[New 2025-26] Chapter 6 - Perimeter and Area NCERT Solution | Worksheet | Extra Questions for Class 6 Maths Ganita Prakash.
Class 6 maths chapter 6 full chapter : Perimeter and Area.
'Perimeter and Area' is an important topic to understand in mathematics. In this chapter, we will explore some of the most basic ideas of geometry including perimeter and area of squares, rectangle, triangle and other regular polygons. These ideas form the building blocks of ‘plane geometry’, and will help us in understanding more advanced topics in geometry such as the construction and analysis of different shapes...
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# THIS PAGE INCLUDES :
# Notes : Perimeter and Area
Perimeter and area class 6 notes
- Perimeter -
The perimeter of any closed plane figure is the distance covered along its boundary when you go around it once.
The perimeter of a polygon = Sum of the lengths of its all sides.
E.g.
# Perimeter of Square :
Let each side of square be 'S'
Perimeter of Square = Sum of all sides
= S + S + S + S
= 4S
Therefore, The formula to find perimeter of Square is 4 x Side.
📌Note : To find perimeter of any regular polygon multiply its number of sides with the length of one side.
# Perimeter of Rectangle :
Let the length of rectangle be 'l'
and the breadth of rectangle be 'b'
Perimeter of Rectangle = Sum of all sides
= l + l + b + b
= 2l + 2b
= 2 (l + b)
Therefore, The formula to find perimeter of Rectangle is 2 (l + b).
📌Note : We can also use (2l + 2b) to find the perimeter of rectangle.
# Perimeter of triangle :
Perimeter of triangle = Sum of all sides
= a + b + c
Therefore, The formula to find perimeter of triangle is Sum of all sides.
📌Note : Perimeter of equilateral triangle is 3 x side as it has all side of equal length.
- Area -
The amount of region enclosed by a closed figure is called its area.
Area of a quadrilateral = length of one Side x length of other Side.
📌Note : We can estimate the area of any simple closed shape by using a sheet of squared paper or graph paper where every square measures 1 unit × 1 unit or 1 square unit.
# Area of Square :
Let the Side of a Square be 'S'
Area of Square = Length of one side x length of other side.
= S x S
= S2
Therefore, Area of square is S2.
# Area of Rectangle :
Let the length and breadth of rectangle be 'l' and 'b' respectively.
Area of rectangle = Length of one side x length of other side.
= length x breadth
= l x b
Therefore, Area of rectangle is l x b.
# Area of Triangle :
Area of rectangle = l x b
Length of rectangle = base of triangle.
breadth of rectangle = height of triangle.
Now,
Area of triangle = 1/2 of area of rectangle.
= 1/2 (length x breadth)
= 1/2 (base x height)
Therefore, Area of triangle is 1/2 (b x h).
# Area of Irregular Shape :
To find the area of an irregular shape, we first need to trace the shape on a piece of transparent paper (tracing paper) and then need to place this paper on a sheet of squared paper (Graph paper) and at last count the number of squares to find the area.
Rules to count the Squares :
- Count full square as 1.
- Count more than half square as 1.
- Count half square as 1/2 (0.5).
- Ignore the squares which are less than half.
----- The End -----
# Key Points :
● The perimeter of a polygon is the sum of the lengths of all its sides.
- The perimeter of a rectangle is twice the sum of its length and width.
- The perimeter of a square is four times the length of any one of its sides.
● The area of a closed figure is the measure of the region enclosed by the figure.
● Area is generally measured in square units.
● The area of a rectangle is its length times its width. The area of a square is the length of any one of its sides multiplied by itself.
● Two closed figures can have the same area with different perimeters, or the same perimeter with different areas.
● Areas of regions can be estimated (or even determined exactly) by breaking up such regions into unit squares, or into more general shaped rectangles and triangles whose areas can be calculated.
# NCERT Solutions :
Ganita Prakash class 6 maths chapter 6 Perimeter and Area figure it out ncert page 132.
ncert class 6 maths chapter 6 perimeter and area
Q1.
Find the missing terms :
a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?.
Solution :
Let the length of rectangle be ' a '.
