Class 6 : Maths ( English ) β Lesson 6. Perimeter and Area
EXPLANATION AND ANALYSIS
πΏ 1. Introduction: Measuring Boundaries and Surfaces
In daily life, we often measure how much space something covers or how long its boundary is. For example, we may want to know how much fencing is needed around a garden π± or how much carpet is required to cover a room π . Mathematics gives us clear ideas and methods to measure such quantities through the concepts of perimeter and area.
π΅ Perimeter deals with the boundary of a shape
π’ Area deals with the surface covered by a shape
π‘ Both are essential for practical planning
π΄ This chapter builds strong measurement sense
π§ 2. What Is Perimeter?
The perimeter of a closed figure is the total length of its boundary.
πΉ It is found by adding the lengths of all sides
πΉ It tells us how far we go around a shape
πΉ Perimeter is measured in units of length (cm, m, km)
π‘ Concept:
Perimeter = sum of all sides
βοΈ Note:
Perimeter is related to the outline, not the inside region.
π± 3. Perimeter of Common Shapes
π΅ Perimeter of a square
πΉ All sides are equal
πΉ If one side = a, then perimeter = 4 Γ a
π’ Perimeter of a rectangle
πΉ Opposite sides are equal
πΉ If length = l and breadth = b
πΈ Perimeter = 2 Γ (l + b)
π‘ Concept:
Equal sides simplify perimeter calculations.
π§ 4. Perimeter of Irregular Shapes
Not all shapes are regular.
πΉ Irregular shapes have sides of different lengths
πΉ Perimeter is found by adding all given sides
πΉ No special formula is needed
βοΈ Note:
Careful addition is important to avoid mistakes.
πΏ 5. Units of Perimeter
Perimeter is always measured in linear units.
π΅ Common units: centimetre, metre, kilometre
π’ Smaller shapes use cm, larger ones use m or km
π‘ Units must be same before adding lengths
π‘ Concept:
Never add lengths with different units directly.
π§ 6. What Is Area?
The area of a figure is the amount of surface it covers.
πΉ It shows how much space lies inside a shape
πΉ Area is measured in square units
πΉ Examples: square centimetre, square metre
π‘ Concept:
Area = surface covered inside a boundary
βοΈ Note:
Area always uses square units.
π± 7. Area of a Rectangle
A rectangle covers space in rows and columns.
π΅ Length shows number of units along one direction
π’ Breadth shows number of units along the other direction
π‘ Area = length Γ breadth
πΉ If l = 6 cm and b = 4 cm
πΈ Area = 6 Γ 4 = 24 square cm
π‘ Concept:
Area of rectangle = l Γ b
π§ 8. Area of a Square
A square is a special rectangle.
π΅ All sides are equal
π’ If side = a
πΈ Area = a Γ a = aΒ²
βοΈ Note:
Square area grows faster than perimeter as side increases.
πΏ 9. Units of Area
Area is measured in square units.
π΅ Small areas β square centimetre
π’ Larger areas β square metre
π‘ Very large areas β hectare or square kilometre
π‘ Concept:
Area unit = (length unit)Β²
π§ 10. Difference Between Perimeter and Area
Understanding the difference is very important.
π΅ Perimeter measures boundary length
π’ Area measures surface covered
π‘ Perimeter uses linear units
π΄ Area uses square units
βοΈ Note:
Two shapes may have same perimeter but different areas.
π 11. Perimeter and Area in Daily Life
These concepts are used everywhere.
π΅ Fencing a field β perimeter
π’ Laying tiles on floor β area
π‘ Painting a wall β area
π΄ Wiring around a park β perimeter
π‘ Concept:
Measurement helps in saving time, money, and resources.
π§ 12. Importance of Perimeter and Area
This chapter helps students:
πΉ Measure boundaries accurately
πΉ Calculate space correctly
πΉ Understand geometry practically
πΉ Prepare for advanced topics like mensuration
π‘ Concept:
Perimeter and area connect mathematics with real life.
π Summary
The chapter Perimeter and Area introduces two important measurement concepts. Perimeter is the total length of the boundary of a closed figure, while area is the measure of the surface enclosed by that boundary. Students learn how to calculate the perimeter of squares, rectangles, and irregular shapes by adding side lengths. Area is understood as the space covered inside a shape and is measured in square units.
