KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area

Students can Download Maths Chapter 9 Perimeter and Area Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 5 Maths Chapter 9 Perimeter and Area

KSEEB Class 5 Maths Perimeter and Area Ex 9.1

I. Answer orally:

Question 1.
What are two-dimensional figures called?
Answer:
Plane figures

Question 2.
Name the two dimensions of a rectangle.
Answer:
Length and breadth

KSEEB Solutions

Question 3.
How many pairs of opposite sides are there in a rectangle?
Answer:
Two pairs

Question 4.
How are the opposite sides in a rectangle?
Answer:
Equal

Question 5.
How many times is the perimeter of a rectangle to the sum of its length and breadth?
Answer:
Twice

KSEEB Solutions

II. Find the perimeter of the following pictures:

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 1

Perimeter of Rectangle = (21 + 2b)
= (2 × 70 + 2 × 90)
= 140 + 180
= 320 cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 2

Perimeter of Rectangle = (21 + 2b)
= (2 × 60 + 2 × 24)
= 120 + 48
= 168 cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 3

Perimeter of Rectangle = (21 + 2b)
= (2 × 153 + 2 × 122)
= 306 + 244
= 550 cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 4

Perimeter of Rectangle = (21 + 2b) units
= (2 × 3 + 2 × 6)
= 6+12
= 18cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 5

Perimeter of Rectangle
= (21 + 2b)
= (2 × 92 + 2 × 80)
= (184 + 160)
= 344 cm

KSEEB Solutions

III. The length and breadth of rectangles are given below. Find their perimeters.

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 6
KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 7

1) The Perimeter of Rectangle = (21 + 2b)
= (2 × 2 + 2 × 3)
= 4 + 6
= 10 cm

KSEEB Solutions

2) The Perimeter of Rectangle = (21 + 2b)
= (2 × 2 + 2 × 4)
= 4 + 8
= 12 cm

3) The Perimeter of Rectangle = (21 + 2b)
= (2 × 3 + 2 × 4)
= 6 + 8
= 14 cm

4) The Perimeter of Rectangle = (21 + 2b)
= (2 × 5 + 2 × 4)
= 10 + 8
= 18 cm

KSEEB Solutions

5) The Perimeter of Rectangle = (21 + 2b)
= (2× 5 + 2 × 2)
= 10 + 4
= 14 cm

6) The Perimeter of Rectangle = (21 + 2b)
= (2 × 3 + 2 × 6)
= 6 + 12
= 18 cm

7) The Perimeter of Rectangle = (21 + 2b)
= (2 × 4 + 2 × 6)
= 8 + 12
= 20 cm

KSEEB Solutions

8) The Perimeter of Rectangle = (21 + 2b)
= (2 × 5 + 2 × 6)
= (10 + 12)
= 22 cm

9) The Perimeter of Rectangle = (21 + 2b)
= (2 × 8 + 2 × 6)
= 16 + 12
= 28 cm

10) The Perimeter of Rectangle = (21 + 2b)
= (2 × 7 + 2 × 9)
= 14 + 18
= 32 cm

IV. Solve the following:

Question 1.
A rectangular room measure 6m in length and 4m in breadth. Find the perimeter of the room.
Answer:
Length = 6m
Breadth = 4m
The Perimeter of a room = (21 + 2b) units
= (2 × 6 + 2 × 4)
= 12 + 8
= 20 meters

KSEEB Solutions

Question 2.
A rectangular field has a length of 150m and breadth 120m. Find the perimeter of the field.
Answer:
The Perimeter of the field = (21 + 2b) units
= (2 × 150 + 2 × 120)
= 300 + 240
= 540 meters

KSEEB Solutions

Question 3.
A rectangular garden measures 80m in length and 50m in breadth. Find its perimeter. If the garden has to be fenced 5 rounds with barbed wire, what is the length of the wire required?
Answer:
The Perimeter of Rectangular Garden
= (21 + 2b) units
= (2 × 80 + 2 × 50)
= 160 + 100 = 260 meters
The Garden has to be fenered 5rounds with barbed wire = 260 × 5
= 1300 metres

