Mensuration 2D

Mensuration 2D mainly deals with problems on perimeter and area. The shape is two dimensional, such as triangle, square, rectangle, circle, parallelogram, etc. This topic does not has many variations and most of the questions are based on certain fixed formulas.

• Perimeter : The length of the boundary of a 2D figure is called the perimeter.
• Area : The region enclosed by the 2D figure is called the area.
• Pythagoras Theorem : In a right angled triangle, (Hypotenuse)² = (Base)² + (Height)^2?

Triangle
Let the three sides of the triangle be a, b and c.

• Perimeter = a + b + c
• Area
  1) 2s = a + b + c
      Area = {s (s - a) (s - b) (s - c) }
  2) Area = 1/2  x Base x Perpendicular Height

Rectangle

• Perimeter = 2 x (length + Breadth)
• Area = Length x Breadth

Square

• Perimeter = 4 x Side Length
• Area = (Side Length)² = 1/2 x (Diagonal Length)²

Parallelogram

• Perimeter = 2 x Sum of adjacent sides
• Area = Base x Perpendicular Height

Rhombus

• Perimeter = 4 x Side Length
• Area =1/2 x Product of diagonals

Trapezium

• Perimeter = Sum of all sides
• Area = 1/2 x Sum of parallel sides x Perpedicular Height

Circle

• Perimeter = 2 π Radius
• Area = π (Radius)²
• Length of an arc that subtends an angle θ at the centre of the circle = (π x Radius x θ) / 180
  HOW ?
  for complete 1 round angle = 360⁰
  then 360⁰ = 2 π R 
  FOR 10 = 2 π R / 360⁰ 
  i.e =  π R / 180⁰
  for a particular angle θ =>   θ  x π R / 180^{0 
  =} θ  π R / 180⁰ 
• Area of a sector that subtends an angle θ at the centere of the circle = (π x Radius² x θ) / 360⁰
  HOW ?
  
  for complete 1 round i.e 360⁰ area will be πR² 
  360⁰ =  πR²    
  1⁰  =   π R² / 360⁰
  for a particular angle θ 
  =>  θ π R² / 360⁰

Some examples and solutions
Example 1: Find the perimeter and area of an isosceles triangle whose equal sides are 5 cm and height is 4 cm.
Solution : Applying Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Height)²
=> (5)² = (0.5 x Base of isosceles triangle)² + (4)²
=> 0.5 x Base of isosceles triangle = 3
=> Base of isosceles triangle = 6 cm
Therefore, perimeter = sum of all sides = 5 + 5 + 6 = 16 cm
Area of triangle = 0.5 x Base x Height = 0.5 x 6 x 4 = 12 cm²

Example 2 : A rectangular piece of dimension 22 cm x 7 cm is used to make a circle of largest possible radius. Find the area of the circle such formed ? 
Solution : In questions like this, diameter of the circle is lesser of length and breadth.
Here, breadth Diameter of the circle = 7 cm
=> Radius of the circle = 3.5 cm
Therefore, area of the circle = π (Radius)² = π (3.5)² = 38.50 cm²

Example 3 : A pizza is to be divided in 8 identical pieces. What would be the angle subtended by each piece at the centre of the circle ?
Solution : By identical pieces, we mean that area of each piece is same.
Total angle of a circular arrangement = 360⁰
i.e now we have to divide this 360⁰ into 8 equal angles 
8 part = 360⁰
1 part = 360⁰/8 = 45⁰
Therefore, angle subtended by each piece at the centre of the circle = 45 degrees
 

Example 4 : Four cows are tied to each corner of a square field of side 7 cm. The cows are tied with a rope such that each cow grazes maximum possible field and all the cows graze equal areas. Find the area of the ungrazed field.
Solution : For maximum and equal grazing, the length of each rope has to be 3.5 cm.
=> Area grazed by 1 cow = (π x Radius² x θ) / 360
=> Area grazed by 1 cow = (π x 3.5² x 90) / 360 = (π x 3.5²) / 4
=> Area grazed by 4 cows = 4 x [(π x 3.5²) / 4] = π x 3.5²
=> Area grazed by 4 cows = 38.5 cm²
Now, area of square field = Side² = 7² = 49 cm²
=> Area ungrazed = Area of field – Area grazed by 4 cows
=> Area ungrazed = 49 – 38.5 = 10.5 cm²

Example 5 : Find the area of largest square that can be inscribed in a circle of radius ‘r’.
Solution : The largest square that can be inscribed in the circle will have the diameter of the circle as the diagonal of the square.
=> Diagonal of the square = 2 r
=> Side of the square = 2 r / 2^1/2
=> Side of the square = 2^1/2 r
Therefore, area of the square = Side² = [2^1/2 r]² = 2 r²