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 -
Answer : (A)
Incentre
While solving questions on circumcentre, you should imagine the above figure. Here two things are quite useful -
∠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
Answer : (C)
Q. 3
Answer : (D)
Centroid
Remember one more property related to centroid,
Area(ΔOBC) = 1/3Area(ΔABC)
We know DG = 1/2AG
Hence DG = 2cm
Answer : (A)
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-
Here is a question based on this fact-
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
Answer : (C)
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
Answer : (B)
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
Answer : (D)
Note : Please note that each and every formula that I have mentioned in this post is extremely important and I can guarantee that after mugging all these formulae you will be able to solve all the questions on orthocentre, incentre, circumcentre, centroid and interior-exterior angles...
Keep reading :)
To buy the super-hit SSC Hack Book, follow the below link-
Buy SSC Hack-Book
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 -
∠AIC = 90 + ∠ABC/2 [Whenever you see the word 'incentre' in the paper, you should immediately recall this formula]
Circumcentre
While solving questions on circumcentre, you should imagine the above figure. Here two things are quite useful -
OA = OB = OC [radius of the circle]. And hence,
∠OAB = ∠OBA
∠OCA = ∠OAC
∠OBC = ∠OCB
Hence ∠AOC = 120
Answer : (C)
Q. 3

Answer : (D)
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
Answer : (A)
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
- Circumradius or outer-radius= 2h/3 = a/√3
- Height(h) = (√3/2)a
3) For a right-angled triangle
- Orthocentre is at the right angle vertex
- Circumcentre is the midpoint of the hypotenuse
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-
AC = 30cm
We have to find AB and BC.
AB = BC [as ∠ACB = ∠ABC = 45]
AB = sin45 * AC = 30/√2 = 15√2
Answer : (C)
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
Answer : (B)
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
Answer : (D)
Note : Please note that each and every formula that I have mentioned in this post is extremely important and I can guarantee that after mugging all these formulae you will be able to solve all the questions on orthocentre, incentre, circumcentre, centroid and interior-exterior angles...
Keep reading :)
To buy the super-hit SSC Hack Book, follow the below link-
Buy SSC Hack-Book













i have shared your blog sir..
ReplyDeletethanks sir ... please provide more...
Thanks a lot �� awsome
ReplyDeleteI have already shared bro
ReplyDeleteThank you so much sir, for thinking of us aspirants. No book can beat your techniques .. Please be with us . Thanks again for all your effort ..
ReplyDeleteThnkew sir
ReplyDeleteSir, in last two formulae you have missed to write "regular".
ReplyDeleteIn last formulare it would be exterior instead of interior
Deleteupdated :)
DeleteThis comment has been removed by the author.
DeleteSir,
DeleteReasoning m 'total no. Of triangle' wale questions ka koi shortcut hota hai kya?
sir please share english also and how to approach to get good marks in english
ReplyDeletesin theta + cos theta
ReplyDelete--------------------- = 3
sin theta - cos theta
then, sin^4 - coz^4 = ?
one more ques
tan2theta.tan4theta = 1
then tan3theta = ?
how to apply 'jugaad' in these ques
1st Ques
Deletesinx^4 - cosx^4 = (sinx^2 + cosx^2)(sinx^2 - cosx^2) = (sinx^2 - cosx^2) = (sinx + cosx)(sinx - cosx)
2 ques :
Note that x = 15
so tan3x = 1
in ques 1
Deleteis there no need of 'if' statement as u' ve solved d ques frm 'then' part only...
and upto (sinx+cosx)(sinx-cosx)is ok...
whts beyond that
as answer is 3/5, how to get to this answer
in second ques !!!
where is x=15
where have you written 'if' and 'then' in the 1st ques? Please write the question again...
DeleteIn 2nd ques... tan2x.tan4x = 1 means x = 15 (hit and trial)
if,
Deletesin theta + cos theta
--------------------------------- = 3
sin theta - cos theta
then, sin^4 - coz^4 = ?
Oh...so that's a 'division' sign in between :D
Deletethen take the denominator on the RHS and solve the equation
you will get tanx = 2
means P = 2, B = 1, H = root(5)
sinx = 2/root(5)
cosx = 1/root(5)
Put it in (sinx)^2 - (cosx)^2 = 4/5 - 1/5 = 3/5
Sir in fourth formula
ReplyDeletegiven formula is applicable if and only if it is a regular polygon
updated :)
DeleteI have shared sir..sir plz give some techniques for mensuration and curcles also...and thanku very much again😀😀
ReplyDeleteHi prashanth is there any trick for solving this :-
ReplyDeleteIf 1+cos^2 Ø = 3SinØCosØ , then the integral value of CotØ is (0<Ø< 180/2 )
A) 1 B) 2 C) 0 D) 3
divide the whole equation with (sinx)^2
DeleteUse 1/(sinx)^2 = (cosecx)^2
And (cosecx)^2 = 1 + (cotx)^2
Now whole equation is in the form of cotx
Solve this quadratic equation, you will get cotx = 1
Thanks a ton bro . Having one more doubt ..