KSEEB Solutions

Question 4.
An auditorium measures 80m in length and 30 min breadth. If the walls of the auditorium have to be decorated with colored buntings 4 times, find the length of buntings required. If the cost of 1m of buntings is Rs. 15, what is the total cost of the buntings used to decorate the auditorium?
Answer:
The Perimeter of Auditorium = (21 + 2b) units = (2 × 80 + 2 × 30)
= 160 + 60
= 220 metres
Auditorium walls have to be decorated with colour buntings = 220 × 4 = 880 metres The cost of the buntings used to decorate the auditorium = 880 × 15 = 13,220
The cost of 1 metre of bunting = Rs 15

KSEEB Solutions

Question 5.
Srilatha, during her morning walk, goes round a rectangular park 3 times. If the length and breadth of the park are 320 m and 210 m respectively, calculate the distance she has covered.
Answer:
The Perimeter of Rectangular Parks
= (21 + 2b) units
= (2 × 320 + 2 × 210)
= 640 + 420
= 1060 metres
Srilatha walks round Rectangular park
= 3 times
= 1060 × 3
= 3180 metres
Srilatha walks 3180 metres of distance

KSEEB Solutions

KSEEB Class 5 Maths Perimeter and Area Ex 9.2

I. Answer orally:

Question 1.
Mention the two dimensions of a square.
Answer:
Length × width

Question 2.
How many equal sides are there in a square?
Answer:
4

KSEEB Solutions

Question 3.
How many times is the perimeter of a square to its length?
Answer:
Four times

Question 4.
What is the perimeter of a square of length 5 cm?
Answer:
The perimeter of a square
= 4 × 1
= 4 × 5
= 20 cm

KSEEB Solutions

II. The rectangle given below are divided into squares of unit length. Find their areas:

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 8

The perimeter of a square
= 4 × 1
= 4 × 14
= 56 cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 9

The perimeter of a square
= 4 × 1
= 4 × 50
= 200cm

KSEEB Solutions

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 10

The perimeter of a square
= 4 × 1
= 4 × 52
= 208 cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 11
The perimeter of a square
= 4 × 1
= 4 × 45
= 180 cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 12

The perimeter of a square
= 4 × 1
= 4 × 35
= 120 cm

KSEEB Solutions

III. Lengths of the squares are given in the following table. Find their perimeters.

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 13
KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 14
a) The perimeter o f a square = 4 × 1 = 4 × 3 = 12 cm
b) The perimeter of a square = 4 × 1 = 4 × 5 = 20 cm
c) The perimeter of a square = 4 × 1 = 4 × 11 = 44 cm
d) The perimeter of a square = 4 × 18 = 72 cm
e) The perimeter of a square = 4 × 25 = 100 cm
f) The perimeter of a square = 4 × 30 = 120 cm
g) The perimeter of a square = 4 × 41 = 164 cm
i) The perimeter of a square = 4 × 55 = 220 cm
j) The perimeter of a square = 4 × 63 = 252 cm
k) The perimeter of a square = 4 × 92 = 368 cm

KSEEB Solutions

IV. Solve the following problems:

Question 1.
The length of a square room is 15m. Find its perimeter.
Answer:
The perimeter of a square room
= 4 × 1
= 4 × 15
= 60 meter

KSEEB Solutions

Question 2.
Rama, runs 4 times around a square park of length 85 m. What is the total distance he covers?
Answer:
The perimeter of a square park
= 4 × 1
= 4 × 85
= 340 mt
Rama runs 4 times around a park
= 340 × 4
= 1360mt

Question 3.
The length of a square room is 16m. The walls of the room should be tied with colored buntings 4 times. Find the total length of buntings required.
Answer:
The perimeter of a square room
= 4 × 1
= 4 × 6
= 64 cm

The room should be tied with colored buntings 4 times
= 64 × 4
= 256 mt

KSEEB Solutions

Question 4.
The length of a square ground is 70m. Find its perimeter. (Additional Que)
Answer:
The perimeter of a square ground
= 4 × 1
= 4 × 70
= 280 mt

Question 5.
The length of a square playground is 30m. Find its perimeter. (Additional Question)
Answer:
The perimeter of a square playground
= 4 × 30
= 120 mt

KSEEB Solutions

KSEEB Class 5 Maths Perimeter and Area Ex 9.3

I. Answer orally:

Question 1.
What is the unit of area?
Answer:
Square units.

