Class 5: Maths Chapter 12 solutions. Complete Class 5 Maths Chapter 12 Notes.
Maharashtra Board Solutions Class 5-Maths (Problem Set 48) - Part 2: Chapter 12- Perimeter and Area
Maharashtra Board 5th Maths Chapter 12, Class 5 Maths Chapter 12 solutions
Important Questions and Answers.
Question 1.
Write the perimeter of each figure in the box given below it.
(1)

Solution:
Perimeter [ ] DABCD
= 20 + 16 + 7 + 14
= 57 cm
∴ 57 cm
(2)

Solution:
Perimeter of the figure
= 12 + 18 + 8 + 8 + 18
= 64m
∴ 64m
(3)

Solution:
Perimeter of the figure
= 10 + 6 + 6 + 10 + 8 + 8
= 48 cm
∴ 48 cm
Question 2.
If a square of side 1 cm is cut out of the corner of a larger square with side 3 cm (see the figure), what will be the perimeter of the remaining shape?

Solution:
Perimeter
= 2 + 3 + 3 + 2 + 1 + 1
= 12 cm
∴ 12 cm
Formula for the perimeter of a rectangle

Perimeter = length + breadth + length + breadth Opposite sides of a rectangle are of the same length.
So, the perimeter of a rectangle
= twice the length + twice the breadth
= 2 × length + 2 × breadth
Perimeter of a rectangle = 2 × length + 2 × breadth
Example : The length of the rectangle below is 7 cm and its breadth, 3 cm. Let us find its perimeter.
Perimeter of rectangle PQRS = 2 × length + 2 × breadth
= 2 × 7 + 2 × 3
= 14 + 6
= 20
Therefore, the perimeter of the rectangle is 20 cm.
Formula for the perimeter of a square
The lengths of all the sides of a square are equal. Therefore, the perimeter of a square = four times the length of one of its sides.
Perimeter of a square = 4 × the length of one side
Example : The length of one side of a square is 6 cm. Find its perimeter. The perimeter of a square is four times the length of one side.
Perimeter of a square = 4 × length of one side
= 4 × 6
= 24
Therefore, the perimeter of the square is 24 cm.
Word problems
Example (1) The length of a rectangular park is 100 m, while its width is 80 m. What is its perimeter?
Perimeter of the rectangle = 2 × length + 2 × breadth
= 2 × 100 + 2 × 80
= 200 + 160
= 360
The perimeter of the rectangular park is 360 m.
Example (2) How much wire will be needed to put a triple fence around a square plot with side 30 m? What will be the total cost of the wire at ₹ 70 per metre ?
To put a single fence around the square plot, we need to find its perimeter.
Perimeter of a square = 4 × length of one side = 4 × 30 = 120
The perimeter of the square plot is 120 metres. Since the fence is to be a triple fence, we must triple the perimeter.
120 × 3 = 360 m of wire will be needed.
Now let us find out how much the wire will cost. One metre of wire costs ₹ 70.
Therefore, the cost of 360 m of wire will be 360 × 70 = 25, 200.
The total cost of wire for putting a triple fence around the plot will be ₹ 25, 200.
Additional Important Questions and Answers
Question 1.
Solution:
Perimeter of the figure
= 2 + 6 + 2 + 6
= 16 cm
∴ 16 cm
Question 2.
Solution:
Perimeter of the figure
= 3 + 3 + 3 + 3
= 12 cm
∴ 12 cm
Question 3.
Solution:
Perimeter of the figure
= 12 + 13 + 5
= 30 cm
∴ 30 cm
Question 4.
If four squares of side 1 cm ¡s cut out of all the corners of a larger square with side 4 cm (see the figure), what will be the perimeter of the remaining shape?

Solution:
Perimeter
= 2 + 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1
= 16 cm
∴ 16 cm
Download PDF
Maharashtra Board Solutions Class 5-Maths (Problem Set 48) - Part 2: Chapter 12- Perimeter and Area
Chapterwise Maharashtra Board Solutions Class 5 Maths :
Part 1
- Chapter 1- Roman Numerals (Problem Set 1)
- Chapter 2- Number Work (Problem Set 2)
- Chapter 2- Number Work (Problem Set 3)
- Chapter 2- Number Work (Problem Set 4)
- Chapter 2- Number Work (Problem Set 5)
- Chapter 2- Number Work (Problem Set 6)
- Chapter 3- Addition and Subtraction (Problem Set 7)
- Chapter 3- Addition and Subtraction (Problem Set 8)
- Chapter 3- Addition and Subtraction (Problem Set 9)
- Chapter 3- Addition and Subtraction (Problem Set 10)
- Chapter 3- Addition and Subtraction (Problem Set 11)
- Chapter 3- Addition and Subtraction (Problem Set 12)
- Chapter 3- Addition and Subtraction (Problem Set 13)
- Chapter 4- Multiplication and Division (Problem Set 14)
- Chapter 4- Multiplication and Division (Problem Set 15)
- Chapter 4- Multiplication and Division (Problem Set 16)
- Chapter 5- Fractions (Problem Set 17)
- Chapter 5- Fractions (Problem Set 18)
- Chapter 5- Fractions (Problem Set 19)
- Chapter 5- Fractions (Problem Set 20)
- Chapter 5- Fractions (Problem Set 21)
- Chapter 5- Fractions (Problem Set 22)
- Chapter 5- Fractions (Problem Set 23)
- Chapter 6- Angles (Problem Set 24)
- Chapter 6- Angles (Problem Set 25)
- Chapter 6- Angles (Problem Set 26)
- Chapter 6- Angles (Problem Set 27)
- Chapter 7- Circles (Problem Set 28)
- Chapter 7- Circles (Problem Set 29)
- Chapter 7- Circles (Problem Set 30)
- Chapter 7- Circles (Problem Set 31)
Part 2.
- Chapter 8- Multiples and Factors (Problem Set 32)
- Chapter 8- Multiples and Factors (Problem Set 33)
- Chapter 8- Multiples and Factors (Problem Set 34)
- Chapter 8- Multiples and Factors (Problem Set 35)
- Chapter 9- Decimal Fractions (Problem Set 36)
- Chapter 9- Decimal Fractions (Problem Set 37)
- Chapter 9- Decimal Fractions (Problem Set 38)
- Chapter 9- Decimal Fractions (Problem Set 39)
- Chapter 9- Decimal Fractions (Problem Set 40)
- Chapter 9- Decimal Fractions (Problem Set 41)
- Chapter 9- Decimal Fractions (Problem Set 42)
- Chapter 10- Measuring Time (Problem Set 43)
- Chapter 10- Measuring Time (Problem Set 44)
- Chapter 10- Measuring Time (Problem Set 45)
- Chapter 11- Problems on Measurement (Problem Set 46)
- Chapter 11- Problems on Measurement (Problem Set 47)
- Chapter 12- Perimeter and Area (Problem Set 48)
- Chapter 12- Perimeter and Area (Problem Set 49)
- Chapter 12- Perimeter and Area (Problem Set 50)
- Chapter 13- Three Dimensional Objects and Nets (Problem Set 51)
- Chapter 14- Pictographs (Problem Set 52)
- Chapter 15- Patterns (Problem Set 53)
- Chapter 16- Preparation for Algebra (Problem Set 54)
- Chapter 16- Preparation for Algebra (Problem Set 55)
- Chapter 16- Preparation for Algebra (Problem Set 56)