needed to pave the floor?
✅ Step-by-Step Solution & Working
Tiling Calculations:
• (a) Floor Area: 5 m × 4 m = 20 sq metres.
• (b) Tile Area: 0.5 m × 0.5 m = 0.25 sq metres.
• (c) Number of Tiles: 20 ÷ 0.25 = 80 tiles.
(Check: 10 tiles along length × 8 tiles along width = 80 tiles).
• (a) Floor Area: 5 m × 4 m = 20 sq metres.
• (b) Tile Area: 0.5 m × 0.5 m = 0.25 sq metres.
• (c) Number of Tiles: 20 ÷ 0.25 = 80 tiles.
(Check: 10 tiles along length × 8 tiles along width = 80 tiles).
CBSE Step-by-Step Marking Scheme
Floor area = 20 m²
1.0 M
Tile area = 0.25 m²
1.0 M
Number of tiles = 80
1.0 M
Q11
Grid Stamp Area • Irregular Shapes
On a 1 cm square grid, a postal stamp covers 6 complete squares, 4 half squares, and 2 squares that are more than half covered.
(a) Estimate the area of the stamp using NCERT grid estimation rules.
(b) What is the perimeter if the stamp is a 4 cm × 2 cm rectangle?
[3 Marks]
(a) Estimate the area of the stamp using NCERT grid estimation rules.
(b) What is the perimeter if the stamp is a 4 cm × 2 cm rectangle?
✅ Step-by-Step Solution & Working
Grid Area Counting Rules:
1. Complete squares: 6 × 1 = 6 sq cm.
2. Half squares: 4 × 0.5 = 2 sq cm.
3. More than half squares count as 1: 2 × 1 = 2 sq cm.
• (a) Estimated Area: 6 + 2 + 2 = 10 square cm.
• (b) Rectangle Perimeter: $2 \times (4 + 2) = 2 \times 6 = \mathbf{12\text{ cm}}$.
1. Complete squares: 6 × 1 = 6 sq cm.
2. Half squares: 4 × 0.5 = 2 sq cm.
3. More than half squares count as 1: 2 × 1 = 2 sq cm.
• (a) Estimated Area: 6 + 2 + 2 = 10 square cm.
• (b) Rectangle Perimeter: $2 \times (4 + 2) = 2 \times 6 = \mathbf{12\text{ cm}}$.
CBSE Step-by-Step Marking Scheme
Grid counting rules applied
1.5 M
Estimated area = 10 cm²; Perimeter = 12 cm
1.5 M
Q12
Square Cutouts from Rectangle
A rectangular cardboard sheet is 30 cm long and 20 cm wide.
(a) How many small squares of side 5 cm can be cut from it without wasting any paper?
(b) What is the total area of the cardboard sheet?
[3 Marks]
(a) How many small squares of side 5 cm can be cut from it without wasting any paper?
(b) What is the total area of the cardboard sheet?
✅ Step-by-Step Solution & Working
Working:
• (a) Squares Cut:
- Along length: 30 ÷ 5 = 6 squares.
- Along width: 20 ÷ 5 = 4 squares.
- Total squares = 6 × 4 = 24 squares.
• (b) Sheet Area: 30 × 20 = 600 square cm (Check: 24 × 25 cm² = 600 cm²).
• (a) Squares Cut:
- Along length: 30 ÷ 5 = 6 squares.
- Along width: 20 ÷ 5 = 4 squares.
- Total squares = 6 × 4 = 24 squares.
• (b) Sheet Area: 30 × 20 = 600 square cm (Check: 24 × 25 cm² = 600 cm²).
CBSE Step-by-Step Marking Scheme
24 squares
1.5 M
Sheet area = 600 cm²
1.5 M
Q13
Changing Dimensions • Effect on Area
(a) If the length of a rectangle is doubled while its width remains the same, how does its area change?
(b) If both the length and width of a rectangle are doubled, by how many times does its area increase?
[3 Marks]
(b) If both the length and width of a rectangle are doubled, by how many times does its area increase?