Breadth of rectangle = 2 cm
Perimeter of rectangle = 14 cm
Perimeter of rectangle = 2 (l + b)
14 = 2 (l + b)
14 = 2 (a + 2)
14 = 2 a + 4
14 - 4 = 2 a
10 = 2 a
10/2 = a
5 = a
Therefore, Length of rectangle is 5 cm.
b. Perimeter of a square = 20 cm; side of a length = ?.
Solution :
Let the length of side of square be 'S'
Perimeter of square = 20 cm
Perimeter of square = 4 x side
20 = 4 x S
20 = 4 S
20/4 = S
5 = S
Therefore, Length of side of square is 5 cm.
c. Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?.
Solution :
Let the breadth of rectangle be ' a '.
Length of rectangle = 3 m
Perimeter of rectangle = 12 m
Perimeter of rectangle = 2 (l + b)
12 = 2 (l + b)
12 = 2 (3 + a)
12 = 6 + 2 a
12 - 6 = 2 a
6 = 2 a
6/2 = a
3 = a
Therefore, Breadth of rectangle is 3 m.
Q2.
A rectangle having side lengths 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?
Solution :
Length of rectangle = 5 m
Breadth of rectangle = 3 m
Perimeter of rectangle = 2 (l + b)
= 2 (l + b)
= 2 (5 + 3)
= 2 (8)
= 2 x 8
= 16 cm
Therefore, Perimeter of rectangle is 16 cm.
Here, Length of wire is equal to the perimeter of rectangle which is 16 cm. and the same wire is re-bent to make a square. So the perimeter of square will be equal to the length of wire which is 16 cm.
Now,
Perimeter of square is 16 cm.
Perimeter of square = 4 x Side
16 = 4 x Side
16/4 = Side
4 = Side
Therefore, The length of each side of square is 4 cm.
Q3.
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, respectively.
Solution :
Let the length of third side of triangle be 'S' cm.
Length of first side of triangle = 20 cm.
Length of second side of triangle = 14 cm.
Perimeter of triangle = 55 cm
Perimeter of triangle = Sum of all sides.
20 + 14 + S = 55
34 + S = 55
S = 55 - 34
S = 21
Therefore, the length of third side of triangle is 21 cm.
Q4.
What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs ₹40 per metre?
Solution :
Length of rectangular park = 150 m
Breadth of rectangular park = 120 m
Perimeter of rectangular park = 2 (l + b)
= 2 (150 + 120)
= 2 (270)
= 2 x 270
= 540 m
Cost of fencing = ₹40 / meter.
= ₹ 40 x 540
= ₹ 21600
Therefore, the cost of fencing the rectangular park is ₹21,600.
Q5.
A piece of string is 36 cm long. What will be the length of each side, if it is used to form :
a. A square.
Solution :
Length of string = 36 cm
Perimeter of Square = 4 x Side
36 = 4 x Side
36/4 = Side
9 = Side.
Therefore, the length of side of a square is 9 cm.
b. A triangle with all sides of equal length.
Solution :
Length of string = 36 cm
Perimeter of equilateral triangle = 3 x Side
36 = 3 x Side
36/3 = Side
12 = Side.
Therefore, the length of side of a square is 12 cm.
c. A hexagon (a six sided closed figure) with sides of equal length.
Solution :
Length of string = 36 cm
Perimeter of regular hexagon = 6 x Side
36 = 6 x Side
36/6 = Side
6 = Side.
Therefore, the length of side of a square is 6 cm.
Q6.
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 ?
Solution :
Length of rectangular field = 230 m
Breadth of rectangular field = 160 m
Perimeter of rectangular field = 2 (l + b)
= 2 (230 + 160)
= 2 (390)
= 2 x 390
= 780 m
Now,
Number of rounds of rope for fencing = 3
Length of rope needed = 3 x 780 m = 2340 m
Therefore, the length of rope needed for 3 round for fencing of rectangular field is 2340 m.
Ganita Prakash class 6 maths chapter 6 Perimeter and Area figure it out ncert page 133.
Chapter 6 perimeter and area class 6 questions
Matha Pachchi!
Each track is a rectangle. Akshi’s track has length 70 m and breadth 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in one round.
Q1.
Find out the total distance Akshi has covered in 5 rounds.
Solution :
Distance covered by Akshi in 1 round is equals to the perimeter of rectangular ground.