Question 2.
What is the space within the boundary of a plane figure called?
Answer:
Area

Question 3.
What is the area of a unit square?
Answer:
Unit square

KSEEB Solutions

II. The rectangle given below are divided into squares of unit length. Find their areas:

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 15
Area of Rectangle
= (1 × b) units
= 3 × 4
= 12 sq.units

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 16
Area of Rectangle
= (1 × b) units
= 5 × 3
= 15 sq.units

KSEEB Solutions

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 17
Area of Rectangle
= (1 × b) units
= 6 × 4
= 24 sq.units

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 18
Area of Rectangle
= (1 × b) units
= 7 × 3
= 21 sq.units

KSEEB Solutions

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 19
Area of Rectangle
= (1 × b) units
= 8 × 4 = 32 sq.units

KSEEB Solutions

III. The length and breadth of the rectangles are given below. Calculate their areas:

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 20
KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 21

(1) Area of Reactangle
= 1 × b
= 2 × 3
= 6 sq.cm

KSEEB Solutions

(2) Area of Reactangle
= 1 × b
= 2 × 4
= 8 sq.cm

(3) Area of Reactangle
= 1 × b
= 3 × 4
= 12 sq.cm

(4) Area ofRectangle
= 1 × b
= 5 × 4
= 20 Sq.cm

KSEEB Solutions

(5) Area ofRectangle
= 1 × b
= 5 × 2
= 10 Sq.cm

(6) Area of Rectangle
= 1 × b
= 3 × 6
= 18 Sq.cm

(7) Area of Rectangle
= 1 × b
= 4 × 6
= 24 Sq.cm

KSEEB Solutions

(8) Area ofRectangle
= 1 × b
= 5 × 6 = 30 Sq.cm

(9) Area of Rectangle
= 1 × b
= 8 × 6
= 48 Sq.cm

(10) Area ofRectangle
= 1 × b
= 7 × 9
= 63 Sq.cm

KSEEB Solutions

IV. Find the area of the figures given below:

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 22

Area ofRectangle
= 1 × b
= 70 × 90
= 6300 Sq.cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 23

KSEEB Solutions

Area of Rectangle
= 1 × b
= 60 × 20
= 1200 Sq.cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 24

Area of Rectangle
= 1 × b
= 153 × 122
= 18,666 Sq.cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 25

Area of Rectangle
= 1 × b
= 3 × 6
= 18 Sq.cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 26

Area of Rectangle
= 1 × b
= 92 × 80
= 7,360 Sq.cm

KSEEB Solutions

V. Solve the following problems:

Question 1.
A farmer has a rectangular land length 250m and breadth 180m. Find the area of the land.
Answer:
Area of the land
= 1 × b
= 250 × 180
= 45,000 Sq.cm

Question 2.
A carpet is needed to cover the entire area of a room. If the length of the room is 16m and breadth 5m, find the area of the carpet required.
Answer:
Area of the carpet
= 1 × b
= 16 × 5
= 80 Sq.cm

KSEEB Solutions

Question 3.
An auditorium measures 25m in length and 18m in breadth. How many slabs of stones of 3m × 1m are required to cover the floor of the auditorium?
Answer:
Area of Auditorium
= 1 × b
= 25 × 18
= 450 Sq.cm

The size of the slab stane
= 3m × 1m
= 3m

The number of slab stanes
KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 27
= 150

