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 = 
  2. Area = 0.5 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 = 0.5 x (Diagonal Length)²

Parallelogram

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

Rhombus 

• Perimeter = 4 x Side Length
• Area = 0.5 x Product of diagonals

Trapezium

• Perimeter = Sum of all sides
• Area = 0.5 x Sum of parallel sides x Perpendicular Height

Circle

• Perimeter = 2 π Radius
• Area = π (Radius)²
• Length of an arc that subtends an angle θ at the center of the circle = (π x Radius x θ) / 180
• Area of a sector that subtends an angle θ at the center of the circle = (π x Radius² x θ) / 360

Sample Problems

Question 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

Question 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

Question 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. 
=> Area of each piece = (π x Radius² x θ) / 360 = (1/8) x Area of circular pizza 
=> (π x Radius² x θ) / 360 = (1/8) x (π x Radius²) 
=> θ / 360 = 1 / 8 
=> θ = 360 / 8 = 45 
Therefore, angle subtended by each piece at the centre of the circle = 45 degrees 

Question 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

Question 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

Question 6 : A contractor undertakes a job of fencing a rectangular field of length 100 m and breadth 50 m. The cost of fencing is Rs. 2 per meter and the labour charges are Re. 1 per meter, both paid directly to the contractor. Find the total cost of fencing if 10 % of the amount paid to the contractor is paid as tax to the land authority. 
Solution : Total cost of fencing per meter = Rs. 2 + 1 = Rs. 3 
Length of fencing required = Perimeter of the rectangular field = 2 (Length + Breadth) 
=> Length of fencing required = 2 x (100 + 50) = 300 meter 
=> Amount paid to the contractor = Rs. 3 x 300 = 900 
=> Amount paid to the land authority = 10 % of Rs. 900 = Rs. 90 
therefore, total cost of fencing = Rs. 900 + 90 = Rs. 990 

Problems on Mensuration 2D | Set 2

Programs on Triangle:

• Find area of a triangle
• Find Perimeter of a triangle
• Find area of triangle if two vectors of two adjacent sides are given
• Calculate area and perimeter of equilateral triangle
• Minimum height of a triangle with given base and area

Programs on Rectangle:

• Program for Area And Perimeter Of Rectangle
• Total area of two overlapping rectangles
• Maximum area of rectangle possible with given perimeter
• Maximum area rectangle by picking four sides from array
• Find minimum area of rectangle with given set of coordinates

Programs on Square:

• Program to find the area of a Square
• Area of a square from diagonal length
• Sum of Area of all possible square inside a rectangle
• Find Perimeter / Circumference of Square and Rectangle

Programs on Parallelogram:

• Program for Circumference of a Parallelogram
• Program to find the Area of a Parallelogram
• Find area of parallelogram if vectors of two adjacent sides are given

Programs on Rhombus and Trapezium:

• Program to calculate area and perimeter of a rhombus whose diagonals are given
• Area of the biggest possible rhombus that can be inscribed in a rectangle
• Program to calculate area and perimeter of Trapezium

Programs on Circle:

• Program to find area of a circle
• Program to find Circumference of a Circle
• Program to calculate area of an Circle inscribed in a Square
• Area of circle inscribed within rhombus
• Area of a Circumscribed Circle of a Square
• Area of a circle inscribed in a regular hexagon
• Program to find the Radius of the incircle of the triangle

This article has been contributed by Nishant Arora
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