You can tackle many tricky geometry questions with a single word 'Construction'. Yes, just precisely draw the diagram as asked in the question and manually measure the unknown side/angle that has been asked...

Since no information about the type of triangle is given, hence we will assume it to be 'Equilateral'.

How to draw an equilateral triangle? Just draw a line AB = 6 cm, and make ∠A = 60 and ∠B = 60, the point where the two angles will intersect, will be point C.

You will find that AO:OG = 3:1

Again, a simple question.

Step 1: Draw a circle of any radius with centre O and diameter AB.

Step 2: Put protractor at A and draw ∠CAB = 34.

Step 3: Join BC

Step 4: Measure ∠CBA with protractor.

Step 2: Draw internal bisector of ∠ABC. Now, be very cautious, a slight mis-measurement give you a wrong answer. Use protractor for accurate measurements (For simplicity, you can take ∠A = 60 and hence use protractor to make ∠CBD = 30)

Step 3: Again with the help protractor, make ∠BDA = 90

Step 4: Draw DE parallel to BC

Step 5: Measure AE with a scale. You will get AE = 3 cm.

Hence actual length of AE = 2*3 = 6 cm

∠ADB = ∠BDF = 90

∠ADB = ∠FDB (BD is the angle bisector)

∠BAD = ∠BFD

Triangles ABD and FBD are congruent. So AD = DF

Triangles ADE and AFC are similar

AE/AC = AD/AF = 1/2

AE = 1/2 * 12 = 6 cm

**Disclaimer:**Use this method only when you are unable to figure out how to solve a question.**Q. 1) 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****(A) 1:1 (B) 1:2 (C) 2:1 (D) 3:1**Since no information about the type of triangle is given, hence we will assume it to be 'Equilateral'.

**Step 1:**Draw an Equilateral triangle of any size(but it should be big enough for the purpose). Take 6 cm.How to draw an equilateral triangle? Just draw a line AB = 6 cm, and make ∠A = 60 and ∠B = 60, the point where the two angles will intersect, will be point C.

**Step 2:**Now draw the medians BE and CF. Make sure that E and F lie exactly in the middle of AC and AB respectively.**Step 3:**Complete the remaining figure.**Step 4:**Measure AO and OG with a ruler.You will find that AO:OG = 3:1

**Answer: (D)**

Step 1: Draw a circle of any radius with centre O and diameter AB.

Step 2: Put protractor at A and draw ∠CAB = 34.

Step 3: Join BC

Step 4: Measure ∠CBA with protractor.

**Answer: (C)**

**This may appear tricky if you go on solving it with conventional methods. But it is way too easy when solved through construction.**

**Step 1:**Draw a triangle ABC with AC = 6 cm. Here we are using the concept of**scaling**. In the exam you can't draw a line 12 cm long. Hence we are reducing AC to 1/2. Hence, when we will measure AE, we will have to multiply it by 2 to get the exact answer.Step 2: Draw internal bisector of ∠ABC. Now, be very cautious, a slight mis-measurement give you a wrong answer. Use protractor for accurate measurements (For simplicity, you can take ∠A = 60 and hence use protractor to make ∠CBD = 30)

Step 3: Again with the help protractor, make ∠BDA = 90

Step 4: Draw DE parallel to BC

Step 5: Measure AE with a scale. You will get AE = 3 cm.

Hence actual length of AE = 2*3 = 6 cm

**Answer: (B)****Method:****AD extended meets BC at F.**

∠ADB = ∠BDF = 90

∠ADB = ∠FDB (BD is the angle bisector)

∠BAD = ∠BFD

Triangles ABD and FBD are congruent. So AD = DF

Triangles ADE and AFC are similar

AE/AC = AD/AF = 1/2

AE = 1/2 * 12 = 6 cm

**Q. 4)**

**Although this is a very simple question and you would get 60 degrees as your answer when you solve it. But, if by any chance, if you are not able to figure out the method, solve it using construction.**

__: The median of an isosceles triangle cuts the opposite side at right angles. Hence ∠AOB = 90.__**Method**

**Similarly you can solve many geometry questions with proper measurements. Try solving some questions on your own...**

did u use this method? never saw someone comming with ruler n pencil in ssc xams :) good method but i doubt it in case of ssc

ReplyDeleteyes i used this method, but I drew a rough diagram (without scale). With a little practice, you won't be needing scale and protractor to construct triangles...