Length of Akshi's rectangular track = 70 m
Breadth of Akshi's rectangular track = 40 m
Perimeter of Akshi's rectangular track = 2 (l + b)
= 2 (70 + 40)
= 2 x 110
= 220 m
Now,
Number of rounds covered by Akshi = 5
Total distance cover by Akshi = 5 x 220 = 1100 m.
Therefore, Total distance cover by Akshi in 5 rounds is 1100 m.
Q2.
Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance?
Solution :
Distance covered by Toshi in 1 round is equals to the perimeter of rectangular ground.
Length of Toshi's rectangular track = 60 m
Breadth of Toshi's rectangular track = 30 m
Perimeter of Toshi's rectangular track = 2 (l + b)
= 2 (60 + 30)
= 2 x 90
= 180 m
Now,
Number of rounds covered by Toshi = 7
Total distance cover by Toshi = 7 x 180 = 1260 m.
Therefore, Total distance cover by Toshi in 7 rounds is 1260 m.
Hence, Toshi Run Longer distance.
Q3.
Think and mark the positions as directed -
a. Mark ‘A’ at the point where Akshi will be after she ran 250 m.
Solution :
Refer the image given above to get the marked point 'A'.
Here,
Each marked point represents 10 meters.
In one complete round Akshi cover 220 meters.
Therefore,
To cover 250 m she need to ran 1 complete round and 30 meters more.
b. Mark ‘B’ at the point where Akshi will be after she ran 500 m.
Solution :
Refer the image given above to get the marked point 'B'.
Here,
Each marked point represents 10 meters.
In one complete round Akshi cover 220 meters.
Therefore,
To cover 500 m she need to ran 2 complete rounds and 60 meters more.
c. Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘C’.
Solution :
Refer the image given above to get the marked point 'C'.
Here,
Each marked point represents 10 meters.
In one complete round Akshi cover 220 meters.
Therefore,
To cover 1000 m she need to ran 4 complete rounds and 120 meters more.
d. Mark ‘X’ at the point where Toshi will be after she ran 250 m.
Solution :
Refer the image given above to get the marked point 'X'.
Here,
Each marked point represents 10 meters.
In one complete round Toshi cover 180 meters.
Therefore,
To cover 250 m she need to ran 1 complete rounds and 70 meters more.
e. Mark ‘Y’ at the point where Toshi will be after she ran 500 m.
Solution :
Refer the image given above to get the marked point 'Y'.
Here,
Each marked point represents 10 meters.
In one complete round Toshi cover 180 meters.
Therefore,
To cover 500 m she need to ran 2 complete rounds and 140 meters more.
f. Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘Z’.
Solution :
Refer the image given above to get the marked point 'Z'.
Here,
Each marked point represents 10 meters.
In one complete round Toshi cover 180 meters.
Therefore,
To cover 1000 m she need to ran 5 complete rounds and 100 meters more.
Ganita Prakash class 6 maths chapter 6 Perimeter and Area figure it out ncert page 138.
Ganita Prakash Class 6 Maths Chapter 6 Solutions Perimeter and Area
Q1.
The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden?
Solution :
Let the breadth of rectangle be 'b'.
Length of rectangle = 25 m
Area of rectangle = 300 m sq.
Area of rectangle = l x b
300 = 25 x b
300/25 = b
12 = b
Therefore, Breadth of rectangle is 12 m.
Q2.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹8 per hundred sq m?
Solution :
Length of rectangle = 500 m
Breadth of rectangle = 200 m
Area of rectangle = l x b
= 500 x 200
= 1,00,000 m sq.
Now,
The cost of tiling 100 sq. m area = ₹ 8
Cost of tiling 1 sq. m area = 8/100
Cost of tiling 1,00,000 sq. m area = 1,00,000 x 8/100 = ₹ 8000
Therefore, The cost of tiling the rectangular plot is ₹ 8000.
Q3.
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?
Solution :
Length of rectangle coconut grove = 100 m
Breadth of rectangle coconut grove = 50 m
Area of rectangle coconut grove = l x b
= 100 x 50
= 5000 m sq.
Now,
Area required by one coconut tree = 25 m sq.
Number of coconut tree can be planted in 5000 m sq. area = 5000/25 = 200.
Therefore, The cost of tiling the rectangular plot is ₹ 200.
Q4.
By splitting the following figures into rectangles, find their areas (all measures are given in metres) :
Solution :
a.