ReplyDeleteTried assuming the angle as 45 for the following question but not working out.. Is there any short cut
* If secø+tanø=2+ root 5 then sinø is (0<=Ø<=90)
you can't assume the angle in this question because the question itself is asking the value of sinx
ReplyDeleteFor such questions I have shared the trick in Part-1 of Geometry
If secx + tanx = p, then secx = (p^2 + 1)2p
From here you will get secx. Reciprocate it to find cosx. From cosx find sinx.
😁 okay . Thanks a ton bro
DeleteHello sir pls guide mein abh ssc cgl ki preparation kar rhi hu bt meine private job kia h approx 4.5 yrs toh koi problem toh ni btana pdta h exam clear krne k baad yaa nii
ReplyDeleteSecond mein MBA kr rhi hu IMT GHAZIABAD s toh ky koi problem toh nii
Pls help and guide me in this regard
Waiting for ur reply sir
MBA karo ya kuch aur karo, CGL dene me koi problem nai aayegi...
DeleteHello sir pls guide mein abh ssc cgl ki preparation kar rhi hu bt meine private job kia h approx 4.5 yrs toh koi problem toh ni btana pdta h exam clear krne k baad yaa nii
ReplyDeleteSecond mein MBA kr rhi hu IMT GHAZIABAD s toh ky koi problem toh nii
Pls help and guide me in this regard
Waiting for ur reply sir
Is there any shorter way to solve this out bro.. Thanks in advance
ReplyDelete* Find the value of 3(sin^4x+cos^4x)+2(sin^6x+cos^6x)+12 sin^2x cos^2x
Pls upload more tricks of non verbal reasoning sir
ReplyDeleteAwesome tricks and all other posts as well..plz upload some more tricks
ReplyDeleteHardcore P...Problem solving
ReplyDeleteThanks sir..for these tricks ..sir pls will you post some tricks related to eligible votes or eligible candidates...
ReplyDeleteHello Sir,you are doing a great job.
ReplyDeletePlease help me solve this
E is the mid point of the median AD of tr triangle ABC.If BE is extended it meets the side AC at F,then CF is equal to?
a)AC/3. b)2AC/3. c)AC/2. d) None
Can you olease suggest any trick to solve such type of questions
Thanks
Thank you sir............Your tricks is really helpful for all aspirants.
ReplyDeleteSir, in last question is that exterior or interior..?
ReplyDeleteIt doesn't matter because when you have 2 polygons, then :
DeleteDifference between their interior angles = Difference between their exterior angles.
Ok Sir, thanks..
ReplyDeleteOk Sir, thanks..
ReplyDeleteSir how to takev pdf prints of all hacks
ReplyDeleteOpen the article by clicking on its title. At the bottom of the article you will see a green "Print PDF" button...
Deletesir after questions to koi" printPDF "nhi h..kha se le pdf print???plz hlp..
DeleteIn a triangle ABC, G is the centroid. If AG=BG, Find angle BGC?
ReplyDeleteExplain this one prashant sir.
I think you mean AG=BC
DeleteLet AG = 2x
Then BG = x (centroid divided the median in 2:1 ratio)
BC = AG = 2x
Let AG when extended cuts BC at D
Then D is the midpoint of BC
BD = DC = x
Now BG = BD = DC = x
That means D is the centre of a circle with diameter BC and one of the radius as BG.
Hence BGC = 90 (angle in a semi-circle)
sir its AG=BG & answer is 120 given in the book
DeleteOh...sorry...AG=BG, this is true for equilateral triangles...Hence angle A = 60 and BGC = 2*A = 120 (G is also the centre of circumcircle)
Deletethank you sir
DeleteIn a triangle ABC, Angle bisector of angle A intersects BC at D and circumcircle of ∆ABC at E. then it is always true that- AB.AC+DE.AE=?
ReplyDelete(a) AD^2 (b)AE^2 (c)CE^2 (d)CD^2
Join CE
DeleteTriangles ABD and ACE are similar. So,
AD/AC = AB/AE
AB.AC = AD.AE
AB.AC = (AE - DE).AE
AB.AC + DE.AE = AE^2
Thank you so much sir... you are genius...
DeleteIf BE and CF are two medians of a triangle ABC and G is the centroid and EF and AG intersects each other at O then find - AO:OG
ReplyDeleteEXPLAIN PRASHANT SIR
options?
Delete(1)1:1
Delete(2)1:2
(3)2:1
(4)3:1
Extend AG to cut BC at D
DeleteE and F are midpoints, hence EF=1/2BC
Triangles AFO and ABD are similar
OF/BD = OA/AD
Now OF/BD = 1/2
Hence OA/AD = 1/2
That means O is the midpoint of AD
OA = OD
And AG:DG = 2:1 (centroid divided in 2:1 ratio)
From these, you will get OA:OD : 3:1
sir OF/BD=1/2 kyu liya hai? EF=1/2BC hai to isse ye kese ho skta hai ki OF/BD=1/2 hoga? D midpoint hai BC ka but OF ka samjh ni aa rha sir?
DeleteSame query I have
ReplyDeleteABC is a triangle with o as its centroid. If AO is equal to BC. Then what is the value of angle AOC. PLEASE SOMEONE PROVIDE THE SOLUTION.THANKS
ReplyDelete