The formulas for the area of a rectangle and a square are explained clearly. The chapter also highlights the difference between perimeter and area and shows their importance in daily life situations like fencing, flooring, and painting. These ideas form the foundation for further study in geometry and mensuration.
π Quick Recap
π’ Perimeter is the total boundary length
π‘ Area is the surface covered
π΅ Square perimeter = 4 Γ side
π΄ Rectangle area = length Γ breadth
β‘ Area uses square units
π§ These concepts are used in daily life
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TEXTBOOK QUESTIONS
πΏ FIGURE IT OUT
πΏ BASED ON PERIMETER
π β Question 1.
Find the missing terms:
π β (a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?
π β
Answer:
π΅ Step 1: Formula of perimeter of rectangle
Perimeter = 2 Γ (length + breadth)
π΅ Step 2: Substitute given values
14 = 2 Γ (length + 2)
π΅ Step 3: Divide both sides by 2
7 = length + 2
π΅ Step 4: Subtract 2
length = 5
βοΈ Final: Length = 5 cm
π β (b) Perimeter of a square = 20 cm; side of a length = ?
π β
Answer:
π΅ Step 1: Formula of perimeter of square
Perimeter = 4 Γ side
π΅ Step 2: Substitute value
20 = 4 Γ side
π΅ Step 3: Divide by 4
side = 5
βοΈ Final: Side = 5 cm
π β (c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?
π β
Answer:
π΅ Step 1: Perimeter = 2 Γ (length + breadth)
π΅ Step 2: Substitute values
12 = 2 Γ (3 + breadth)
π΅ Step 3: Divide both sides by 2
6 = 3 + breadth
π΅ Step 4: Subtract 3
breadth = 3
βοΈ Final: Breadth = 3 m
π β Question 2.
A rectangle having sidelengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?
π β
Answer:
π΅ Step 1: Find perimeter of rectangle
Perimeter = 2 Γ (5 + 3) = 2 Γ 8 = 16 cm
π΅ Step 2: Same wire forms square, so perimeter remains same
Perimeter of square = 16 cm
π΅ Step 3: Side of square = Perimeter Γ· 4
Side = 16 Γ· 4 = 4
βοΈ Final: Side of square = 4 cm
π β Question 3.
Find the length of the third side of a triangle having a perimeter of 55 cm and having two sides of length 20 cm and 14 cm.
π β
Answer:
π΅ Step 1: Perimeter of triangle
Perimeter = sum of all three sides
π΅ Step 2: Substitute values
55 = 20 + 14 + third side
π΅ Step 3: Add known sides
55 = 34 + third side
π΅ Step 4: Subtract 34
third side = 21
βοΈ Final: Third side = 21 cm
π β Question 4.
What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs βΉ40 per metre?
π β
Answer:
π΅ Step 1: Find perimeter of the park
Perimeter = 2 Γ (150 + 120)
= 2 Γ 270
= 540 m
π΅ Step 2: Cost per metre = βΉ40
π΅ Step 3: Total cost
= 540 Γ 40
= 21600
βοΈ Final: Cost = βΉ21,600
π β Question 5.
A piece of string is 36 cm long. What will be the length of each side, if it is used to form:
π β (a) A square
π β
Answer:
π΅ Step 1: Perimeter of square = 36 cm
π΅ Step 2: Side = 36 Γ· 4 = 9
βοΈ Final: Each side = 9 cm
π β (b) A triangle with all sides of equal length
π β
Answer:
π΅ Step 1: Perimeter of triangle = 36 cm
π΅ Step 2: Number of sides = 3
π΅ Step 3: Side = 36 Γ· 3 = 12
βοΈ Final: Each side = 12 cm
π β (c) A hexagon with sides of equal length
π β
Answer:
π΅ Step 1: Perimeter of hexagon = 36 cm
π΅ Step 2: Number of sides = 6
π΅ Step 3: Side = 36 Γ· 6 = 6
βοΈ Final: Each side = 6 cm
π β Question 6.
A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed?
