# 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**

Area = length × breadth

Perimeter = 2(length+breadth)

**Square**

Area = (side)2

Perimeter = 4(side)

**Circle**

Diameter = 2 × radius

Area = π × (radius)

^{2}

**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 = 4/3πr

^{3}Surface area = 4πr

^{2}

**Cone**

Volume = 1/3πr

^{2}hTotal 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.