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How many different rectangles can be made with a perimeter of 24 cm?

  • Last Updated : 21 Nov, 2021

Rectangle is a closed two-dimensional figure composed of four sides and four vertices. All angles of the rectangle are 90 degrees. A rectangle with all sides equal is equivalent to a square. A rectangle is composed of two pairs of parallel sides, length, and width respectively.

Perimeter of Rectangle

The perimeter of a rectangle is the length of the outer boundary of a rectangle. It is also calculated by the summation of the total measure of both the lengths and breadths of the rectangle.

Perimeter of Rectangle Formula

Let us assume a rectangle of perimeter P, whose length and width are ‘l’ and ‘w’ respectively is 2(l + w).

Perimeter of a Rectangle Formula = 2 (Length + Width) units

How many different rectangles can be made with a perimeter of 24 cm?

Solution:

There may be multiple rectangles having a perimeter equivalent to 24 cm. For example, one rectangle of sides 5 cm and 7cm have perimeter 24, similarly, a rectangle of sides 4 cm and 8 cm have perimeter 24. So, there can be many rectangles of dimensions such as (5,7), (4,8), (3,9), (3.6,8.4), etc. 

If a and b are the sides of the rectangle then the perimeter is 2a+2b. You are given a string 24 cm long (which will be the perimeter) from which to make the rectangles. So,

2a + 2b = 24

2(a + b) = 24

a + b = 12

If the sides must be integers, then the possibilities are:



a = 1, b = 11

a = 2, b = 10

a = 3, b = 9

a = 4, b = 8

a = 5, b = 7

a = 6, b = 6

If a=7, then b=5 which is the same as the fifth rectangle listed, i.e., a=5, b=7.

So there are only 6 possible rectangles (the square a=6 & b=6, being a special case of a rectangle).

Complete step by step solution:

We know, 

The length of the string forming the rectangle is 24cm. Therefore, it can be concluded that the perimeter of the rectangle is equivalent to 24 cm.

The perimeter of a rectangle is given by 2(a+b)=P, which implies that the perimeter is equal to double of the sum of the sides of the rectangle. 

Substituting, we get, 

∴ 2(a+b) = 24

Therefore,

 (a+b) = 12

Starting with a = 6, b = 6.

On increasing the value by 1, we conclude that, following are the other sides of the rectangle

a = 7, b = 5

a = 8, b = 4



a = 9, b = 3

a = 10, b = 2

a = 11, b = 1

Thus all we can say 6 rectangles are possible and the following are those rectangles:

  1. 6, 6, 6, 6
  2. 5, 7, 5, 7
  3. 4, 8, 4, 8
  4. 3, 9, 3, 9
  5. 2, 10, 2, 10
  6. 1, 11, 1, 11

Sample Questions

Question 1. Assume that the sides of the rectangle are in the ratio of 6:4 and its perimeter is 100 cm then find its dimensions?

Solution:

Here we have,

Perimeter of rectangle = 100 cm

Ratio of length and breadth is 6 : 4

We have to find length and breadth of the rectangle

Assume the common ratio be x

Thus, the sides will be 6x and 4x

As we know that,

Perimeter of the rectangle = 2(length + breadth)

Perimeter of the rectangle = 2(6x + 4x)

Perimeter of the rectangle = 2 × 10x

Perimeter of the rectangle = 20x

100 = 20x

x = 100/20

x = 5



Therefore,

Length = 6x = 6 × 5 = 30 cm

Breadth = 4x = 4 × 5 = 20 cm

Therefore, 

Length is 30 cm and breadth is 20 cm.

Question 2. If the sides of the rectangle are in the ratio of 4:2 and its perimeter is 600 cm then find its dimensions?

Solution:

Here we have,

Perimeter of rectangle = 600 m

Ratio of length and breadth is 4 : 2

We have to find length and breadth of the rectangle

Assume the common ratio be x

Thus, the sides will be 4x and 2x

As we know that,

Perimeter of the rectangle = 2(length + breadth)

Perimeter of the rectangle = 2(4x + 2x)

Perimeter of the rectangle = 2 × 6x

Perimeter of the rectangle = 12x

600 = 12x

x = 600/12

x = 50

Therefore,

Length = 4x = 4 × 50 = 200 m

Breadth = 2x = 2 × 50 = 100 m

Therefore,

Length is 200 m and breadth is 100 m.

Question 3. Find how many tiles of dimensions length 26 cm and breadth 14 cm will be required to cover a rectangular godown floor of dimensions length 1040 cm and breadth 280 cm?  

Solution:

To find the number of tiles

First we need to find the area of tile and godown

Area of rectangle = Length × Breadth

Area of rectangle tile = 26 × 14

Area of rectangle tile = 364 cm2

Area of rectangle godown = 1040 × 280

Area of rectangle godown = 291200 cm2

Now,

Finding the number of tiles required

Number of tiles = Area of godown/ Area of a tile

Number of tiles = 291200/364

Number of tiles = 800 tiles



Therefore,

We will be needing 800 tiles to cover the godown floor.

Question 4. Find the cost of fencing a rectangular playground with an area of 3200 m2 at the rate of ₹20 per meter? If the length is double the breadth.

Solution:

Here we have to find the cost of fencing the rectangular playground

As given in the question, length is double the breadth

Breadth = b

Length = 2 × b = 2b

First we need to find the length and breadth of the playground

Area of rectangle = Length × Breadth

Area of rectangular playground = 2b × b

3200 = 2b2

b2 = 3200/2

b2 = 1600

b = √1600

b = 40

Breadth = 40 m

Length = 2 × breadth = 2 × 40 = 80 m

Further,

Perimeter of the rectangle = 2(length + breadth)

Perimeter of the rectangle = 2(80 + 40)

Perimeter of the rectangle = 2 × 120

Perimeter of the rectangle = 240 m

Now,

Finding the cost of fencing

Cost of fencing the playground =   ₹20 × perimeter of the playground

Cost of fencing the playground =   ₹20 × 240

Cost of fencing the playground =   ₹480

Therefore,

Cost of fencing the rectangular playground is ₹480.

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