π β
Answer:
π΅ Step 1: Perimeter of the field
Perimeter = 2 Γ (230 + 160)
= 2 Γ 390
= 780 m
π΅ Step 2: Number of rounds = 3
π΅ Step 3: Total rope needed
= 3 Γ 780
= 2340
βοΈ Final: Total length of rope = 2340 m
πΏ MATHA PACHCHI!
π β Question 1.
Find out the total distance Akshi has covered in 5 rounds.
π β
Answer:
π΅ Step 1: Akshi runs on the outer rectangular track
π΅ Step 2: Length = 70 m, Breadth = 40 m
π΅ Step 3: Perimeter of rectangle
Perimeter = 2 Γ (length + breadth)
= 2 Γ (70 + 40)
= 2 Γ 110
= 220 m
π΅ Step 4: Distance covered in 5 rounds
= 5 Γ 220
= 1100 m
βοΈ Final: Akshi covered 1100 m in 5 rounds
π β Question 2.
Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance?
π β
Answer:
π΅ Step 1: Toshi runs on the inner rectangular track
π΅ Step 2: Length = 60 m, Breadth = 30 m
π΅ Step 3: Perimeter of inner track
Perimeter = 2 Γ (60 + 30)
= 2 Γ 90
= 180 m
π΅ Step 4: Distance covered in 7 rounds
= 7 Γ 180
= 1260 m
π΅ Step 5: Compare distances
Akshi = 1100 m
Toshi = 1260 m
βοΈ Final: Toshi ran a longer distance
π β Question 3.
Think and mark the positions as directedβ
π β (a) Mark βAβ at the point where Akshi will be after she ran 250 m.
π β
Answer:
π΅ Step 1: One full round of Akshi = 220 m
π΅ Step 2: After 250 m, extra distance = 250 β 220 = 30 m
π΅ Step 3: Starting from Akshiβs starting point, move 30 m along the track
βοΈ Final: Point A is 30 m ahead of Akshiβs starting point on the outer track
π β (b) Mark βBβ at the point where Akshi will be after she ran 500 m.
π β
Answer:
π΅ Step 1: One round = 220 m
π΅ Step 2: Distance after 2 rounds = 2 Γ 220 = 440 m
π΅ Step 3: Extra distance = 500 β 440 = 60 m
βοΈ Final: Point B is 60 m ahead of Akshiβs starting point on the outer track
π β (c) Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as βCβ.
π β
Answer:
π΅ Step 1: One round = 220 m
π΅ Step 2: Number of full rounds
1000 Γ· 220 = 4 full rounds (remainder left)
π΅ Step 3: Distance after 4 rounds
= 4 Γ 220 = 880 m
π΅ Step 4: Extra distance = 1000 β 880 = 120 m
βοΈ Final:
β’ Akshi completed 4 full rounds
β’ Point C is 120 m ahead of her starting point
π β (d) Mark βXβ at the point where Toshi will be after she ran 250 m.
π β
Answer:
π΅ Step 1: One round of Toshi = 180 m
π΅ Step 2: Extra distance after one round
= 250 β 180 = 70 m
βοΈ Final: Point X is 70 m ahead of Toshiβs starting point on the inner track
π β (e) Mark βYβ at the point where Toshi will be after she ran 500 m.
π β
Answer:
π΅ Step 1: Two rounds = 2 Γ 180 = 360 m
π΅ Step 2: Extra distance = 500 β 360 = 140 m
βοΈ Final: Point Y is 140 m ahead of Toshiβs starting point on the inner track
π β (f) Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as βZβ.
π β
Answer:
π΅ Step 1: One round = 180 m
π΅ Step 2: Number of full rounds
1000 Γ· 180 = 5 full rounds
π΅ Step 3: Distance after 5 rounds
= 5 Γ 180 = 900 m
π΅ Step 4: Extra distance = 1000 β 900 = 100 m
βοΈ Final:
β’ Toshi completed 5 full rounds
β’ Point Z is 100 m ahead of her starting point
πΏ AREA
π β Question 1.
The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden?
π β
Answer:
π΅ Step 1: Formula for area of a rectangle
Area = length Γ width
π΅ Step 2: Substitute given values
300 = 25 Γ width
π΅ Step 3: Divide both sides by 25
width = 300 Γ· 25
π΅ Step 4: Calculate
width = 12
βοΈ Final: Width of the garden = 12 m
π β Question 2.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of βΉ8 per hundred sq m?
