CBSE Class 4 Mathematics Perimeter Assignment

TOPIC: PERIMETER - Part (1)

Sub topic: Perimeter

Rectilinear figure
A plane figure made of line segments is called a rectilinear figure. For example, a triangle, a quadrilateral etc. are rectilinear figures.

Perimeter of a Rectilinear figure:
Definition: The sum of all sides of a rectilinear figure is called its perimeter.
OR
The distance around the edge of a figure is its perimeter.

Explanation: Suraj needs to put a frame around the picture he has made. To find how much he needs to buy, he takes a string and wraps it around the picture with a knot in every corner. He then measures the string and finds that it is 90 cm long. The string shows the length of the frame or the distance around the picture. Thus the total length or the distance around the picture is its perimeter.

Note: Perimeter must always be expressed along with its units of length like cm, m etc. We use 1 unit to measure perimeter.

Question 1. Find the perimeter of the given triangle with sides 4 cm, 7 cm, and 6 cm.
Answer: Perimeter = 4 cm + 7 cm + 6 cm = 17 cm
In simple words: The perimeter of a triangle is found by adding the lengths of all three of its sides together.

Exam Tip: Always write the unit of length (cm) at the end of your final answer.

Question 2. Find the perimeter of the figure formed by shaded squares where each square has sides of 1 cm.
Answer: There are 18 squares that are shaded, and counting the outer boundary length gives a total perimeter of 18 cm.
In simple words: Count the outer edge lengths of the shaded grid region to find the total distance around it.

Exam Tip: When counting grid perimeters, count the outer boundary segments rather than just the area inside.

Question 3. Find the perimeter of the cross-shaped figure with side lengths 3 cm, 1 cm, 1 cm, 3 cm, 1 cm, 1 cm, 3 cm, 1 cm, 1 cm, 3 cm, 1 cm, and 1 cm.
Answer:
Perimeter = Sum of the length of all sides
Perimeter = \(3\text{cm} + 1\text{cm} + 1\text{cm} + 3\text{cm} + 1\text{cm} + 1\text{cm} + 3\text{cm} + 1\text{cm} + 1\text{cm} + 3\text{cm} + 1\text{cm} + 1\text{cm} = 20\text{cm}\)
In simple words: Add up every individual outer side length of the shape to get the total perimeter.

Exam Tip: Tick off each side as you add it to make sure you do not miss any side or count any side twice.

Question 4. Find the missing length of the figure given that its total perimeter is 20 cm, and four of its known sides measure 4 cm, 1 cm, 3 cm, and 5 cm.
Answer:
Sum of the lengths of the four known sides = \(4\text{cm} + 1\text{cm} + 3\text{cm} + 5\text{cm} = 13\text{cm}\)
Length of the missing side = Perimeter - Sum of the lengths of the four sides
Length of the missing side = \(20\text{cm} - 13\text{cm} = 7\text{cm}\)
In simple words: Subtract the sum of the known sides from the total perimeter to find the missing side length.

Exam Tip: If the total perimeter is given, finding a missing side is always done through subtraction.

Question 5. How much wire does Ahmad need to fence a triangular garden with side lengths 9 m, 12 m, and 10 m?
Answer:
Perimeter of the garden = \(9\text{m} + 12\text{m} + 10\text{m} = 31\text{m}\)
Therefore, Ahmad needs 31 m of wire to fence the garden.
In simple words: Fencing a garden requires enough wire to cover its entire outer boundary, which is its perimeter.

Exam Typ: Boundary-related word problems like fencing, framing, or running around a park always ask for the perimeter.

Question 6. How much distance did Rajesh cover if he took 5 rounds around a rectangular playground with length 7 cm and breadth 5 cm?
Answer:
Perimeter of the playground = \(7\text{cm} + 5\text{cm} + 7\text{cm} + 5\text{cm} = 24\text{cm}\)
Number of rounds taken = 5
Distance covered in 1 round = 24 cm
Distance covered in 5 rounds = \(24 \times 5 = 120\text{cm}\)
Therefore, Rajesh covered 120 cm in 5 rounds.
In simple words: Multiply the perimeter of one round by the total number of rounds to get the total distance covered.

Exam Tip: For multiple rounds, first calculate the perimeter for one round and then multiply by the number of rounds.