KSEEB Solutions

Question 4.
A rectangular plot’s length is 25m and breadth is 15m. Find the area of the plot. If 1 square meter of the plot costs Rs. 250, what is the total value of the plot?
Answer:
Area of the plot = 1 × b
= 25 × 15
= 375 Sq.cm

1 Sq.mt of the plot Rs 250
375 Sq.mt of the plot = 375 × 250
= 93,750

KSEEB Solutions

Question 5.
A rectangular room’s length is 20m and breadth is 11m. How many tiles of 2m × 1m are required to cover the floor of the room?
Answer:
Area of room
= 1 × b
= 20 × 11
= 220 Sq.cm

The number of tiles to cover the floor of the room
= 2m × 1m
= 2m
KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 28

KSEEB Class 5 Maths Perimeter and Area Ex 9.4

I. Answer orally:

Question 1.
What is the total space within the boundary of a square called?
Answer:
Area

KSEEB Solutions

Question 2.
What is the unit of area?
Answer:
(1 × 1)

Question 3.
What is the area of a square?
Answer:
Sq.unit

KSEEB Solutions

II. Find the area of the figures given below:

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 29
Area of a square
= 1 × 1
= 52 × 52
= 2,704 Sq.cm

KSEEB Solutions

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 30
Area of a square
= 1 × 1
= 50 × 50
= 2,500 Sq.cm

KSEEB Solutions

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 31

Area of a square
= 1 × 1
= 14 × 14
= 136 Sq.cm

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 32

Area of a square
= 1 × 1
= 45 × 45
= 2,025 Sq.cm

KSEEB Solutions

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 33

Area of a square
= 1 × 1
= 35 × 35
= 1,225 Sq.cm

KSEEB Solutions

III. The measure of one side of the squares are given below. Find their areas.

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 34
KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 35

(1) Area of a Square
= 1 × 1
= 3 × 3 Sq.cm

(2) Area of a Square
= 1 × 1
= 5 × 5
= 25 Sq.cm

KSEEB Solutions

(3) Area of a Square
= 1 × 1
= 11 × 11
= 121 Sq.cm

(4) Area of a Square
= 1 × 1
= 18 × 18
= 324 Sq.cm

(5) Area of a Square
= 1 × 1
= 25 × 25
= 625 Sq.cm

KSEEB Solutions

(6) Area of a Square
= 1 × 1
= 30 × 30
= 900 Sq.cm

(7) Area of a Square
= 1 × 1
= 41 × 41
= 1,681 Sq.cm

(8) Area ofa Square
= 1 × 1
= 55 × 55
= 3,025 Sq.cm

KSEEB Solutions

(9) Area of a Square
= 1 × 1
= 63 × 63
= 3,969 Sq.cm

(10) Area ofa Square
= 1 × 1
= 92 × 92
= 8464 Sq.cm

KSEEB Solutions

IV. Solve the following problems:

Question 1.
The length of a square room is 6m. What is the area of the floor of the room?
Answer:
Area of the floor of the room
= 1 × 1
= 6 × 6
= 36 sq.mt

Question 2.
The length of a square paper is 21 cm. What is its area?
Answer:
Area of a square paper
= 1 × 1
= 21 × 21
= 441 sq.cm

KSEEB Solutions

Question 3.
The length of a square canvas cloth is 15m. Find the area of the cloth?
Answer:
Area of the cloth
= 1 × 1
= 15 × 15
= 225 Sq.mt

Question 4.
The length of a square room is 8m. How many granite stones of 2m × 1m are required to cover the floor of the room?
Answer:
Area of a square room
= 1 × 1
= 8 × 8
= 64 Sq.mt
The number of granites
= 2m × 1m
= 2m

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 36

KSEEB Solutions

Question 5.
The length of a square room is 400cm. How many square tiles of 10cm in length, are required to cover the floor of the room.
Answer:
Area of a square
= 1 × 1
= 400 × 400
= 1600 sq.cm
The length of a square tile = 10cm
The number of tiles to cover 1,60,000

KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area 37

Area of tiles = 10 × 10
= 100 sq.cm

KSEEB Solutions

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