Geometry Hacks – SSC CGL 2018

Geometry, an extremely important topic under SSC CGL 2018, is known to either make or break your Quantitative section. To ensure that you make the productive choice, Team SSCHacks is here with the best Geometry Hacks for SSC CGL 2018. Take a look.
(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)
Note : Such questions are rare in geometry and in most of the questions you will have to pick up your pen, but stillwe shared these questions just to unleash the 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, our 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.

Q. 1

In the above question you can assume that the triangle is equilateral. Then angles A, B and C will be 60. Hence sin2A + sin2B + sin2C = (√3/2)2 + (√3/2)2 + (√3/2)2 = 9/4
Answer : (B)
Note : 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 is solved, some additional equations were given, but then too it is supposed that the triangle is equilateral only because the equation was symmetrical. Had the equation been unsymmetrical, it could not have been able to assume a =b =c

Q. 2

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 is:

(√3/4)s² = (√3/4) (2√3a)²  = 3√3a²
Now put a = b = c in all the options and check which one will give 3√3a²
A) √3a
B) 3√2a2
C) √2a
D) 3√3a2
Answer : (D)

Moving on…
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 were asked from 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.
Note : All the below questions are taken from a single paper [Tier2, 2014].

Formula 1 : (Applicable only for RightAngled Triangle)
Q. 3 

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
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..

Q. 4

Again apply the same formula
Answer : (C)

Formula 2 : ½*b*c*sinϴ
This formula is only applicable when ϴ lies between the sides ‘b’ and ‘c’.

Q. 5

Apply the formula, area = 1/2 * 10 * 10 * sin45
Answer : (D)

Formula 3 : Area of a triangle = r * S
where r = inradius
S = semiperimeter
Q. 6

Given, perimeter = 50, hence semiperimeter = 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)a²
Area of a triangle = inradius * semiperimeter
= a * (3s)/2
Now, a * (3s)/2 = 3√3a²
or s = 2√3a [Same result]

Since, we could not cover all the Superhacks in a single post, we will update more posts containing the other important Geometry Hacks. Happy Learning.

All the Best !         

-Team SSCHacks 

Kickstart the SSC CGL 2018 preparation with these AMAZING SUPERHACKS !!