By splitting the following figure into rectangles we get 4 rectangles A, B, C and D.
⏺ Rectangle A :
Length of rectangle A = 4 m
Breadth of rectangle A = 3 m
Area of rectangle A = l x b
= 4 x 3
= 12 m
⏺ Rectangle B :
Length of rectangle B = 4 m
Breadth of rectangle B (4 - 3) = 1 m
Area of rectangle B = l x b
= 4 x 1
= 4 m
⏺ Rectangle C :
Length of rectangle C (4 + 3 - 2) = 5 m
Breadth of rectangle C = 2 m
Area of rectangle C = l x b
= 5 x 2
= 10 m
⏺ Rectangle D :
Length of rectangle D (7 - 5) = 2 m
Breadth of rectangle D = 1 m
Area of rectangle D = l x b
= 2 x 1
= 2 m
Therefore, Total area of the given figure is 28 m.
b.
By splitting the following figure into rectangles we get 3 rectangles A, B and C.
⏺ Rectangle A :
Length of rectangle A = 5 m
Breadth of rectangle A (3 - 2) = 1 m
Area of rectangle A = l x b
= 5 x 1
= 5 m
⏺ Rectangle B :
Length of rectangle B = 2 m
Breadth of rectangle B (3 - 2) = 1 m
Area of rectangle B = l x b
= 2 x 1
= 2 m
⏺ Rectangle C :
Length of rectangle C = 2 m
Breadth of rectangle C = 1 m
Area of rectangle C = l x b
= 2 x 1
= 2 m
Therefore, Total area of the given figure is 9 m.
Ganita Prakash class 6 maths chapter 6 Perimeter and Area figure it out ncert page 139.
Perimeter and Area Class 6 NCERT Solutions Ganita Prakash Maths Chapter 6
Cut out the tangram pieces given at the end of your textbook.
Q1.
Explore and figure out how many pieces have the same area.
Solution :
Triangles (A & B) and Triangles (C & E) have same area.
Q2.
How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D and E?
Solution :
Shape 'D' is big as two times the Shape 'C'. The relationship between shape C, D and E is that Shape D has the area equals to the combined area of shape 'C' and 'E'. (C + E = D)
Q3.
Which shape has more area: Shape D or F? Give reasons for your answer.
Solution :
Both Shapes (D and F) has same area because Shape 'C' and 'E' together fitted perfectly in both the shapes.
Q4.
Which shape has more area: Shape F or G? Give reasons for your answer.
Solution :
Both Shapes (F and G) has same area because Shape 'C' and 'E' together fitted perfectly in both the shapes.
Q5.
What is the area of Shape A as compared to Shape G? Is it twice as big? Four times as big?
Solution :
The area of Shape 'A' is twice as big as compared to Shape 'G' because Shape 'C' and 'E' together fitted in shape 'G' only one time but in Shape 'A' Shape 'C' and 'E' together fitted two times.
Q6.
Can you now figure out the area of the big square formed with all seven pieces in terms of the area of Shape C?
Solution :
The area of big square is = Sum of area of triangle A, B, C, D, E, F and G.
Area of triangle 'A' = 4C
Area of triangle 'B' = 4C
Area of triangle 'C' = 1C
Area of triangle 'D' = 2C
Area of triangle 'E' = 1C
Area of triangle 'F' = 2C
Area of triangle 'G' = 2C
The area of big square is = 4C + 4C + C + 2C + C + 2C + 2C = 16C.
Therefore, Area of Big Square is 16 C.
Q7.
Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the area of Shape C now? Give reasons for your answer.
Solution :
Area of rectangle will be same as the area of square. This is because the total area of all the pieces will remain same, regardless of how they are aligned.
Q8.
Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Give an explanation for your answer.
Solution :
No, The perimeter of square and rectangle formed from these 7 pieces will be different because the length of boundaries of each piece is different.
Ganita Prakash class 6 maths chapter 6 Perimeter and Area figure it out ncert page 144.
perimeter and area class 6 questions
Q1.
Find the areas of the figures below by dividing them into rectangles and triangles.
Solution :
(a) Area of 1 Rectangle and 2 Triangles.
Length of rectangle = 5 units
Breadth of rectangle = 4 units
Area of rectangle = l x b
= 4 x 5
= 20 sq. units.
