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.

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.

Over the years SSC has asked many questions based on this simple formula. Like -

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

Remember one more property related to centroid,

Area(ΔOBC) = 1/3Area(ΔABC)

We know DG = 1/2AG

Hence DG = 2cm

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

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

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

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

__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 mediansThe 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__**∠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__
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**

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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?

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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 :)

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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

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ReplyDeleteSecond mein MBA kr rhi hu IMT GHAZIABAD s toh ky koi problem toh nii

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Waiting for ur reply sir

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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

This comment has been removed by the author.

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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..

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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