π β
Answer:
π΅ Step 1: Find area of the plot
Area = 500 Γ 200 = 100000 sq m
π΅ Step 2: Rate given is βΉ8 per 100 sq m
π΅ Step 3: Number of 100 sq m units
100000 Γ· 100 = 1000
π΅ Step 4: Find total cost
Cost = 1000 Γ 8
βοΈ Final: Cost of tiling = βΉ8000
π β Question 3.
A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove?
π β
Answer:
π΅ Step 1: Find area of the grove
Area = 100 Γ 50 = 5000 sq m
π΅ Step 2: Area required per tree = 25 sq m
π΅ Step 3: Number of trees
Number of trees = 5000 Γ· 25
π΅ Step 4: Calculate
Number of trees = 200
βοΈ Final: Maximum number of trees = 200
π β Question 4.
By splitting the following figures into rectangles, find their areas (all measures are given in metres).
π β (a)
π β
Answer:
π΅ Step 1: Split the figure into three rectangles
π΅ Rectangle 1 (bottom left):
Length = 3 m, Breadth = 4 m
Area = 3 Γ 4 = 12 sq m
π΅ Rectangle 2 (middle):
Length = 4 m, Breadth = 3 m
Area = 4 Γ 3 = 12 sq m
π΅ Rectangle 3 (top right):
Length = 3 m, Breadth = 1 m
Area = 3 Γ 1 = 3 sq m
π΅ Step 2: Add all areas
Total area = 12 + 12 + 3
βοΈ Final: Area of figure (a) = 27 sq m
π β (b)
π β
Answer:
π΅ Step 1: Treat the figure as a big rectangle minus a small rectangle
π΅ Big rectangle:
Length = 5 m, Breadth = 3 m
Area = 5 Γ 3 = 15 sq m
π΅ Inner cut rectangle:
Length = 3 m, Breadth = 2 m
Area = 3 Γ 2 = 6 sq m
π΅ Step 2: Subtract inner area
Required area = 15 β 6
βοΈ Final: Area of figure (b) = 9 sq m
FIGURE IT OUT
π β Question 1
Explore and figure out how many pieces have the same area.
π β
Answer:
πΉ To compare areas, we place one tangram piece over another.
πΉ If two pieces completely cover each other, their areas are equal.
πΉ On careful observation:
πΈ Shape A and Shape B cover each other exactly.
πΈ Shape C and Shape E also cover each other exactly.
βοΈ Conclusion:
πΉ A and B have the same area.
πΉ C and E have the same area.
π β Question 2
How many times bigger is Shape D as compared to Shape C?
What is the relationship between Shapes C, D and E?
π β
Answer:
πΉ When Shape C and Shape E are placed together, they fit perfectly into Shape D.
πΉ This shows that Shape D is made of two equal smaller pieces.
βοΈ Therefore:
πΉ Area of D = Area of C + Area of E
πΉ Area of D = 2 Γ Area of C
βοΈ Relationship:
πΉ C and E have equal area.
πΉ D has twice the area of C and E.
π β Question 3
Which shape has more area: Shape D or Shape F? Give reasons.
π β
Answer:
πΉ Shape D is formed using two small pieces (C and E).
πΉ Shape F is a larger piece and cannot be fully covered by Shape D.
βοΈ Conclusion:
πΉ Shape F has more area than Shape D.
π β Question 4
Which shape has more area: Shape F or Shape G? Give reasons.
π β
Answer:
πΉ When Shape F and Shape G are rearranged and placed over each other, they cover the same region.
πΉ Rotation or flipping does not change area.
βοΈ Conclusion:
πΉ Shape F and Shape G have equal area.
π β Question 5
What is the area of Shape A as compared to Shape G?
Is it twice as big? Four times as big?
π β
Answer:
πΉ By placing Shape G repeatedly on Shape A, we observe that two Shapes G exactly cover Shape A.
βοΈ Therefore:
πΉ Area of A = 2 Γ Area of G
βοΈ Shape A is twice as big as Shape G.
π β Question 6
Find the area of the big square formed with all seven pieces in terms of the area of Shape C.
