Many students asked me to share some geometry tricks. So here is the Part-1 :

(1) Be it algebra or geometry, such questions are always there that don't deserve your rough space.

Like the above questions asks you the value of x. Now value of x can't be negative in this context. Because CO = x - 3 and any negative value of 'x' will make CO negative, like if x = -8, then CO = -11. We know that side can never be negative. Hence all the options that have negative value for x are wrong.

In the above question a certain area is asked. Now see the options and observe that in options A, C and D if we take r=7, 2 and 1 respectively, then the area will become zero, and area can never be zero in this case, although value of 'r' can be 7, 2 or 1.

In the above question you can assume that the triangle is equilateral. Then angles A, B and C will be 60. Hence sin

Now put a = b = c in all the options and check which one will give 3√3a

A) √3a

Moving on...

I have found that SSC has some real love with 'Area of a triangle' and it keeps on asking it again and again under different contexts. Like in Tier 2 (2014) around 5 questions asked area of triangles.

Therefore it is very important that you memorize all the possible formulas to calculate it. This will save you a lot of time.

In the above question you may struggle to calculate the area. You can try this question yourself (with a timer), to see how much time you are taking...

There is a direct formula for such questions -

Apply this formula, area = (100^2 * sin30)/4 = 1250

Again apply the same formula

where r = inradius

S = semi-perimeter

Given, perimeter = 50, hence semi-perimeter = 25

Area = 6 * 25 = 150

Area of an equilateral triangle = (√3/4)s

Area of a triangle = inradius * semi-perimeter = a * (3s)/2

(1) Be it algebra or geometry, such questions are always there that don't deserve your rough space.

Like the above questions asks you the value of x. Now value of x can't be negative in this context. Because CO = x - 3 and any negative value of 'x' will make CO negative, like if x = -8, then CO = -11. We know that side can never be negative. Hence all the options that have negative value for x are wrong.

**Answer : (B)**

In the above question a certain area is asked. Now see the options and observe that in options A, C and D if we take r=7, 2 and 1 respectively, then the area will become zero, and area can never be zero in this case, although value of 'r' can be 7, 2 or 1.

**Answer : (B)****Such questions are rare in geometry and in most of the questions you will have to pick up your pen, but still I shared these questions just to unleash the**__Note :__*jugaad*within you. Moreover, if you are able to solve even a single question with such approach, you will save at least 1 crucial minute in the examination hall.

**(2) In geometry too, my beloved concept of 'symmetry' plays an important role. If in any question you find that some symmetrical expressions/equations are given, you can assume the triangle to be equilateral.**In the above question you can assume that the triangle is equilateral. Then angles A, B and C will be 60. Hence sin

^{2}A + sin^{2}B + sin^{2}C = (√3/2)^{2}+ (√3/2)^{2}+ (√3/2)^{2}= 9/4**Answer : (B)****If in any question, the sides of a triangle are given (like a, b and c), then you can assume a = b = c, but make sure any additional detail is not given. Like in the above question that I solved, some additional equations were given, but then too I supposed the triangle to be equilateral only because the equation was symmetrical. Had the equation been unsymmetrical, I could not have been able to assume a =b =c**__Note :__
Here the lengths of perpendiculars are given to be a, b and c. Note that no additional information is given, hence it is safe to assume a = b = c. Let the side of the triangle be 's'. The figure will look like this -

We have to establish a relation between 'a' and 's'. In an equilateral triangle, the incentre, orthocentre, circumcentre and centroid, all coincide. So you can calculate 'a', by which ever method you like.

a = inradius = s/2√3 [The inradius of an equilateral triangle is s/2√3 and the circumradius is s/√3]

Hence s = 2√3a

We know the formula for calculating the area of an equilateral triangle = (√3/4)s

^{2}
= (√3/4) (2√3a)

^{2}
= 3√3a

^{2}Now put a = b = c in all the options and check which one will give 3√3a

^{2}A) √3a

B) 3√2a

C) √2a^{2}**D) 3√3a**

^{2}

**Answer : (D)**

^{}

I have found that SSC has some real love with 'Area of a triangle' and it keeps on asking it again and again under different contexts. Like in Tier 2 (2014) around 5 questions asked area of triangles.

Therefore it is very important that you memorize all the possible formulas to calculate it. This will save you a lot of time.

**: All the below questions are taken from a single paper [Tier-2, 2014].**

__Note__

__Formula 1 : (Applicable only for Right-Angled Triangle)__There is a direct formula for such questions -

**Answer : (D)**

__Note :__You can choose any of the angles of the triangle (except the 90 degree one) and you will get the same result. Like in the above question, the 3 angles of the triangle are 15, 75 and 90.**sin2*15 = sin30**

**sin2*75 = sin150**

**And we know that sin150 = sin30**

**Hence it doesn't matter which angle you take. But to avoid any confusion,**

__always take the smaller angle (15 in this case)__.

**Next question -**

**Answer : (C)**

**Formula 2 :**½*b*c*sinϴ__This formula is only applicable when ϴ lies between the sides 'b' and 'c'.__

**Apply the formula, area = 1/2 * 10 * 10 * sin45**

**Answer : (D)**

__Formula 3 : Area of a triangle = r * S__where r = inradius

S = semi-perimeter

Given, perimeter = 50, hence semi-perimeter = 25

Area = 6 * 25 = 150

**Answer : (C)**

**Now when you know this formula, in Q. 2 above, where we had to establish relation between 's' and 'a', you can establish it by applying this formula too.**

