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# If the length of a rectangle is reduced by 20% by what percent would the width have to be increased to maintain the original area?

Mathematics is a subject that is responsible for calculations. And, according to the type of calculation or operation to be performed mathematics is divided into different branches like algebra, geometry, arithmetic, etc.

Mensuration is a branch that deals with the calculation of parameters like perimeter, area, volume, etc. of various shapes whether it be two-dimensional or three-dimensional.

In two-dimensional shapes, objects comprise of length and width or any two dimensions that can be represented on a plane surface. While, three-dimensional shapes, objects are placed in the real world and have three dimensions that are length, width, and height.

### Some Basic formulas for 2D and 3D shapes

2D Shapes

• Rectangle

• Square

Area = (side)2

Perimeter = 4(side)

• Circle

• Triangle

Area = 1/2 breadth × height

3D shapes

• Cube

Volume = (side)3

Lateral surface area = 4 × (side)2

Total surface area = 6 × (side)2

• Cuboid

Volume = length × breadth × height

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

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

• Sphere

Volume = 4/3πr3

Surface area = 4πr2

• Cone

Volume = 1/3πr2h

Total surface area = πr (l+radius)

### If the length of a rectangle is reduced by 20% by what percent would the width have to be increased to maintain the original area?

Solution:

Let x and y represent the length and breadth of the rectangle respectively.

As we know the original area of the rectangle by the standard formula

Area of rectangle(A) = l × b

Area of given rectangle = xy

According to the question the length of rectangle is reduced by 20%. So, the new length would be

=>x-20/100x

=>x(1-1/5)

=>4/5x

And, k% be the increase in breadth to maintain original area.

=>y+k/100y

=>y(1+k/100)

As it is stated that the original and new area needs to be same.

=>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 original area.

### Sample Problems

Problem.1. When the length and breadth  of a rectangle is increased by 40% by what percent would the area will increase.

Solution:

Let x and y be the length and breadth of the rectangle respectively.

The area of rectangle by standard formula will be

=> area of rectangle(A)= xy

According to the question,

Length of rectangle is increased by 40% = x+40/100x

=>x+40/100x

=>x(1+40/100)

=>140/100x

=>7/5x=1.4x

Breadth of rectangle is increased by 40%=y+40/100y

=>y+40/100y

=>y(1+40/100)

=>140/100y

=>7/5y=1.4y

Now, the new area of rectangle will be =1.4x x 1.4y =1.96xy

And, increase in area of rectangle =1.96xy-xy = 0.96xy

Increase in percentage of area of rectangle = 0.96xy/xy x 100%

= 96%

Problem.2. When the length is increased by 20% and breadth  of a rectangle is increased by 40% by what percent would the area will increase.

Solution:

Let x and y be the length and breadth of the rectangle respectively.

The area of rectangle by standard formula will be

=> area of rectangle(A)= xy

According to the question,

Length of rectangle is increased by 20%=x+20/100x

=>x+20/100x

=>x(1+20/100)

=>120/100x

=>6/5x=1.2x

Breadth of rectangle is increased by 40%=y+40/100y

=>y+40/100y

=>y(1+40/100)

=>140/100y

=>7/5y=1.4y

Now, the new area of rectangle will be =1.2x x 1.4y =1.68xy

And, increase in area of rectangle =1.68xy-xy = 0.68xy

Increase in percentage of area of rectangle=0.68xy/xy x 100%

=68%

Problem.3. The length of a rectangle is triple its breadth. Its perimeter is 40m. Find the length and breadth.

Solution:

Let the breadth of the rectangle be x

As per the question the length is double the value of breadth so it will be 3x

perimeter(P) = 40cm

Now, by the formula,

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

=>40 = 2(x+3x)

=>40 = 2.4x

=>x=40/8

=>x=5cm

=>3x=3 . 5=15cm

Hence, the length and breadth of the rectangle are 15cm and 5cm respectively.

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