# Geometry Tricks – 2 [Centres of a Triangle]

In this post I will share some very important formulae for Geometry. Geometry is all about theorems and properties and there are endless things to mug in it, but I will only mention those formulae that are asked by SSC every year and hence are indispensable. Make sure you absorb all these well.

Orthocentre, Incentre, Circumcentre and Centroid :

This topic is very important from SSC point of view. You all must be knowing the theory of these terms, but I will reiterate –
Orthocentre – Point of intersection of altitudes of a triangle
Incentre – Point of intersection of angle bisectors
Circumcentre – Point of intersection of perpendicular bisectors
Centroid – Point of intersection of medians
The above information, although important, is not sufficient enough to solve the questions. Each of these terms require different approach to solve the questions based on them.

Orthocentre ∠AHB + ∠ACB = 180  [Whenever you see the word ‘orthocentre’ in the paper, you should immediately recall this formula]
Over the years SSC has asked many questions based on this simple formula. Like –

Incentre

∠AIC = 90 + ∠ABC/2 [Whenever you see the word ‘incentre’ in the paper, you should immediately recall this formula]
Circumcentre

∠AOB = 2∠ACB
OA = OB = OC [radius of the circle]. And hence,
∠OAB = ∠OBA
∠OCA = ∠OAC
∠OBC = ∠OCB

Each angle of an equilateral triangle is 60 and we have seen earlier that the angle at the centre is twice that of the triangle.
Hence AOC = 120

Q. 3 Centroid

Centroid divides the medians in the ratio 2:1. Hence for the median AE,
OA = 2/3AE
OE = 1/3AE
OA = 2OE
Similarly for the medians CD and BF.

Remember one more property related to centroid,
Area(ΔOBC) = 1/3Area(ΔABC)

We know DG = 1/2AG
Hence DG = 2cm

Some additional things to remember :

1) The orthocentre, incentre, circumcentre and centroid of an equilateral traingle coincide, i.e., a single point acts as all the centres.

2) For an equilateral triangle-

• Inradius = h/3 = a/2√3
• Height(h) = (√3/2)a

where a = side of the equilateral triangle

3) For a right-angled triangle

Here is a question based on this fact-

Two angles are 45 degrees and hence the third angle is 90 degrees. The figure will look like this-

Radius of the circumcircle = 15, hence diameter = 30cm
AC = 30cm
We have to find AB and BC.
AB = BC [as ∠ACB = ∠ABC = 45]
AB = sin45 * AC = 30/√2 = 15√2

Interior and Exterior Angles
Another important topic of geometry is “Internal and External Angles” of polygon.

Now remember few formulae for such questions-
1) Sum of a interior angle and its corresponding exterior angle adds up to 180 degrees.
2) Sum of all exterior angles of a regular polygon is always 360.
3) Sum of interior angles = (n-2)*180
4) Each interior angle of a regular polygon is equal to [(n-2)*180]/n
5) Each exterior angle of a regular polygon is equal to 360/n

Mug all these formulae well and you will be able to solve all questions on interior and exterior angles easily.
Now let’s solve some CGL questions

Given
(n – 2)*180 = 2*360
So n – 2 = 4
or n = 6

Note : Here the difference between the “angles” of two polygons is given. You might be wondering that they haven’t mentioned the type of angle, whether interior or exterior. Well, it doesn’t matter! Because as far as two polygons are concerned-
Difference between their interior angles = Difference between their exterior angles
But I will take the angles to be exterior, because that will make our calculations simple
Let the sides be 5x and 4x.
Measure of each exterior angle of Polygon 1 = 360/5x = 72/x
Measure of each exterior angle of Polygon 1 = 360/4x = 90/x
Given, 90/x – 72/x = 6
so x = 3
The two angles are 5x and 4x
5x = 5*3 = 15
4x = 4*3 = 12

Had I taken the angles to be interior –
We have seen earlier that the measure of each angle of a polygon is [(n – 2)*180]/n.
Given [(5x – 2)*180]/5x – [(4x – 2)*180]/4x = 6
Solve it and you will get x = 3
So the angles are 15 and 12