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 = 4/3πr3
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.