✅ Step-by-Step Solution & Working
Area Scaling Factor:
• (a) Doubling Length Only: Original area = $L \times B$. New area = $(2L) \times B = 2 \times (L \times B)$. The area doubles (2 times).
• (b) Doubling Both Length and Width: New area = $(2L) \times (2B) = 4 \times (L \times B)$. The area quadruples (4 times larger).
• (a) Doubling Length Only: Original area = $L \times B$. New area = $(2L) \times B = 2 \times (L \times B)$. The area doubles (2 times).
• (b) Doubling Both Length and Width: New area = $(2L) \times (2B) = 4 \times (L \times B)$. The area quadruples (4 times larger).
CBSE Step-by-Step Marking Scheme
Part (a): Doubles (2 times)
1.5 M
Part (b): Quadruples (4 times)
1.5 M
Q14
Triangular Patch Area inside Rectangle
A triangular patchwork piece is made by cutting a 10 cm × 6 cm rectangular cloth diagonally into two equal halves.
(a) What is the area of the original rectangle?
(b) What is the area of each triangular patch?
[3 Marks]
(a) What is the area of the original rectangle?
(b) What is the area of each triangular patch?
✅ Step-by-Step Solution & Working
Diagonal Division of Rectangle:
• (a) Rectangle Area: $L \times B = 10 \times 6 = \mathbf{60\text{ sq cm}}$.
• (b) Triangular Patch Area: The diagonal divides the rectangle into two identical triangles.
- Area of triangle = Half of rectangle area = 60 ÷ 2 = 30 square cm.
• (a) Rectangle Area: $L \times B = 10 \times 6 = \mathbf{60\text{ sq cm}}$.
• (b) Triangular Patch Area: The diagonal divides the rectangle into two identical triangles.
- Area of triangle = Half of rectangle area = 60 ÷ 2 = 30 square cm.
CBSE Step-by-Step Marking Scheme
Rectangle area = 60 cm²
1.5 M
Triangle area = 30 cm²
1.5 M
Q15
Walkway Pathway around a Lawn
A square lawn of side 20 metres is surrounded on the outside by a 2-metre-wide concrete walking path.
(a) Find the side length of the outer square including the path.
(b) Calculate the total area of the concrete walking path.
[3 Marks]
(a) Find the side length of the outer square including the path.
(b) Calculate the total area of the concrete walking path.
✅ Step-by-Step Solution & Working
Concentric Squares Area:
• (a) Outer Square Side: 20 m + 2 m (left) + 2 m (right) = 24 metres.
• (b) Pathway Area:
- Outer square area = 24 × 24 = 576 sq m.
- Inner lawn area = 20 × 20 = 400 sq m.
- Path area = Outer area - Inner area = 576 - 400 = 176 square metres.
• (a) Outer Square Side: 20 m + 2 m (left) + 2 m (right) = 24 metres.
• (b) Pathway Area:
- Outer square area = 24 × 24 = 576 sq m.
- Inner lawn area = 20 × 20 = 400 sq m.
- Path area = Outer area - Inner area = 576 - 400 = 176 square metres.
CBSE Step-by-Step Marking Scheme
Outer side = 24 m
1.0 M
Outer area = 576, Inner area = 400
1.0 M
Path area = 176 sq m
1.0 M
Q16
Perimeter of Regular Shapes
Find the perimeter of:
(a) An equilateral triangle of side 14 cm
(b) A regular pentagon of side 9 cm
(c) A regular octagon of side 6 cm
[3 Marks]
(a) An equilateral triangle of side 14 cm
(b) A regular pentagon of side 9 cm
(c) A regular octagon of side 6 cm
✅ Step-by-Step Solution & Working
Perimeter Formulas:
• (a) Equilateral Triangle (3 sides): 3 × 14 = 42 cm.
• (b) Regular Pentagon (5 sides): 5 × 9 = 45 cm.
• (c) Regular Octagon (8 sides): 8 × 6 = 48 cm.
• (a) Equilateral Triangle (3 sides): 3 × 14 = 42 cm.
