Divisibility Rules: Quantitative Aptitude Section There are some specific rules by which we can determine the divisor of the given number. Today I wil

Polynomial division
asked 2021-02-26
Divisibility Rules: Quantitative Aptitude Section There are some specific rules by which we can determine the divisor of the given number. Today I will discuss divisibility rules from 2 to 19. Using these rules you can easily determine a divisor of given number, however large it may be. Let me tell you the rules of divisibility from 2 to 19.

Answers (1)



Triangle = Figure with three sides.  Study the following triangle: A,B,C to points. a, b, c to sides. x, y, z to angles.
Perimeter of triangle = a + b + c
Remember that, Sum of all the angles is always \(180^{\circ}\)
i.e. x + y + z = \(180^{\circ}\)

Classification of Triangles

Basically there are three types of triangles excluding right angle triangle. Let me tell you how they vary from each other.

  • Scalene Triangle
  • Isosceles Triangle
  • Equilateral Triangle

Scalene Triangle: No side of triangle is equal.
Isosceles Triangle: Two sides of triangle are equal.
Equilateral Triangle: All sides of triangle are equal.


  Scalene Isosceles Equilateral
Definition a ne b ne c ane b = c a = b = c
Area   A \((b/4) \sqrt{4a^2 – b^2}\) \((3)^{(1/2)(a^2)/4}\)
Height - \(\sqrt{(4a^2-b^2) /2}\) \((3)^{(1/2)a/2}\)

\(A = {s(s-a)(s-b)(s-c)}^{1/2}\)
where, s = (a+ b+ c)/2





Properties of external angles of Triangle:

1. Sum of all exterior angles is \(360^{\circ}\)
Study the following set of triangles and their exterior angles,
a, b, c to Interior angles. p, q, r and s, t, u to Exterior angles.
So, sum of exterior angles = \(360^o\) i.e. p + q + r = 360^o and s + t + u = \(360^o\)
2. Next property of exterior angle which is important in paper point of view:

External angle = Sum of two internal angles.

For example: In above figures,
r = a + b
q = a + c
s = b + c and so on.


Right angle Triangle

Following triangle is a right angle triangle i.e. a triangle with one out of three 90^o angle. image




Area of right angle triangle 

Area = 1/2 times Base times Perpendicular 



Example with Solution

 Example: In following figure, CE is perpendicular to AB, angle ACE = 20^o and angle ABD = 50^o. Find angle BDA:
Solution: To Find: angle BDA
For this what we need --- angle BAD  Because, Sum of all angles = \(180 ^o\)
Consider, triangle ECA,
CEA + EAC + ACE = \(180^o\) i.e. \(90^o\)\(20^o\) + EAC = \(180^o\) Therefore, EAC = \(70^o\)
Now, come to triangle ABD,
ABD + BDA + BAD = \(180^o\)  \(70^o\) + \(50^o\) + BAD = \(180^o\) Therefore, BAD = \(60^o\)
Example: In given figure. BC is produced to D and angle BAC = \(40^o\) and angle ABC = \(70^o\). Find angle ACD:
 Solution: In above figure, ACD is an exterior angle, and according to property, Exterior angle = Sum of interior angles Therefore, ACD = \(70^o\) + \(40^o\) =\(110^o\)
This is not the end of this chapter. These are just the basics. In next session, I will discuss some important results, properties (congruency, similarity) and much more. Always remember, Geometry needs practice and time.


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