Base of triangle = 4 units
Height of triangle = 1 unit.
Area of triangle = 1/2 (b x h)
= 1/2 (1 x 4)
= 1/2 (4)
= 2 sq. units.
Area of figure 'a' = area of rectangle + area of both triangles
= 20 + 2 + 2
= 24 sq. units.
Therefore, Area of figure 'a' is 24 sq. units.
(b) Area of 1 Rectangle and 2 Triangles.
Length of rectangle = 5 units
Breadth of rectangle = 4 units
Area of rectangle = l x b
= 4 x 5
= 20 sq. units.
Base of first triangle = 4 units
Height of first triangle = 3 unit.
Area of first triangle = 1/2 (b x h)
= 1/2 (3 x 4)
= 1/2 (12)
= 6 sq. units.
Base of Second triangle = 4 units
Height of Second triangle = 2 unit.
Area of Second triangle = 1/2 (b x h)
= 1/2 (2 x 4)
= 1/2 (8)
= 4 sq. units.
Area of figure 'b' = area of rectangle + area of both triangles
= 20 + 6 + 4
= 30 sq. units.
Therefore, Area of figure 'b' is 30 sq. units.
(c) Area of 1 Rectangle and 3 Triangles.
Length of rectangle = 8 units
Breadth of rectangle = 3 units
Area of rectangle = l x b
= 3 x 8
= 24 sq. units.
Base of first triangle = 3 units
Height of first triangle = 2 unit.
Area of first triangle = 1/2 (b x h)
= 1/2 (3 x 2)
= 1/2 (6)
= 2 sq. units.
Base of Second triangle = 12 units
Height of Second triangle = 3 unit.
Area of Second triangle = 1/2 (b x h)
= 1/2 (12 x 3)
= 1/2 (36)
= 18 sq. units.
Area of figure 'c' = area of rectangle + area of all 3 triangles
= 24 + (2 + 2) + 18
= 46 sq. units.
Therefore, Area of figure 'c' is 46 sq. units.
(d) Area of 1 Rectangle and 2 Triangles.
Length of rectangle = 4 units
Breadth of rectangle = 3 units
Area of rectangle = l x b
= 4 x 3
= 12 sq. units.
Base of first triangle = 1 units
Height of first triangle = 2 unit.
Area of first triangle = 1/2 (b x h)
= 1/2 (1 x 2)
= 1/2 (2)
= 1 sq. units.
Base of Second triangle = 3 units
Height of Second triangle = 2 unit.
Area of Second triangle = 1/2 (b x h)
= 1/2 (3 x 2)
= 1/2 (6)
= 3 sq. units.
Area of figure 'd' = area of rectangle + area of both triangles
= 12 + 1 + 3
= 18 sq. units.
Therefore, Area of figure 'd' is 18 sq. units.
(e) Area of 2 Triangles.
Base of first triangle = 4 units
Height of first triangle = 2 unit.
Area of first triangle = 1/2 (b x h)
= 1/2 (4 x 2)
= 1/2 (8)
= 4 sq. units.
Base of Second triangle = 4 units
Height of Second triangle = 4 unit.
Area of Second triangle = 1/2 (b x h)
= 1/2 (4 x 4)
= 1/2 (16)
= 8 sq. units.
Area of figure 'e' = area of both triangles
= 4 + 8
= 12 sq. units.
Therefore, Area of figure 'e' is 12 sq. units.
Ganita Prakash class 6 maths chapter 6 Perimeter and Area figure it out ncert page 149.
Perimeter and area Class 6 Solutions
Q1.
Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m.
Solution :
Length of first rectangle = 10 m
Breadth of first rectangle = 5 m
Area of first rectangle = l x b
= 10 x 5
= 50 sq. m.
Length of second rectangle = 7 m
Breadth of second rectangle = 2 m
Area of second rectangle = l x b
= 7 x 2
= 14 sq. m.
Now,
The sum area of two rectangle = 50 + 14 = 64 sq. m.
Therefore, the dimensions of rectangle whose area is 64 sq. m. = 16 m x 4 m.
Hint : (Prime factors of 64 = 2 x 2 x 2 x 2 x 2 x 2)
Q2.
The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Solution :
Let the width of the rectangular garden be 'b'.