• (b) Regular Pentagon (5 sides): 5 × 9 = 45 cm.
• (c) Regular Octagon (8 sides): 8 × 6 = 48 cm.
CBSE Step-by-Step Marking Scheme
Triangle = 42 cm
1.0 M
Pentagon = 45 cm
1.0 M
Octagon = 48 cm
1.0 M
Q17
Comparing Ribbon Perimeter to String Length
A girl has a 60 cm piece of string. She bends it to form:
(a) An equilateral triangle
(b) A square
(c) A regular hexagon
Find the length of each side for each shape.
[3 Marks]
(a) An equilateral triangle
(b) A square
(c) A regular hexagon
Find the length of each side for each shape.
✅ Step-by-Step Solution & Working
Side Lengths from Fixed Perimeter (P = 60 cm):
• (a) Equilateral Triangle (3 sides): 60 ÷ 3 = 20 cm.
• (b) Square (4 sides): 60 ÷ 4 = 15 cm.
• (c) Regular Hexagon (6 sides): 60 ÷ 6 = 10 cm.
• (a) Equilateral Triangle (3 sides): 60 ÷ 3 = 20 cm.
• (b) Square (4 sides): 60 ÷ 4 = 15 cm.
• (c) Regular Hexagon (6 sides): 60 ÷ 6 = 10 cm.
CBSE Step-by-Step Marking Scheme
Triangle side = 20 cm
1.0 M
Square side = 15 cm
1.0 M
Hexagon side = 10 cm
1.0 M
Q18
Cost of Fabric per Square Metre
A tailor makes 5 curtains, each measuring 2 metres by 1.5 metres.
(a) Find the area of one curtain.
(b) Find the total cloth area required for all 5 curtains.
(c) If the curtain cloth costs ₹160 per square metre, what is the total cost of fabric?
[3 Marks]
(a) Find the area of one curtain.
(b) Find the total cloth area required for all 5 curtains.
(c) If the curtain cloth costs ₹160 per square metre, what is the total cost of fabric?
✅ Step-by-Step Solution & Working
Working:
• (a) One Curtain Area: 2 m × 1.5 m = 3 sq metres.
• (b) Total Fabric Area: 5 × 3 sq m = 15 sq metres.
• (c) Fabric Cost: 15 sq m × ₹160 = ₹2,400.
• (a) One Curtain Area: 2 m × 1.5 m = 3 sq metres.
• (b) Total Fabric Area: 5 × 3 sq m = 15 sq metres.
• (c) Fabric Cost: 15 sq m × ₹160 = ₹2,400.
CBSE Step-by-Step Marking Scheme
One curtain = 3 m²
1.0 M
Total cloth = 15 m²
1.0 M
Total cost = ₹2,400
1.0 M
Q19
HOTS • Area & Perimeter Puzzle
A wire bent in the shape of a rectangle of 18 cm × 12 cm is straightened and then bent into the shape of a square. Which shape has a larger area, and by how much?
[3 Marks]
✅ Step-by-Step Solution & Working
Isoperimetric Problem:
1. Wire Length (Perimeter): 2 × (18 + 12) = 2 × 30 = 60 cm.
2. Square Side: 60 ÷ 4 = 15 cm.
3. Areas:
• Rectangle Area = 18 × 12 = 216 sq cm.
• Square Area = 15 × 15 = 225 sq cm.
4. Comparison: The Square has a larger area by: 225 - 216 = 9 sq cm.
1. Wire Length (Perimeter): 2 × (18 + 12) = 2 × 30 = 60 cm.
2. Square Side: 60 ÷ 4 = 15 cm.
3. Areas:
• Rectangle Area = 18 × 12 = 216 sq cm.
• Square Area = 15 × 15 = 225 sq cm.
4. Comparison: The Square has a larger area by: 225 - 216 = 9 sq cm.
CBSE Step-by-Step Marking Scheme
Perimeter = 60 cm, Square side = 15 cm
1.0 M
Areas: 216 cm² and 225 cm²
1.0 M
Square is larger by 9 cm²
1.0 M
Q20
Table Top Varnish • Rate per Unit
A rectangular dining table top measures 2.2 m by 1.1 m.
