Sunday, 20 March 2016

[Q] 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 -
Q. 1

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

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

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)
Q. 4

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
where a = side of the equilateral triangle

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-
Q. 5

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
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
Q. 6
Given
(n - 2)*180 = 2*360
So n - 2 = 4
or n = 6
Answer : (B)
Q. 7

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 :)

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59 comments:

  1. i have shared your blog sir..
    thanks sir ... please provide more...

    ReplyDelete
  2. Thank 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 ..

    ReplyDelete
  3. Sir, in last two formulae you have missed to write "regular".

    ReplyDelete
    Replies
    1. In last formulare it would be exterior instead of interior

      Delete
    2. This comment has been removed by the author.

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    3. Sir,
      Reasoning m 'total no. Of triangle' wale questions ka koi shortcut hota hai kya?

      Delete
  4. sir please share english also and how to approach to get good marks in english

    ReplyDelete
  5. sin theta + cos theta
    --------------------- = 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

    ReplyDelete
    Replies
    1. 1st Ques
      sinx^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

      Delete
    2. in ques 1
      is 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

      Delete
    3. where have you written 'if' and 'then' in the 1st ques? Please write the question again...

      In 2nd ques... tan2x.tan4x = 1 means x = 15 (hit and trial)

      Delete
    4. if,
      sin theta + cos theta
      --------------------------------- = 3
      sin theta - cos theta

      then, sin^4 - coz^4 = ?

      Delete
    5. Oh...so that's a 'division' sign in between :D
      then 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

      Delete
  6. Sir in fourth formula
    given formula is applicable if and only if it is a regular polygon

    ReplyDelete
  7. regarding your complaint
    nobody wants to raise the level of competition by sharing ur blog
    haha :)

    ReplyDelete
  8. I have shared sir..sir plz give some techniques for mensuration and curcles also...and thanku very much again😀😀

    ReplyDelete
  9. Hi prashanth is there any trick for solving this :-
    If 1+cos^2 Ø = 3SinØCosØ , then the integral value of CotØ is (0<Ø< 180/2 )

    A) 1 B) 2 C) 0 D) 3

    ReplyDelete
    Replies
    1. divide the whole equation with (sinx)^2
      Use 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

      Delete
  10. Thanks a ton bro . Having one more doubt ..
    Tried 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)

    ReplyDelete
  11. you can't assume the angle in this question because the question itself is asking the value of sinx
    For 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.

    ReplyDelete
  12. Hello 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
    Second 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

    ReplyDelete
    Replies
    1. MBA karo ya kuch aur karo, CGL dene me koi problem nai aayegi...

      Delete
  13. Hello 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
    Second 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

    ReplyDelete
  14. Is there any shorter way to solve this out bro.. Thanks in advance

    * Find the value of 3(sin^4x+cos^4x)+2(sin^6x+cos^6x)+12 sin^2x cos^2x

    ReplyDelete
    Replies
    1. This comment has been removed by the author.

      Delete
  15. Pls upload more tricks of non verbal reasoning sir

    ReplyDelete
  16. Awesome tricks and all other posts as well..plz upload some more tricks

    ReplyDelete
  17. Thanks sir..for these tricks ..sir pls will you post some tricks related to eligible votes or eligible candidates...

    ReplyDelete
  18. Hello Sir,you are doing a great job.
    Please 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

    ReplyDelete
  19. Thank you sir............Your tricks is really helpful for all aspirants.

    ReplyDelete
  20. Sir, in last question is that exterior or interior..?

    ReplyDelete
    Replies
    1. It doesn't matter because when you have 2 polygons, then :
      Difference between their interior angles = Difference between their exterior angles.

      Delete
  21. Sir how to takev pdf prints of all hacks

    ReplyDelete
    Replies
    1. Open the article by clicking on its title. At the bottom of the article you will see a green "Print PDF" button...

      Delete
    2. sir after questions to koi" printPDF "nhi h..kha se le pdf print???plz hlp..

      Delete
  22. In a triangle ABC, G is the centroid. If AG=BG, Find angle BGC?
    Explain this one prashant sir.

    ReplyDelete
    Replies
    1. I think you mean AG=BC
      Let 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)

      Delete
    2. sir its AG=BG & answer is 120 given in the book

      Delete
    3. Oh...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)

      Delete
    4. thank you sir

      Delete
  23. In 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=?
    (a) AD^2 (b)AE^2 (c)CE^2 (d)CD^2

    ReplyDelete
    Replies
    1. Join CE
      Triangles 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

      Delete
    2. Thank you so much sir... you are genius...

      Delete
  24. If 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

    EXPLAIN PRASHANT SIR

    ReplyDelete
    Replies
    1. (1)1:1
      (2)1:2
      (3)2:1
      (4)3:1

      Delete
    2. Extend AG to cut BC at D
      E 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

      Delete
    3. 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?

      Delete
  25. Same query I have

    ReplyDelete
  26. ABC 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