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
Let the three sides of the triangle be a, b and c.
- Perimeter = a + b + c
- 2s = a + b + c
- Area = 0.5 x Base x Perpendicular Height
- 2s = a + b + c
- Perimeter = 2 x (length + Breadth)
- Area = Length x Breadth
- Perimeter = 4 x Side Length
- Area = (Side Length)2 = 0.5 x (Diagonal Length)2
- Perimeter = 2 x Sum of adjacent sides
- Area = Base x Perpendicular Height
- Perimeter = 4 x Side Length
- Area = 0.5 x Product of diagonals
- Perimeter = Sum of all sides
- Area = 0.5 x Sum of parallel sides x Perpedicular Height
- 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
- Area of a sector that subtends an angle θ at the centere of the circle = (π x Radius2 x θ) / 360
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 / 21/2
=> Side of the square = 21/2 r
Therefore, area of the square = Side2 = [21/2 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 metre and the labour charges are Re. 1 per metre, 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 metre = 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 metre
=> 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
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
Please write comments if you have any doubts related to the topic discussed above, or if you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
- Mensuration 3D
- Mensuration 2D | Set 2
- Mensuration 3D | Set-2
- Mensuration 3D- Hollow sphere
- Barclays Interview Experience | On-Campus ( Virtual ) September 2020
- Minimize splits to generate monotonous Substrings from given String
- Minimize Sum of an Array by at most K reductions
- Add and Remove Edge in Adjacency Matrix representation of a Graph
- Check if two nodes are on same path in a tree | Set 2
- Minimum count of numbers required with unit digit X that sums up to N
- Print the longest path from root to leaf in a Binary tree
- Largest number M less than N such that XOR of M and N is even
- Shortest path in a complement graph
- First element of every K sets having consecutive elements with exactly K prime factors less than N