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If the length of a rectangle is decreased by 5% and its breadth is increased by 5%, then find the percentage change in the area

  • Last Updated : 02 Dec, 2021

Mensuration is a branch of mathematics associated with geometry that is concerned with the measurements, analysis, and calculation of parameters of geometrical shapes with the help of standard derived formulas. The parameters of geometrical shapes that mensuration deals with are area, volume, lateral surface area, total surface area, etc. Mensuration involves the calculation of parameters of two-dimensional and three-dimensional shapes.

  • 2D (two-dimensional) shapes are geometrical figures with two dimensions are length and breadth. They do not deal with the thickness or height of the object as they are represented on a plane surface.
  • 3D(three-dimensional) shapes are geometrical figures with three dimensions that are height, width, and depth. These kinds of objects are found in real surroundings.

The given article covers a detailed discussion under the topic of mensuration, explaining its subtopics and standard formulas for the calculation of various parameters as per the shapes. The article also includes mathematical problems along with their solution for a better understanding of the calculation process.

The standard formula of mensuration

Mensuration and its calculations are conducted with the help of some standard set formulas. Each shape has its own formula for determining different parameters like area, volume, surface area, etc based on their dimensions. Some of the formulas for different shapes are given in the below table.

Shapes Formulas
Rectangle

Perimeter = 2(l + b)

Area = l x b

Square

Area = (side)2

Perimeter = 4 x side

Circle

Diameter = 2 x radius

Area = π x (radius)2

Triangle Area = 1/2 b x h
Cube

Volume = (side)3

Lateral surface area = 4 x (side)2

Total surface area = 6 x (side)2

Cuboid

Volume = l x b x h

Lateral surface area = 2 x height(l + b)

Total surface area = 2(lb + bh + lh)

Sphere

Volume = 4/3πr3

Surface area = 4πr2

Cone

Volume = 1/3πr2h

Total surface area = πr(l + radius)

Rectangle

A rectangle is a geometrical shape having four enclosed sides. The opposite sides of rectangles are equal and parallel to each other. In a rectangle, every pair of adjacent sides form an interior angle of 90 degrees and the diagonals bisect each other.

Properties of rectangle

A rectangle as a geometrical shape has a number of properties and some of them are:

  • A rectangle has four sides.
  • The opposite sides of a rectangle are always parallel and equal to each other.
  • Each pair of adjacent sides form an interior angle measuring 90 degrees.
  • The sum of all interior angles of a rectangle equals 360 degrees.
  • Its diagonals are equal in length and bisect each other.
  • As the sides of a rectangle are parallel, it is considered as a parallelogram with an interior angle of 90 degrees.

If the length of a rectangle is decreased by 5% and its breadth is increased by 5%, then find the percentage change in the area

Solution:

Let us assume that x and y be the length and breadth.

Here, the original area of the rectangle by the area formula of the rectangle will be

Area of rectangle(A) = l x b

Area of given rectangle = xy

Now, according to the question the length of given rectangle is reduced by 5%, and the new length will be 

=> x – 5/100x

=> x(1 – 5/100)

=> x(100 – 5/100)

=> 95x/100

=> 0.95x

Then, the breadth of given rectangle is increased by 5% and the new breadth will be

=> y + 5/100y

=> y(1 + 5/100)

=> y(100 + 5/100)

=> 105y/100

=> 1.05y

The new area of rectangle will be = 0.95x x 1.05y = 0.9975xy

The decrease in area of rectangle = xy – 0.9975xy = 0.0025xy

Now, as per the question,

Decrease in percentage of the area of rectangle = 0.0025xy/xy x 100

                                                                                  = 0.25%

Sample Problems

Problem 1. When the length is increased by 5% and the breadth of a rectangle is increased by 10% by what percent the area will increase.

Solution:

Let us assume that x and y be the length and breadth. 

The area of a rectangle by the standard formula will be

=> area of rectangle(A)= xy

According to the question,

Length of rectangle is increased by 5% = x + 5/100x

=> x + 5/100x

=> x(1 + 5/100)

=> 105/100x

=> 1.05x

Breadth of rectangle is increased by 10%=y+10/100y

=> y + 10/100y

=> y(1 + 10/100)

=> 110/100y

=> 1.1y

Now, the new area of rectangle will be = 1.05x x 1.1y = 1.155xy

And, increase in area of rectangle = 1.155xy – xy = 0.155xy

Increase in percentage of area of rectangle = 0.155xy/xy x 100%

                                                                          = 15.5%

Problem 2. If the length of a rectangle is reduced by 40% by what percent would the width have to be increased to maintain the original area?

Solution:

Let us assume x and y represent the length and breadth of the rectangle

As we know the original area of the rectangle is: 

Area of rectangle(A) = l × b

Area of given rectangle = xy

According to the question, the length of the rectangle is reduced by 20%. 

So, the new length would be

=> x – 20/100x

=> x(1 – 1/5)

=> 4/5x

Here, k% represents the increase in breadth 

=> y + k / 100y

=> y(1 + k / 100)

It is given that the original and new area needs to be same.So, 

=> Original area = new area

=> xy = (4/5) x (1 + k/100)y

=> 1 = (4/5)(100 + k/100)

=> 100 + k/100 = 5/4

=> 100 + k = 125

=> k = 125 – 100

=> k = 25

Hence, the breadth needs to be increased by 25% to maintain the original area.

Problem 3. If the length of a rectangle is triple its breadth and its perimeter is 80m then find the length and breadth.

Solution:

Let us assume that the breadth of the rectangle be x

According to the question the length is double the value of breadth. Hence it will be 3x

perimeter(P) = 80cm

Now, by the formula,

Perimeter of rectangle(P) = 2(l + b)

=> 80 = 2(x + 3x)

=> 80 = 2.4x

=> x = 80/8

=> x = 10cm

=> 3x = 3 x 10 = 30cm

Hence, the length and breadth of the rectangle are 30cm and 10cm respectively.


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