Deletedid u use this method? never saw someone comming with ruler n pencil in ssc xams :) good method but i doubt it in case of ssc

ReplyDeletebhai,will the officials permit scale,protractor,compass etc inside the exam hall?will they debar us if we get caught using materials other than pen & pencil?Genuine reply awaited.anyway great job sir!

ReplyDeleteI took protractor and there wasn't any problem. I would recommend you to practise questions. See what a 45 degree, 60 degree angle looks like. With time, you won't be needing anything...

DeleteSir I have an doubt in qus plz given a req solution

ReplyDelete2p+1/p=4 the value of p(cub)+1/8(cub)

The approx value of p is 1.7 [read algebra tricks - 3]

Deletesolve it now

Thnk u sir

DeleteSir I have an doubt in qus plz given a req solution

ReplyDelete2p+1/p=4 the value of p(cub)+1/8(cub)

Hello Sir..Any updates regarding the exam? Speculations going around whether the exam will be in Aug itself or still get extended..and regarding the exam pattern too

ReplyDeletemost bakwasss method...

ReplyDeleteI am sorry for that...

DeleteSir, (assuming ssc cgl to be held in the month of august) with 1.5 months left for ssc cgl exam should I prepare reasoning from previous year question papaers or should I study topic wise from rs aggarwal- verbal and non verbal reasoning book. please help me sir. thanks in advance.??

ReplyDeleteOne of the angles of a Right angled triangle is 15, and the hypotenuse is 1 m. the area of the triangle is (in sq. cm)

ReplyDelete[SSC 2014 Tier 2]

This comment has been removed by the author.

Delete0.5* hypotenuse*(1/2 of hypotenuse)

Delete= 2500 square cm

ssc hack is awesome but how to hack whatsapp

ReplyDeleteSir, can you plz tell me if there will be sectional cutoff in tier-1??

ReplyDeleteI am weak in General Awareness Section. If there is no sectional cut off, I will try to skip it.

Also Plz make it clear if tier-1 is only of qualifying nature...

For 2nd Question, image in below link

ReplyDeletehttp://imgur.com/0udJK65

The angle subtended by diameter will be 90 deg

therefore,

angle(CAB) + angle(CBA) + 90 = 180

=> angle(CBA)=56

Yes brother...all these questions are easily solvable with conventional methods. But as I have mentioned earlier, use construction only if you are not able to figure out how to solve a particular question. Many candidates find geometry tricky and hence take time to solve the questions with conventional methods, but construction can be done by anyone...

DeleteSir, can you plz tell me if there will be sectional cutoff in tier-1??

DeleteI am weak in General Awareness Section. If there is no sectional cut off, I will try to skip it.

Also Plz make it clear if tier-1 is only of qualifying nature...

Q1) since equilateral is assumed

ReplyDeletehttp://imgur.com/28qBImk

sir,apne kha ki ap rough fig bnate the..but rough figure k sath correct values kese niklegi???

ReplyDeleteSir, I don't have any government issued identity address proof other than bankpassbook and pan card. Can, I use these documents for the police verification. I have same addresse on attestation form and bank passbook.

ReplyDeleteNice info.. Checkout the SSC CGL Admit Card 2016

ReplyDeletesir page no 11 there is a option how can I calculate easily 3.2^4 or 3.3.^4 .there is a note please clik here..at pdf this is not available.please send me thricks...mail id :atanu1604@gmail.com

ReplyDeleteSir can u tell the strategy to follow in the last 20 days before tier 2

ReplyDelete