π β
Answer:
πΉ Let the area of Shape C be 1 unit.
πΉ From comparisons:
πΉ C = 1 unit
πΉ E = 1 unit
πΉ D = 2 units
πΉ A = 2 units
πΉ B = 2 units
πΉ F = 2 units
πΉ G = 2 units
πΉ Total area of big square
= 1 + 1 + 2 + 2 + 2 + 2 + 2
= 12 units
βοΈ Area of the big square = 12 times the area of Shape C.
π β Question 7
Arrange the 7 pieces to form a rectangle.
What will be the area of this rectangle in terms of the area of Shape C? Give reasons.
π β
Answer:
πΉ Rearranging pieces changes the shape, not the area.
πΉ All seven pieces are still used completely.
βοΈ Therefore:
πΉ Area of rectangle = 12 Γ area of Shape C.
π β Question 8
Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Explain.
π β
Answer:
πΉ Area depends on how many pieces are used.
πΉ Perimeter depends on the outer boundary of the shape.
πΉ A square and a rectangle have different outer boundaries.
βοΈ Conclusion:
πΉ Their perimeters are different, even though their areas are the same.
πΏ AREA OF TRIANGLE
π β Question 1
Find the areas of the figures below by dividing them into rectangles and triangles.
π β
Answer (Teacher-style explanation)
πΉ Each small square on the grid represents 1 square unit
πΉ Full squares are counted directly
πΉ Two half-squares together make 1 square unit
πΉ Add areas of all rectangles and triangles formed after splitting
π΅ Figure (a)
π Split into one rectangle and two triangles
πΉ Rectangle area = 20 square units
πΉ Two triangles together = 4 square units
βοΈ Total area = 24 square units
π΅ Figure (b)
π Split into one rectangle and one triangle
πΉ Rectangle area = 24 square units
πΉ Triangle area = 6 square units
βοΈ Total area = 30 square units
π΅ Figure (c)
π Split into one large rectangle and two triangles
πΉ Rectangle area = 36 square units
πΉ Two triangles together = 8 square units
βοΈ Total area = 44 square units
π΅ Figure (d)
π Split into one rectangle and two triangles
πΉ Rectangle area = 20 square units
πΉ Two triangles together = 6 square units
βοΈ Total area = 26 square units
π΅ Figure (e)
π Split into two equal triangles
πΉ Each triangle = 6 square units
βοΈ Total area = 12 square units
βοΈ Final Answers (in square units)
πΉ (a) 24
πΉ (b) 30
πΉ (c) 44
πΉ (d) 26
πΉ (e) 12
πΏ AREA MAZE PUZZLES
π β Question
Area Maze Puzzles
In each figure, find the missing value of either the length of a side or the area of a region.
π β Answer (Teacher-style classroom explanation)
π β (a)
(13 sq cm, 26 sq cm, 15 sq cm, ? sq cm)
π β
Answer
πΉ The top-left and top-right rectangles have the same height.
πΉ So their areas are proportional to their widths.
πΉ 26 sq cm is double of 13 sq cm
πΉ So, top-right width is double of top-left width
πΉ Bottom-left area = 15 sq cm
πΉ Bottom-right will also be double (same widths as above)
πΉ Step 1: ? area = 2 Γ 15
πΉ Step 2: ? area = 30
βοΈ Final: 30 sq cm
π β (b)
(10 sq cm at bottom, 10 sq cm vertical, top pink ? sq cm, with 3 cm and 2 cm markings)
π β
Answer
πΉ Bottom rectangle has height = 2 cm (given).
πΉ Its area = 10 sq cm.
πΉ Step 1: Bottom length = Area Γ· height
πΉ Step 2: Bottom length = 10 Γ· 2 = 5 cm
πΉ The marked 3 cm is only the left part, so the remaining right part = 5 β 3 = 2 cm
πΉ That 2 cm is the width of the vertical rectangle above (same strip).
πΉ Vertical rectangle area = 10 sq cm, width = 2 cm
πΉ Step 3: Vertical height = 10 Γ· 2 = 5 cm
πΉ The vertical gap shown next to the pink rectangle is 2 cm (given).