Length of rectangular garden = 50 m
Area of rectangular garden = 1000 sq. m.
l x b = 1000
50 x b = 1000
b = 1000/50
b = 20 m.
Therefore, the width of the rectangular garden is 20 m.
Q3.
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.
Solution :
Length of rectangular room = 5 m
Breadth of rectangular room = 4 m
Area of rectangular room = l x b
= 5 x 4
= 20 sq. m.
Length of side of square carpet = 3m
Area of square carpet = Side x Side
= 3 x 3
= 9 sq. m.
Area that is not carpeted = Area of rectangular room - Area of square carpet.
= 20 - 9
= 11 sq. m
Therefore, the area of room which is not carpeted is 11 sq. m.
Q4.
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?
Solution :
Length of flower bed = 2 m
Breadth of flower bed = 1 m
Area of flower bed = l x b
= 2 x 1
= 2 sq. m
Total area of 4 flower bed = 4 x 2 = 8 sq. m.
Length of rectangular garden = 15 m.
Breadth of rectangular garden = 12 m.
Area of rectangular garden = l x b
= 15 x 12
= 180 sq. m
Therefore, Area left for laying down a lawn is (180 - 8) 172 sq. m
Q5.
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.
Solution :
Q6.
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?
Solution :
Lets assume that we have a note book of size 20 cm x 15 cm.
First we draw border 1 cm away from top and bottom.
So, Length of border = 20 - (1 + 1) = 18 cm.
Second we draw border 1.5 cm away from left and right side.
So, Breadth of border = 15 - (1.5 + 1.5) = 12 cm.
Now,
The perimeter of rectangular border = 2 (l + b)
= 2 (18 + 12)
= 2 x 30
= 60 cm.
Therefore, the perimeter of border is 60 cm.
Q7.
Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area.
Solution :
Length of outer rectangle = 12 units
Breadth of outer rectangle = 8 units
Area of outer rectangle = l x b = 12 x 8 = 96 sq. units.
Now,
Half of area 96 sq. unit is 48 sq. units.
Dimensions of inner rectangle = 8 unit x 6 units.
Hint : (Prime factor of 48 = 2 x 2 x 2 x 2 x 3)
Q8.
A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here?
- a. The area of each rectangle is larger than the area of the square.
- b. The perimeter of the square is greater than the perimeters of both the rectangles added together.
- c. The perimeters of both the rectangles added together is always 1 + (1 / 2) times the perimeter of the square.
- d. The area of the square is always three times as large as the areas of both rectangles added together.
Solution :
Statement 'C' is always True. The perimeter of both the rectangle added together is always 1.5 times the perimeter of the square.
E.g.
In the given figure :
The Side of square = 8 cm.
Perimeter of square = 4 x side
= 4 x 8
= 32 cm.
Length of rectangle = 8 cm
Breadth of rectangle = 4 cm
Perimeter of rectangle = 2 (l + b)
= 2 (8 + 4)
= 2 x 12
= 24 cm
Now, Perimeter of both the rectangles = 2 x 24 = 48 cm.
Therefore, The perimeter of rectangle (48 cm) is 1.5 times the perimeter of square (32 cm).
# Worksheet - Extra Questions :
Chapter 6 perimeter and area class 6 worksheet
perimeter and area class 6 extra questions
# Fill in the blanks :
1. Perimeter of a triangle with sides 4.5cm, 6.02cm and 5.38cm is _______ .
2. The amount of region enclosed by a plane closed figure is called its _______ .
3. Area of a rectangle with length 5cm and breadth 3cm is _______ .
4. 1 sq. cm. = _______ cm × 1cm.
5. 1 sq. m. = _______ sq. cm.
# True - False :
1. If length of a rectangle is halved and breadth is doubled then the area of the rectangle obtained remains same.
2. Area of a square is doubled if the side of the square is doubled.
3. Perimeter of a regular octagon of side 6cm is 36cm.
4. A farmer who wants to fence his field, must find the perimeter of the field.
5. An engineer who plans to build a compound wall on all sides of a house must find the area of the compound.
# Multiple Choice Questions :
Q1. A square of side 1 cm is joined to a square of side 3 cm. The perimeter of the new figure is ______.
(a) 13 cm
(b) 14 cm
(c) 15 cm
(d) 16 cm
Q2. A square shaped park ABCD of side 100m has two equal rectangular flower beds each of size 10m × 5m. Length of the boundary of the remaining park is _____ .