Area of an equilateral triangle = (√3/4)s

^{2}

Area of a triangle = inradius * semi-perimeter = a * (3s)/2

Now,

a * (3s)/2 = (√3/4)s

^{2}
or s = 2√3a [Same result]

**Geometry Part-2 to follow soon...**

**To buy the super-hit SSC Hack Book, follow the below link-**

**Buy SSC Hack-Book**

Sir marvellous..u have done a great thing for all of us.generally after qualifying no body helps in such a great way.thank u so much sir.

ReplyDeleteI request u if u could give us all the important formulae for geometry yhen it would of much help.waiting for ur next post☺☺

This comment has been removed by the author.

ReplyDeleteNow my daily routine to refresh SSC hack blog to check updates. You are doing great sir. Thanks a lot

ReplyDeleteA sum of Rs.1550 was lent partly at 5% and partly at 8% simple interest. The total interest received after 3 years is Rs.300. The ratio of money lent at 5% to that at 8% is ?

ReplyDeleteA)5:8

B)8:5

C)31:6

D)16:15

If an amount P1 is lent out at simple interest of R1% per annum and another amount P2 at simple interest rate of R2% per annum, then the rate of interest for the whole sum can be given by

DeleteR = (P1.R1 + P2.R2)/(R1 + R2)

SI=300, time=3 and P=1550

calculate R from above data.

Now apply the formula, you will get the answer 16:15

Thanks sir

DeleteThanks sir

DeleteWhat annual installment will discharge a debt of Rs.6450 due in 4 years at 5 person simple interest?

ReplyDeleteA) Rs. 1500

B) Rs. 1400

C) Rs. 1600

D) Rs. 1550

The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum

Delete= 100D/[100T + RT(T-1)/2]

D = 6450, T = 4, R = 5%

Put the values

Answer : 1500

Thanks sir

DeleteSir. If u have no problem,kindly send ur WhatsApp No.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteSin 20°sin40°sin60°sin80°

ReplyDeleteKoi shortcut hah isko solv krne k sir ... pls rply

Put the value of sin60

Deletenow write sin40.sin80 = sin(60-20).sin(60+20)

Apply the formula sin(a+b).sin(a-b) = sin^2a - sin^2b

Hence sin40.sin80 = sin60^2 - sin20^2

Then multiply and divide by 4

Now apply the formula 3sinx - 4sin^3x = sin3x

you will get the answer : 3/16

Sir is SSC is difficult as upse

ReplyDeleteSir is SSC is difficult as upse

ReplyDeleteSir is SSC is difficult as upse

ReplyDeletethanks sir

ReplyDeleteAbcd is a parallelogram . p is a point on BC such that pb:pc is 1:2. Do produced meets an produced at q. If the area of the triangle bpq is 20 sq units. What is the area of triangle dcp(

ReplyDeleteAbcd is a parallelogram . p is a point on BC such that pb:pc is 1:2. Do produced meets an produced at q. If the area of the triangle bpq is 20 sq units. What is the area of triangle dcp(

ReplyDeleteThere are typos in this question. Please post again...

DeleteDear Prashanth , you are doing such a great service by helping all the aspirants out here. Please post such tricks for Algebra too when you get time . Wishing you to attain great heights and a happy life good health . Thanks a ton

ReplyDeleteI have already shared algebra tricks-

Deletehttp://sschacks.blogspot.in/2016/03/algebra-tricks-for-ssc-cgl-algebra-is.html

Yaar apke marks kitne the? T1+T2+interview

ReplyDeleteSorry, as I'm browsing through the mobile, I was unable to find it then . Got it. Thanks a ton :) thanks again for spending your quality time

ReplyDeleteteen admi A B C milkar ek kam krne par ek karya ko akele A se 6 hr kam me, akele B se 1 hr kam me aur C k akele kam krne k liye avshyak samay se adhe samya me pura kr skte h. A aur B milkar us karya ko kitne time me kr skte h??

ReplyDeleteindebited

ReplyDeleteif circumference of a circle is decreased by50% then the pecentage decrease in its area is????

ReplyDeletecircumference is getting reduced by 50%, that means radius is getting reduced by 50%

DeleteNew area = pi * (0.5r)^2 = 0.25*pi*r^2

From pi*r^2, the area has reduced to 0.25*pi*r^2

So area has been reduced by 75%

Gr work prashant...every articles of urs is thougroly researched and is indeed a masterpiece...gr8 work

ReplyDeleteQ. chords AC and BD of a circle with centre O intersect at right angles at e. if angle OAB=25 degree then the value of angle EBC. sir please explain the concept of intersection of chords at right angle.

ReplyDeletePALLAVI9918@GMAIL.COM

ReplyDeletePLZ SENT THE BOOKOF MNATH SIR