πΉ So, pink height = total height β gap = 5 β 2 = 3 cm
πΉ Pink top length = 3 cm (given)
πΉ Step 4: Pink area = 3 Γ 3 = 9 sq cm
βοΈ Final: 9 sq cm
π β (c)
(Total height 15 cm, middle 42 sq cm with height 6 cm, bottom 60 sq cm, extra 5 cm on bottom, and 3 cm marking near top)
π β
Answer
πΉ Middle rectangle area = 42 sq cm and its height = 6 cm (given).
πΉ Step 1: Middle width = 42 Γ· 6 = 7 cm
πΉ Bottom rectangle extends 5 cm more than the middle (given).
πΉ Step 2: Bottom width = 7 + 5 = 12 cm
πΉ Bottom rectangle area = 60 sq cm
πΉ Step 3: Bottom height = 60 Γ· 12 = 5 cm
πΉ Total height on left = 15 cm
πΉ Heights already used = bottom 5 cm + middle 6 cm = 11 cm
πΉ Step 4: Top (pink) height = 15 β 11 = 4 cm
πΉ The 3 cm mark shows the middle is 3 cm wider than the top pink part.
πΉ So, top pink width = 7 β 3 = 4 cm
πΉ Step 5: Pink area = 4 Γ 4 = 16 sq cm
βοΈ Final: 16 sq cm
π β (d)
(Left rectangle 38 sq cm with top ? cm, right rectangle 18 sq cm with width 5 cm, and height difference 4 cm)
π β
Answer
πΉ Right rectangle area = 18 sq cm and its width = 5 cm (given).
πΉ Step 1: Right height = 18 Γ· 5 = 3.6 cm
πΉ The left rectangle is 4 cm taller than the right one (given).
πΉ Step 2: Left height = 3.6 + 4 = 7.6 cm
πΉ Left rectangle area = 38 sq cm
πΉ Step 3: Left top length = Area Γ· height = 38 Γ· 7.6 = 5 cm
βοΈ Final: 5 cm
βοΈ Final Answers
πΉ (a) 30 sq cm
πΉ (b) 9 sq cm
πΉ (c) 16 sq cm
πΉ (d) 5 cm
πΏ FIGURE IT OUT
π β Question 1
Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m Γ 10 m and 2 m Γ 7 m.
π β
Answer
Area of first rectangle = 5 Γ 10 = 50 sq m
Area of second rectangle = 2 Γ 7 = 14 sq m
Total area = 50 + 14 = 64 sq m
One rectangle having area 64 sq m can have
Length = 8 m
Breadth = 8 m
βοΈ Final Answer
8 m Γ 8 m
π β Question 2
The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
π β
Answer
Area = Length Γ Breadth
1000 = 50 Γ Breadth
Breadth = 1000 Γ· 50
Breadth = 20
βοΈ Final Answer
Width = 20 m
π β Question 3
The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted.
π β
Answer
Area of floor = 5 Γ 4 = 20 sq m
Area of carpet = 3 Γ 3 = 9 sq m
Area not carpeted = 20 β 9
Area not carpeted = 11
βοΈ Final Answer
11 sq m
π β Question 4
Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
π β
Answer
Area of garden = 15 Γ 12 = 180 sq m
Area of one flower bed = 2 Γ 1 = 2 sq m
Area of four flower beds = 4 Γ 2 = 8 sq m
Area available for lawn = 180 β 8
Area available for lawn = 172
βοΈ Final Answer
172 sq m
π β Question 5
Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions.
π β
Answer
Take Shape A as a rectangle of size 1 Γ 18
Area of Shape A = 18 square units
Perimeter of Shape A = 2(1 + 18) = 38 units
Take Shape B as a rectangle of size 4 Γ 5
Area of Shape B = 20 square units
Perimeter of Shape B = 2(4 + 5) = 18 units
Shape A has smaller area but larger perimeter
βοΈ Final Answer
Shape A: 1 Γ 18
Shape B: 4 Γ 5
π β Question 6
On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border?
π β
Answer
Let the page length be L cm and breadth be B cm
Border length = L β 2
Border breadth = B β 3
Perimeter of border = 2[(L β 2) + (B β 3)]
βοΈ Final Answer
Perimeter = 2(L + B β 5) cm
π β Question 7
Draw a rectangle of size 12 units Γ 8 units. Draw another rectangle inside it, without touching the outer rectangle, that occupies exactly half the area.
