# Mensuration Notes for Defence Exams

By Naveen Singh|Updated : June 16th, 2020

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.

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.

## Mensuration Notes for Defence Exams

Rectangle

Let d1 and d2 are diagonals of the given rectangle ABCD.

then, both diagonals are equal but not perpendicular to each other.

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

Circle

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

 Radius Perimeter (2πr) Area (πr2) 7 44 154 14 88 616 21 132 1386 28 176 2464 35 220 3850 42 264 5544

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

Sector

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

Segment

Area of minor segment

Area of major segment

There is a square of side ‘a’; ‘r’ is the inradius and ‘R’ is the circumradius.

Triangle:

Let ABC is a triangle and M1, M2 and M3 are medians of the given triangle.

Then,

Given, ABC is a triangle and a, b and c are the sides of given triangle. Let ‘r’ is the inradius 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.

Equilateral triangle

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.

Isosceles triangle

Regular Polygon

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

or

# External angle of regular polygon
# sum of all external angle = 360º

#### For Regular Hexagon

Parallelogram

Let a and b are the sides, h is the height and d1 and d2 are the diagonals of parallelogram

then,

Area of parallelogram = (i) Base × height

(ii)

(iii)

Imp. Relation

Imp. Note: In rectangle, parallelogram, square and Rhombus diagonals bisect other.

Rhombus

In Rhombus, diagonals are not equal to each other but they bisect each other at 90 degree.

Area = Base × height = a x h

Or Area

Trapezium

Case 1: If AD = BC, then DM = CN

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