The CDS exam consists of 3 papers who appear for IMA, INA and AFA entries. One of the major portions of there preparation is Elementary Mathematics. To assist the students and boost their preparation, we are here with PDF notes on Mensuration Topic. This would give you an idea about what type of question can be asked in the exam.
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Mensuration Notes for Defence Exams
Let d1 and d2 are diagonals of the given rectangle ABCD.
then, both diagonals are equal but not perpendicular to each other.
Area of rectangle = length x breadth and perimeter = 2(length+breadth)
Path outside the rectangle
Suppose there is a park having length l and breadth b. There is a road of width x outside of it.
Then, Area of path = 2x (l + b + 2x)
The path inside the rectangle
Suppose there is a park having length l and breadth b. There is a road of width x inside of it.
Then, Area of path = 2x (l + b – 2x)
When there is a road along both the length and breadth of the park
Then, Remaining area of Rectangle (shaded region) = (l–x) (b-x)
Area of the path = lx + bx – x2
Given a circle of radius ‘r’
We recommend you learn this table as it will save your time in calculating these all.
If radius is ‘r’, then perimeter = 2πr and Area = πr2
Length of Rope
Let ‘d’ is the diameter of pulley and ‘r’ is the radius, then d = 2r. All pulleys are similar.
Length of rope = 2d + 2pr
Length of rope = 3d + 2pr
Length of rope = 4d + 2pr
Note: Trick to remember these formulas: number of pulleys x diameter + Perimeter of one pulley
In this circle, ‘r’ is the radius, θ is the angle made by the arc of length ‘l’
Length of arc
Area of sector
Area of sector when ‘l’ is given
Area of minor segment
Area of major segment
Area of shaded portion
Inradius and Circumradius of Square:
There is a square of side ‘a’; ‘r’ is the inradius and ‘R’ is the circumradius.
Let ABC is a triangle and M1, M2 and M3 are medians of the given triangle.
Inradius of triangle:
Given, ABC is a triangle and a, b and c are the sides of given triangle. Let ‘r’ is the inradius of triangle.
Circumradius of triangle
Given, ABC is a triangle and a, b and c are the sides of given triangle. Let ‘R’ is the circumradius of triangle.
Right angle triangle
Given ‘a’ is the base, ‘b’ is the perpendicular and ‘c’ is the hypotenuse of triangle ABC.
Where, h is the height of triangle,
Hence, we can say that height of equilateral triangle is equal to the sum of side perpendicular of the triangle.
Let, n = no. of sides of regular polygon and a = length of side of regular polygon
# Internal angle of regular polygon =
# Sum of internal angle of regular polygon
# Angle made by centre =
#Area of Regular polygon
# External angle of regular polygon
# sum of all external angle = 360º
For Regular Hexagon
Circumradius R = a
Let a and b are the sides, h is the height and d1 and d2 are the diagonals of parallelogram
Area of parallelogram = (i) Base × height
Imp. Note: In rectangle, parallelogram, square and Rhombus diagonals bisect other.
In Rhombus, diagonals are not equal to each other but they bisect each other at 90 degree.
Area = Base × height = a x h
Case 1: If AD = BC, then DM = CN
You can download these notes both in Hindi & English:
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