π β
Answer
Area of outer rectangle = 12 Γ 8 = 96 sq units
Half of the area = 96 Γ· 2 = 48 sq units
One possible inner rectangle is 6 Γ 8
βοΈ Final Answer
Inner rectangle area = 48 sq units
π β Question 8
A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Which statement is always true?
π β
Answer
Let side of square = a
Area of square = aΒ²
After folding, each rectangle has area = aΒ² Γ· 2
Perimeter of square = 4a
Each rectangle has dimensions a and a/2
Perimeter of one rectangle = 2(a + a/2) = 3a
Perimeter of two rectangles together = 6a
6a = 1Β½ Γ 4a
βοΈ Final Answer
Correct option is (c)
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OTHER IMPORTANT QUESTIONS
(CBSE MODEL QUESTION PAPER)
ESPECIALLY MADE FROM THIS CHAPTER ONLY
π΅ Section A β Very Short Answer (1 mark each)
π β Question 1
What is meant by the perimeter of a figure?
π β
Answer:
πΉ Perimeter is the total length of the boundary of a closed figure
π β Question 2
In which units is perimeter measured?
π β
Answer:
πΉ Perimeter is measured in units of length like cm or m
π β Question 3
What is the perimeter of a square of side 5 cm?
π β
Answer:
πΉ Perimeter of square = 4 Γ side
πΈ Perimeter = 4 Γ 5 = 20 cm
π β Question 4
Name the shape whose opposite sides are equal.
π β
Answer:
πΉ Rectangle
π β Question 5
What is meant by area?
π β
Answer:
πΉ Area is the amount of surface covered by a figure
π β Question 6
True or False:
Area is measured in square units.
π β
Answer:
πΉ Area always uses square units
βοΈ Final: True
π’ Section B β Short Answer I (2 marks each)
π β Question 7
Write the formula for the perimeter of a rectangle.
π β
Answer:
πΉ Perimeter of rectangle = 2 Γ (length + breadth)
π β Question 8
Find the perimeter of a rectangle whose length is 6 cm and breadth is 4 cm.
π β
Answer:
πΉ Perimeter = 2 Γ (6 + 4)
πΈ Perimeter = 2 Γ 10 = 20 cm
π β Question 9
What is the formula for the area of a square?
π β
Answer:
πΉ Area of square = side Γ side
π β Question 10
Find the area of a square of side 7 cm.
π β
Answer:
πΉ Area = 7 Γ 7
πΈ Area = 49 square cm
π β Question 11
Write one difference between perimeter and area.
π β
Answer:
πΉ Perimeter measures boundary length
πΈ Area measures surface covered
π β Question 12
Mention one daily-life situation where perimeter is used.
π β
Answer:
πΉ Perimeter is used to find fencing required around a field
π‘ Section C β Short Answer II (3 marks each)
π β Question 13
Define perimeter and write its formula for a square.
π β
Answer:
πΉ Perimeter is the total length of the boundary of a closed figure
πΉ A square has four equal sides
πΈ Perimeter of a square = 4 Γ side
π β Question 14
Find the perimeter of a square whose side is 9 cm.
π β
Answer:
πΉ Perimeter of square = 4 Γ side
πΉ Perimeter = 4 Γ 9
πΈ Perimeter = 36 cm
π β Question 15
Write the formula for the area of a rectangle and explain its terms.
π β
Answer:
πΉ Area of rectangle = length Γ breadth
πΉ Length shows the longer side
πΈ Breadth shows the shorter side
π β Question 16
Find the area of a rectangle whose length is 8 cm and breadth is 5 cm.
π β
Answer:
πΉ Area = length Γ breadth
πΉ Area = 8 Γ 5
πΈ Area = 40 square cm
π β Question 17
State two differences between perimeter and area.
π β
Answer:
πΉ Perimeter measures boundary length, area measures surface covered
πΈ Perimeter uses linear units, area uses square units
π β Question 18
Why are the units of area written as square units?
π β
Answer:
πΉ Area is measured using squares of unit length
πΉ It shows how many unit squares cover a surface
πΈ Hence, units are written as square units
π β Question 19
Can two figures have the same perimeter but different areas? Explain.
