SSC CGL Geometry Questions & Concepts: Important Theorems and Properties of Circle

SSC CGL Geometry Syllabus

Despite the fact that SSC CGL geometry questions are crucial, due to the extensiveness of the syllabus, candidates sometimes ignore geometry. Additionally, Geometry for the SSC CGL exam is only an extension of what we learned up till Class 10. Consequently, if handled systematically, this subject can be easily prepared in a few of months. Let us take look at the SSC CGL Geometry topics which constitute SSC CGL Geometry Syllabus.

1. Basics of Geometry
2. Polygons
3. Triangle
4. Circle
5. Quadrilateral
6. Mensuration 2D and 3D
7. Miscellaneous

SSC CGL Geometry Questions

As per the SSC CGL Previous Year Question Paper trends, the commission generally asks around 5 to 6 questions from this topic. Hence, candidates must not skip this topic and should familiarise themselves with the SSC CGL Geometry questions. Below, we have explained the concepts of geometry questions which will help you clear your doubts and improve your performance. 

SSC CGL Geometry Concepts: Important Theorems and Properties of Circle

Let the radius of circle is ‘R’, diameter = D

(1)      Maximum area of triangle = R²

If the given triangle is right angle triangle and its base is the diameter of the circle. Then, the maximum possible area of the triangle will be R².

(2) If AB and CD are two chords intersecting each other at P.

(3) If PA and PB are two tangents and LM is the tangent intersecting both tangents PA and PB as given in the figure.

Then, PA = PB = 1/2 Perimeter of triangle PLM.

(4) If BA is a chord intersecting tangent T at P as in the below figure.

Then, PA x PB = PT²

5. This property is an important property of circle. It will help you to save your time in the exam. But don’t confuse between figure 5(a) and 5(b).

If angle AOC is x⁰ and BOD is y⁰, then angle APC and BPD will be equal to half of the sum of both angle x⁰ and y⁰.

                            Fig. 5 (a)

If angle AOC is x⁰ and BOD is y⁰, then angle APC and BPD will be equal to half of difference of the angle x⁰ and y⁰.

            Fig. 5 (b)

(6) Quadrilateral formed by angle bisectors of a cyclic quadrilateral will be always a cyclic quadrilateral.

If given quadrilateral ABCD is a cyclic quadrilateral and RA, RB, PC and PD are angle bisectors of angle A, B, C and D. Then, quadrilateral PQRS formed by these angle bisectors will also be a cyclic quadrilateral.

Properties of Cyclic Quadrilateral:

1. The sum of diagonally opposite angle will always be equal to 180⁰.

PQRS is also a cyclic quadrilateral.

(7) In the given diagram, ABCD is a cyclic quadrilateral.

If AB is parallel to CD, and AD is parallel to BC then, diagonals AC and BD will also equal.

(8) Tangents

(a) If there are two circles which are at a distance apart from each other, then the maximum possible number of common tangents are 4.

 Fig. 8 (a)

(b) If there are two circles touching each other externally, then the maximum number of possible common tangents are 3.

Fig. 8 (b)

(c) If there are two circles intersecting each other, then the maximum number of possible common tangents are 2.

Fig. 8 (c)

(d) If there are two circles touching each other internally, then the maximum number of possible common tangents are 1.

Fig. 8 (d) No. of common tangents = 1

(e) If there are two circles, either concentric or one is inside the another and do not touch.
then, the number of common tangents will be zero.

Fig. 8 (e) No. of common tangents = 0

(9)

(a) If PA and PB are two tangents from P on the circle at A and B.
then, PA = PB

(b) The line joining the centre and the point on the circle at which tangent meets will always be perpendicular to the tangent.

(10) If the tangent PAQ and chord AB makes an angle θ, then the angle made by chord AB on the circle will always be θ.
Similarly, If the tangent PAQ and chord AC makes an angle ϕ, then the angle made by chord AC on the circle will always be ϕ.

(10) If AB, BC, CD and AD are tangents on circle as given in the diagram,

Then AB + CD = BC + AD

(11) The distance between centres of two circles.

(a)

distance b/w centres = r₁ - r₂

(b)

distance b/w centres = r₁ + r₂

(12) Length of Direct Common Tangent.

If ‘d’ is the distance between centres of two circles and r₁ and r₂ are the radius of given two circles.

or

Fig. 12 (a)                                                                          Fig. 12 (b)

Transversal common tangent

Note: Do not confuse between transversal and direct common tangent.

(12). This question is asked many times in SSC Exams. Remember the diagram as it is given, all three circles have a direct common tangent.

(13)

If PQ is diameter and PR||QS

Then, PQ = RS and

(14) If ABCD is a cyclic quadrilateral and angle APC is x⁰, then angle ADC = 90⁰ – x/2 and angle ABC = 90⁰+x/2.

(15) If AB, BC, CD and AD are tangents on the given circle, then x⁰+y⁰ = 180 degree.

(16) If you find such diagram and angle ADB = x⁰ then angle ACB = 180⁰ -x⁰.

(17) If AB is the diameter of the bigger circle and ‘r’ is the radius of the smaller circle.

 A                                                                 B

(18) If AB is the tangent of two circles at A and B, P is the point at which both circles meet. Then, angle APB will be 90⁰.

(19) If there are two circles of same radius ‘R’ and the distance between their centres is ‘R’.

Then, the length of common chord CD = √ 3 R.

20. How to find the distance between two chords?

(a) If chords are on the opposite sides of centre

Let ‘a’ and ‘b’ are the length of chords and ‘d’ is the distance between them.
If it follows this relation,

Then, we can say that

(b) If chords are on the same side of centre

Let ‘a’ and ‘b’ are the length of chords and ‘d’ is the distance between them.
If it follows this relation,

Then, we can say that

Solving SSC CGL Geometry questions on the basis of these concepts will help you score the maximum marks from the quantitative aptitude section. You can also