If the length of a rectangle is increased by 50% and its breadth is decreased by 25%, then find the change percent in its area?

In our daily life, we notice different things around us, which covers some space and volume. Some things might be in one dimension like rope, thread, etc, something might be in two dimensions like plains, floor, wall or something might be in three-dimension like a ball, cylinder, etc. In mensuration when we talk about one dimension it means that we are calculating perimeter when we talk about two dimensions we are talking about the area, and when we talk about three means that we are calculating area. So we can conclude that mensuration is the branch of mathematics that deals with the geometric figures calculation and their area, perimeter, surface area, or volume.

Let discuss in detail the terms related to mensuration.

• Perimeter: Perimeter is the summation of the length of the boundary. It is a one-dimensional quantity. Its unit is the unit itself.
• Area: Area is the space within a perimeter. It is a two-dimensional quantity. Its unit is (unit)².
• Volume: Volume is the amount of space a body cover. It is a three-dimensional quantity. Its unit is (unit)³.

Area of rectangle

The area of the rectangle is the multiplication of length and width. It is extended in two dimensions. For example: If we paint a sheet of paper and we move our hands two dimensions, either horizontal or vertical. 

If the length of a rectangle is ‘l’ and the width of the rectangle is ‘b’ then the formula to calculate the area of the rectangle is:

Area = l × b

If the length of a rectangle is increased by 50% and its breadth is decreased by 25%, then find the change percent in its area? 

Step to solve the problem:

Step 1: Suppose the length and width of the rectangle.

Let the length be ‘l’ and width be ‘b’.

Step 2: Find out the initial area.

Initial Area = l×b

Step 3: According to the question, increase or decrease the length and width of the rectangle.

It is given that length is increased by 50% and the width of the rectangle is decreased by 25%.

New Length = l × (1+50/100)

                    = l × (1+1/2)

                    = (3l)/2

New Width = b × (1-25/100)

                   = b × (1-1/4)

                   = 3b/4

Step 4: Find out the new area.

New area = new length × new width

                = {(3l)/2} × (3b/4)

                = (9lb)/8

Step 5: Find out the change in percent.

Change percent = {(Final Area – Initial Area)/(Initial Area)} × 100

                         = [{(9lb)/8 – (lb)}/(lb)] × 100

                         = {(lb/8)/lb} × 100

                         = (1/8) × 100

                         =  12.5%

Step 5: If the final answer is negative, it means a decrease in area and if the final answer is positive it means an increase in area.

The answer is 12.5%, which means that area is increased by 12.5%.

Similar Questions

Question 1: If the length of a rectangle is increased by 50% and its breadth is decreased by 50%, what is the change percent in its area? 

Solution:

Let the length be ‘l’ and width be ‘b’.

Initial Area = l×b

It is given that length is increased by 50% and the width of the rectangle is decreased by 50%.

New Length = l×(1+50/100)

                    = l×(1+1/2)

                    = (3l)/2

New Width = b×(1-50/100)

                  = b×(1-1/2)

                  = b/2

New area = new length × new width

               = {(3l)/2}×(b/2)

               = (3lb)/4

Change percent = {(Final Area – Initial Area)/(Initial Area)} × 100

                         =[{(3lb)/4 – (lb)}/(lb)] × 100

                         = {(-lb/4)/lb} × 100

                         = (-1/4) × 100

                         = – 25%

The answer is -25%, which means that area is decreased by 25%.

Question 2: If the length of a rectangle is increased by 75% and its breadth is decreased by 25%, what is the change percent in its area? 

Solution:

Let the length be ‘l’ and width be ‘b’.

Initial Area = l×b

It is given that length is increased by 75% and the width of the rectangle is decreased by 25%.

New Length = l×(1+75/100)

                   = l×(1+3/4)

                   = (7l)/4

New Width = b × (1-25/100)

                  = b × (1-1/4)

                  = 3b/4

New area = new length × new width

                = {(7l)/4} × (3b/4)

                = (21lb)/16

Change percent = {(Final Area – Initial Area)/(Initial Area)} × 100

                         = [{(21lb)/16 – (lb)}/(lb)] × 100

                         = {(5lb/16)/lb} × 100

                        = (5/16) × 100

                        = 31.25%

The answer is 31.25%, which means that area is increased by 31.25%.