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...

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

Delete= 2500 square cm

It's wrong. Correct answer is 1250 Sq cm

Delete(1/4)*sin^2(2*15) Sq m

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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

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