Sub topic: Perimeter of Rectangle and Square

Perimeter of a Rectangle:
Let us consider a rectangle with length = 6 cm and breadth = 4 cm.
Perimeter of Rectangle = 2 times (sum of length and breadth)
\(= 2 \times (l + b)\)
\(= 2 \times (6\text{cm} + 4\text{cm})\)
\(= 2 \times 10\text{cm} = 20\text{cm}\)
In simple words: The perimeter of a rectangle is twice the sum of its length and breadth.

Exam Tip: Formula for rectangle perimeter is \(2 \times (l + b)\).

Perimeter of a Square:
Let us consider a square with side = 3 cm.
Perimeter of square = 4 times (length of each side)
\(= 4 \times 3\text{cm}\)
\(= 12\text{cm}\)
In simple words: Since all 4 sides of a square are equal, multiply the side length by 4.

Exam Tip: Formula for square perimeter is \(4 \times \text{side}\).

DELHI PUBLIC SCHOOL, SRINAGAR
(Worksheet)
Topic: Perimeter

Q.1. Find the perimeter of the following figures.
a. Arrow-like rectilinear figure with outer side lengths 4 cm, 8 cm, 2 cm, 3 cm, 3 cm, 2 cm, 8 cm.
Answer:
Perimeter = Sum of all outer sides
\(= 4\text{cm} + 8\text{cm} + 2\text{cm} + 3\text{cm} + 3\text{cm} + 2\text{cm} + 8\text{cm} = 30\text{cm}\)
In simple words: Add all the given side lengths around the arrow figure to find the total perimeter.

Exam Tip: Carefully list each labeled edge from the diagram to avoid missing any segment.

Q.1. b. Rectangle with length 15 cm and breadth 12 cm.
Answer:
Perimeter = \(2 \times (\text{length} + \text{breadth})\)
\(= 2 \times (15\text{cm} + 12\text{cm})\)
\(= 2 \times 27\text{cm} = 54\text{cm}\)
In simple words: Add length and breadth, then multiply the sum by 2.

Q.1. c. Triangle with side lengths 4 cm, 4 cm, and 7 cm.
Answer:
Perimeter = Sum of all three sides
\(= 4\text{cm} + 4\text{cm} + 7\text{cm} = 15\text{cm}\)
In simple words: Add the three side lengths of the triangle together.

Exam Tip: For any triangle, the perimeter is simply side 1 + side 2 + side 3.

Q. 2. Find the missing length if the perimeter of the given figure is 51 cm, and its known outer sides are 10 cm, 16 cm, 13 cm, 5 cm.
Answer:
Sum of known sides = \(10\text{cm} + 16\text{cm} + 13\text{cm} + 5\text{cm} = 44\text{cm}\)
Missing length = Total Perimeter - Sum of known sides
Missing length = \(51\text{cm} - 44\text{cm} = 7\text{cm}\)
In simple words: Subtract the sum of all the given outer sides from the total perimeter of 51 cm to find the missing side.

Exam Tip: Always double-check your addition of the known sides before subtracting from the total perimeter.

Q. 3. What length of the wire will be needed to fence the garden if the length of a garden is 14 m and breadth is 8 m?
Answer:
Length (\(l\)) = 14 m
Breadth (\(b\)) = 8 m
Length of wire needed = Perimeter of the rectangular garden
Perimeter = \(2 \times (l + b)\)
\(= 2 \times (14\text{m} + 8\text{m})\)
\(= 2 \times 22\text{m} = 44\text{m}\)
Therefore, 44 m of wire will be needed to fence the garden.
In simple words: Find the perimeter of the rectangular garden using its length and breadth to determine the wire length.

Exam Tip: State the formula clearly before substituting the values of length and breadth.

Rectilinear figure
A plane figure made of line segments is called a rectilinear figure. For example, a
triangle, a quadrilateral etc. are rectilinear figures.
Perimeter of a Rectilinear figures:
Definition: The sum of all sides of a rectilinear figure is called its perimeter.
OR
The distance around the edge of a figure is its perimeter.
Explanation: Suraj needs to put a frame around the picture he has made.

To find how much he needs to buy, he takes a string and wraps it around the picture with a knot in every corner.

This is how the string looks when he takes it off.

(String)

He then measures the string and finds that it is 90cm long. The string shows the length of the frame or the distance around the picture.

Please click the link below to download CBSE Class 4 Mathematics-Perimeter