(a) 360 m
(b) 400 m
(c) 340 m
(d) 460 m
Q3. The side of a square is 10cm. How many times will the new perimeter become if the side of the square is doubled?
(a) 2 times
(b) 4 times
(c) 6 times
(d) 8 times
Q4. Two regular Hexagons of perimeter 30cm each are joined as shown in Fig. 6.7. The perimeter of the new figure is ____ .
(a) 65 cm
(b) 60 cm
(c) 55 cm
(d) 50 cm
Q5. Length and breadth of a rectangular sheet of paper are 20cm and 10cm, respectively. A rectangular piece of size 2cm x 5 cm is cut from the sheet. Which of the following statements is correct for the remaining sheet?
(a) Perimeter remains same but area changes.
(b) Area remains the same but perimeter changes.
(c) Both area and perimeter are changing.
(d) Both area and perimeter remain the same.
# Numerical Question :
Q1.
Bhavna runs 10 times around a square field of side 80m. Her sister Sushmita runs 8 times around a rectangular field with length 150m and breadth 60m. Who covers more distance? By how much?
Q2.
The length of a rectangular field is thrice its breadth. If the perimeter of this field is 800m, what is the length of the field?
Q3.
There is a rectangular lawn 10m long and 4m wide in front of Meena’s house. It is fenced along the two smaller sides and one longer side leaving a gap of 1m for the entrance. Find the length of fencing.
Q4.
The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of 6km. What is the length of the field?
Q5.
Three squares are joined together as shown in Fig. Their sides are 4cm, 10cm and 3cm. Find the perimeter of the figure.
Perimeter and area Class 6 Worksheets with answers
perimeter and area class 6 worksheets
# Fill in the Blanks :
1. 15.9 cm
2. Area
3. 16 sq. cm.
4. 1
5. 10,000
# True - False :
1. True
2. False
3. False
4. True
5. False
# MCQ's :
1. 14 cm (b)
2. 400 m (b)
3. 2 times (a)
4. 50 cm (d)
5. Perimeter remains same but area changes. (a)
# Numerical Questions :
Solution Q1 :
Side of square field = 80 m
Perimeter of square field = 4 x side
= 4 x 80
= 320 m
Now,
Number of times Bhavna run around the square field = 10
Total distance covered by Bhavna = 10 x 320 = 3200 m
Length of rectangular field = 150 m
Breadth of rectangular field = 60 m
Perimeter of rectangular field = 2 (l + b)
= 2 (150 + 60)
= 2 x 210
= 420 m
Now,
Number of times Sushmita run around the rectangular field = 8
Total distance covered by Sushmita = 8 x 420 = 3360 m
Therefore, Sushmita covered more distance than Bhavna by (3360 - 3200) 160 m.
Solution Q2 :
Let the Breadth of rectangular field be 'x'.
So, Length of the rectangular field is 3x.
Perimeter of rectangular field = 800 m
2 (x + 3x) = 800
2 x 4x = 800
8x = 800
x = 800/8
x = 100 m
Therefore, Length of rectangular field is (3 x 100) 300m.
Solution Q3 :
Length of rectangular lawn = 10 m
Breadth of rectangular lawn = 4 m
Length of fencing required = 2 small side + (front side - 1m)
= 2 x 4 + (10 -1)
= 8 + 9
= 17 m
Therefore, length of fencing required is 17 m.
Solution Q4 :
Let the breadth of rectangular field be 'x'.
So, Length of rectangular field is 2x
Distance covered by Jamal in 4 rounds = 6 km = 6000 m
Distance covered by Jamal in 1 round = 6000/4 = 1500 m
Perimeter of rectangular field = Distance covered by Jamal in one round i.e. 1500 m.
Perimeter of rectangular field = 2 (l + b)
1500 = 2 x (2x + x)
1500 = 2 x (3x)
1500 = 6x
1500/6 = x
250 = x
Therefore, The length of rectangular field is (2 x 250) 500 m.
Solution Q5 :
Length of side of left square = 4 cm
Length of side of center square = 10 cm
Length of side of right square = 3 cm
Perimeter of figure = length of base + Length of sides
= (4 + 10 + 3) + (4 + 4 + 6 + 10 + 7 + 3 + 3)
= 17 + 37
= 54 cm
Therefore, The perimeter of the given figure is 54 cm.