(a) Find the area of the table top.
(b) If polishing the surface costs ₹150 per square metre, calculate the total cost of polishing.
[3 Marks]
(a) Find the area of the table top.
(b) If polishing the surface costs ₹150 per square metre, calculate the total cost of polishing.
✅ Step-by-Step Solution & Working
Calculations:
• (a) Table Area: 2.2 m × 1.1 m = 2.42 sq metres.
• (b) Polishing Cost: 2.42 × ₹150 = ₹363.
• (a) Table Area: 2.2 m × 1.1 m = 2.42 sq metres.
• (b) Polishing Cost: 2.42 × ₹150 = ₹363.
CBSE Step-by-Step Marking Scheme
Area = 2.42 m²
1.5 M
Polishing cost = ₹363
1.5 M
Q21
Square Tile Border Framing
A picture measuring 24 cm by 18 cm is pasted on a sheet of cardboard such that there is a margin of 2 cm along each side.
(a) Find the total length and breadth of the cardboard.
(b) Find the area of the margin.
[3 Marks]
(a) Find the total length and breadth of the cardboard.
(b) Find the area of the margin.
✅ Step-by-Step Solution & Working
Calculations:
• (a) Cardboard Dimensions:
- Length = 24 + 2 + 2 = 28 cm.
- Breadth = 18 + 2 + 2 = 22 cm.
• (b) Margin Area:
- Cardboard area = 28 × 22 = 616 sq cm.
- Picture area = 24 × 18 = 432 sq cm.
- Margin area = 616 - 432 = 184 square cm.
• (a) Cardboard Dimensions:
- Length = 24 + 2 + 2 = 28 cm.
- Breadth = 18 + 2 + 2 = 22 cm.
• (b) Margin Area:
- Cardboard area = 28 × 22 = 616 sq cm.
- Picture area = 24 × 18 = 432 sq cm.
- Margin area = 616 - 432 = 184 square cm.
CBSE Step-by-Step Marking Scheme
Cardboard: 28 cm × 22 cm
1.5 M
Margin area = 184 cm²
1.5 M
Q22
Mental Arithmetic • Perimeter Shortcuts
Find the perimeter of:
(a) A square of side 25 cm
(b) A rectangle of 35 m by 15 m
(c) A rhombus of side 12.5 cm
[3 Marks]
(a) A square of side 25 cm
(b) A rectangle of 35 m by 15 m
(c) A rhombus of side 12.5 cm
✅ Step-by-Step Solution & Working
Shortcuts:
• (a) Square: 4 × 25 = 100 cm (1 metre).
• (b) Rectangle: 2 × (35 + 15) = 2 × 50 = 100 metres.
• (c) Rhombus: 4 × 12.5 = 50 cm.
• (a) Square: 4 × 25 = 100 cm (1 metre).
• (b) Rectangle: 2 × (35 + 15) = 2 × 50 = 100 metres.
• (c) Rhombus: 4 × 12.5 = 50 cm.
CBSE Step-by-Step Marking Scheme
Part (a): 100 cm
1.0 M
Part (b): 100 m
1.0 M
Part (c): 50 cm
1.0 M
Q23
True or False with Geometric Reasoning
State True or False with reasoning:
(a) Two shapes with the same perimeter must have the same area.
(b) The perimeter of any rectangle is always greater than its area.
(c) A 1 cm² square fits inside a 1 cm² circle.
[3 Marks]
(a) Two shapes with the same perimeter must have the same area.
(b) The perimeter of any rectangle is always greater than its area.
(c) A 1 cm² square fits inside a 1 cm² circle.
✅ Step-by-Step Solution & Working
Justifications:
• (a) False: For example, rectangles with dimensions 5×1 (P=12, A=5) and 3×3 (P=12, A=9) have the same perimeter but different areas.
• (b) False: Perimeter and area have different units (cm vs cm²) and cannot be directly compared; moreover, a rectangle of 100×100 has area 10,000 cm² and perimeter only 400 cm.