π β
Answer:
πΉ Yes, figures can have the same boundary length
πΉ But their shapes may be different
πΈ Hence, the surface covered can be different
π β Question 20
Write one example each where perimeter and area are used in daily life.
π β
Answer:
πΉ Perimeter is used to calculate fencing around a garden
πΈ Area is used to calculate carpet needed for a room
π β Question 21
Find the perimeter of a rectangle whose length is 10 m and breadth is 6 m.
π β
Answer:
πΉ Perimeter = 2 Γ (length + breadth)
πΉ Perimeter = 2 Γ (10 + 6)
πΈ Perimeter = 32 m
π β Question 22
Why is it important to use the same units while finding perimeter or area?
π β
Answer:
πΉ Different units cannot be added or multiplied directly
πΉ Using same units avoids calculation errors
πΈ It gives correct and meaningful results
π΄ Section D β Long Answer (4 marks each)
π β Question 23
Explain how to find the perimeter of an irregular shape.
π β
Answer:
πΉ An irregular shape has sides of different lengths
πΉ Measure the length of each side carefully
πΉ Add the lengths of all sides together
πΈ The total sum gives the perimeter of the irregular shape
βοΈ Final: Perimeter = sum of all side lengths
π β Question 24
Find the perimeter of a rectangle whose length is 12 cm and breadth is 8 cm. Show all steps.
π β
Answer:
πΉ Perimeter of rectangle = 2 Γ (length + breadth)
πΉ Perimeter = 2 Γ (12 + 8)
πΉ Perimeter = 2 Γ 20
πΈ Perimeter = 40 cm
βοΈ Final: Perimeter of the rectangle is 40 cm
π β Question 25
Explain the method of finding the area of a rectangle with an example.
π β
Answer:
πΉ A rectangle is covered by rows and columns of unit squares
πΉ Length shows number of squares in one row
πΉ Breadth shows number of rows
πΈ Area of rectangle = length Γ breadth
πΈ Example: If length = 6 cm and breadth = 5 cm
πΈ Area = 6 Γ 5 = 30 square cm
βοΈ Final: Area is found by multiplying length and breadth
π β Question 26
Explain the difference between perimeter and area using a suitable example.
π β
Answer:
πΉ Perimeter measures the boundary length of a figure
πΉ Area measures the surface covered inside the figure
πΉ Example: A square of side 4 cm
πΉ Perimeter = 4 Γ 4 = 16 cm
πΈ Area = 4 Γ 4 = 16 square cm
βοΈ Final: Perimeter and area measure different quantities
π β Question 27
OR
Can two figures have the same area but different perimeters? Explain.
π β
Answer:
πΉ Yes, two figures can cover the same surface area
πΉ Their shapes may be different
πΉ Different shapes can have different boundary lengths
πΈ Hence, perimeters can be different even if area is same
βοΈ Final: Same area does not guarantee same perimeter
π β Question 28
Explain why area is measured in square units.
π β
Answer:
πΉ Area is found by counting unit squares that cover a surface
πΉ Each unit square has length and breadth of one unit
πΉ This gives square units like cmΒ² or mΒ²
πΈ Square units show two-dimensional measurement
βοΈ Final: Area uses square units because it measures surface
π β Question 29
Explain with examples how perimeter and area are used in daily life.
π β
Answer:
πΉ Perimeter is used to calculate fencing around a field
πΉ Area is used to find the amount of tiles needed for flooring
πΉ Area is used in painting walls
πΈ Perimeter is used in placing wires or boundaries
βοΈ Final: These measurements help in planning and saving resources
π β Question 30
OR
A room is 7 m long and 5 m wide.
(i) Find the perimeter of the room
(ii) Find the area of the room
π β
Answer:
πΉ (i) Perimeter = 2 Γ (length + breadth)
πΉ Perimeter = 2 Γ (7 + 5)
πΉ Perimeter = 2 Γ 12 = 24 m
πΉ (ii) Area = length Γ breadth
πΉ Area = 7 Γ 5 = 35 square m
βοΈ Final:
πΉ Perimeter = 24 m
πΈ Area = 35 square m
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