# MCQ Test :
Perimeter and Area mcq :
# Still have some Doubts ?
-- Feel FREE to Ask your Doubts Here --
# Learn More :
CBSE Class 6th Mathematics teaches students about the basics, which have a wide range of application in their higher studies. All the chapters given below includes Solutions for all the question available in Class 6th Mathematics NCERT Textbook Ganita Prakash. It also includes some Important Extra Questions related to the Chapters and we have also provided Free Quiz Based Test which is consist of Objective Type Questions which help the students to test their Understanding about the given Chapters. This material is available for Free For the Students So that they can prepare & score good marks in their upcoming exams.
| Chapters | Class 6 Maths Syllabus |
|---|---|
| Chapter 01 : | Patterns in Mathematics |
| Chapter 02 : | Lines and Angles |
| Chapter 03 : | Number Play |
| Chapter 04 : | Data Handling and Presentation |
| Chapter 05 : | Prime Time |
| Chapter 06 : | Perimeter and Area |
| Chapter 07 : | Fractions |
| Chapter 08 : | Playing with Constructions |
| Chapter 09 : | Symmetry |
| Chapter 10 : | The Other Side of Zero |
Curiosity, Textbook of Science for Grade 6, comprises twelve chapters. As the name of the textbook suggests, there are numerous opportunities for the learners to explore the world of science and its nature. Through the chapters, learners will embark on a journey that will connect them to the world around and spark curiosity for further exploration. The hands-on activities embedded within each chapter engages the learners and provide them an opportunity to reflect on learning. The primary aim of Curiosity is to prepare the children for becoming the responsible members of the society, and therefore efforts have been made to raise awareness about various issues, such as gender, region, environment, health and hygiene, water scarcity and energy conservation.
| Chapters | Class 6 Science Syllabus |
|---|---|
| Chapter 01 : | The Wonderful World of Science |
| Chapter 02 : | Diversity in the Living World |
| Chapter 03 : | Mindful Eating: A Path to a Healthy Body |
| Chapter 04 : | Exploring Magnets |
| Chapter 05 : | Measurement of Length and Motion |
| Chapter 06 : | Materials Around Us |
| Chapter 07 : | Temperature and its Measurement |
| Chapter 08 : | A Journey through States of Water |
| Chapter 09 : | Methods of Separation in Everyday Life |
| Chapter 10 : | Living Creatures: Exploring their Characteristics |
| Chapter 11 : | Nature’s Treasures |
| Chapter 12 : | Beyond Earth |
Class 6th English NCERT Poorvi has five thematic units that comprise stories, poems, conversation, narrative and descriptive pieces. Themes such as friendship, wellness, sports, nature, art and culture, etc. have been included. Cross-cutting themes, such as Indian Knowledge Systems, values, heritage, gender sensitivity and inclusion have been integrated in all the units. Each unit has three literary pieces― story or conversation, poem and non-fiction. There are intext questions, ‘Let us discuss’ to assess comprehension of the text. The end-of-the-text questions given in ‘Let us think and reflect’ are designed to encourage critical thinking, reasoning, responding, analyzing, etc. These literary pieces are not only entertaining but also instill valuable life lessons, fostering personal growth and helping children navigate social situations with confidence. The selected pieces will resonate with children’s daily experiences and encourage positive values like resilience, empathy and emotional intelligence that can have a profound impact on their development.
| No. | Class 6 English Syllabus |
|---|---|
| Unit 1 : | Fables and Folk Tales |
| A Bottle of Dew | |
| The Raven and The Fox | |
| Rama to the Rescue | |
| Unit 2 : | Friendship |
| The Unlikely Best Friends | |
| A Friend's Prayer | |
| The Chair | |
| Unit 3 : | Nurturing Nature |
| Neem Baba | |
| What a Bird Thought | |
| Spices that Heal Us | |
| Unit 4 : | Sports and Wellness |
| Change of Heart | |
| The Winner | |
| Yoga - A Way of Life | |
| Unit 5 : | Culture and Tradition |
| Hamara Bharat - Incredible India! | |
| The Kites | |
| Ila Sachani : Embroidering Dreams with her Feet | |
| National War Memorial |