• (c) False: A 1 cm² circle has diameter ≈ 1.13 cm, while a 1 cm² square has a diagonal of $\sqrt{2} \approx 1.41\text{ cm}$ which exceeds the diameter of the circle.
• (a) False: For example, rectangles with dimensions 5×1 (P=12, A=5) and 3×3 (P=12, A=9) have the same perimeter but different areas.
• (b) False: Perimeter and area have different units (cm vs cm²) and cannot be directly compared; moreover, a rectangle of 100×100 has area 10,000 cm² and perimeter only 400 cm.
• (c) False: A 1 cm² circle has diameter ≈ 1.13 cm, while a 1 cm² square has a diagonal of $\sqrt{2} \approx 1.41\text{ cm}$ which exceeds the diameter of the circle.
CBSE Step-by-Step Marking Scheme
Part (a): False with counter-example
1.0 M
Part (b): False with dimension explanation
1.0 M
Part (c): False with diagonal reasoning
1.0 M
Q24
Comprehensive Quilt Masterpiece Problem
Grandmother designs a royal quilt of 2.5 m × 2.0 m.
• Central medallion: 1.5 m × 1.0 m silk brocade.
• Surrounding border: Printed cotton patchwork.
(a) Find the total area of the quilt in square metres.
(b) What is the area of the silk central medallion?
(c) What is the area of the patchwork border?
(d) If satin lace costs ₹25 per metre, how much will it cost to border the outer edge of the quilt?
[3 Marks]
• Central medallion: 1.5 m × 1.0 m silk brocade.
• Surrounding border: Printed cotton patchwork.
(a) Find the total area of the quilt in square metres.
(b) What is the area of the silk central medallion?
(c) What is the area of the patchwork border?
(d) If satin lace costs ₹25 per metre, how much will it cost to border the outer edge of the quilt?
✅ Step-by-Step Solution & Working
Comprehensive Geometry of Quilt:
• (a) Total Quilt Area: 2.5 m × 2.0 m = 5.0 sq metres.
• (b) Central Medallion Area: 1.5 m × 1.0 m = 1.5 sq metres.
• (c) Patchwork Border Area: Total area - Medallion area = 5.0 - 1.5 = 3.5 sq metres.
• (d) Outer Lace Cost:
- Perimeter = 2 × (2.5 + 2.0) = 2 × 4.5 = 9.0 metres.
- Cost = 9.0 m × ₹25 = ₹225.
• (a) Total Quilt Area: 2.5 m × 2.0 m = 5.0 sq metres.
• (b) Central Medallion Area: 1.5 m × 1.0 m = 1.5 sq metres.
• (c) Patchwork Border Area: Total area - Medallion area = 5.0 - 1.5 = 3.5 sq metres.
• (d) Outer Lace Cost:
- Perimeter = 2 × (2.5 + 2.0) = 2 × 4.5 = 9.0 metres.
- Cost = 9.0 m × ₹25 = ₹225.
CBSE Step-by-Step Marking Scheme
Total area = 5.0 m²
0.75 M
Medallion = 1.5 m²
0.75 M
Border area = 3.5 m²
0.75 M
Lace cost = ₹225
0.75 M
🌟 NEP 2020 Competency-Based Case Study
Competency Question: School Garden Lawn Turfing
A school garden has a square lawn of side 15 metres. Grass turf tiles of size 50 cm by 50 cm cost ₹40 each.
Analytical Task: (a) What is the area of the lawn in square metres?
(b) How many turf tiles are needed to cover the entire lawn?
(c) What is the total cost of turfing?
(b) How many turf tiles are needed to cover the entire lawn?
(c) What is the total cost of turfing?
Evaluation & Model Answer:
Suggested Model Solution:
(a) Lawn Area: 15 × 15 = 225 sq metres.
(b) Number of tiles: Tile = 0.25 sq m ⇒ 225/0.25 = 900 tiles.
(c) Total Cost: 900 × ₹40 = ₹36,000.
Rubric: 1M concept identification + 1M step execution + 1M conclusion