How to solve difficult SSC CGL Geometry problems in a few steps 1

Shape analysis, deductive reasoning and use of basic concepts result in quick solution

How to solve difficult SSC CGL geometry problems in a few steps 1

Geometry is no different from Algebra or Profit and loss with respect to solving a difficult SSC CGL problem in a few steps by applying Basic subject concepts together with Problem solving strategies and techniques. In this first episode of Geometry problem solving in a few steps, we will analyze solution process of two chosen problems.

For refreshing the basic geometry concepts you may refer to the concept tutorials, Basic concepts on Geometry 1 - points, lines and triangles, Basic concepts on Geometry 2 - quadrilaterals and polygons, and Basic concepts on Geometry 3 - Circles.

Chosen Problem 1.

In a circle with centre at $O$, the $\angle OAC = 15^0$ and $\angle OBC=50^0$ where $A$, $B$ and $C$ are the points on the circle periphery. The $\angle AOB$ is then,

How to solve difficult SSC CGL geometry problems in a few steps 1-1

  1. $30^0$
  2. $70^0$
  3. $20^0$
  4. $40^0$

Problem analysis

The two angles given belong to the two triangles $\triangle OAC$ and $\triangle OBC$. The crucial aspect of these two triangles are, in each, the pair of sides OA, OC and OB, OC are the radii of the circle and so are equal. As a result the two triangles of interest turn out to be isosceles triangles.

This is the key information discovery based on basic Geometry concepts.

Note: wherever you encounter triangles in circles look for sides as radii. Often this will be the key information contributing towards quick elegant solution. By using this property of circles (equal radii), often the triangles in circles can be identified as isosceles triangles and sometimes with a bit more favorable information, to even equilateral ones. We can term this as a rich geometric concept classified under Triangle Sides as equal radii concept.

End state analysis

On analyzing the end requirement, we find the required $\angle AOB = \angle AOC - \angle BOC$, both these angles being the vertex angle of the two triangles of interest.

Last reasoning

The two triangles are then isosceles with base angles equal in each. Also for each triangle value of one of the base angles is given. So for each of the two triangles, the vertex angle can be found out. Final step will then be just one step away.

Problem solving execution

In $\triangle AOC$, at the vertex,

$\angle AOC = 180^0 - 2\times{\angle OAC} $

$\hspace{15mm}= 180^0 - 2\times{15^0}$

$\hspace{15mm}=180^0 -30^0$


Similarly in $\triangle BOC$, at the vertex,

$\angle BOC = 180^0- 2\times{50^0} = 80^0$.


$\angle AOB = \angle AOC - \angle BOC $

$\hspace{15mm}= 150^0 - 80^0 $

$\hspace{15mm}= 70^0$

Answer: Option b: $70^0$.

Key concepts used: Identifying two target triangles from two given angles -- on end state analysis identifying the end requirement as the difference of two vertex angles of the two triangles of interest -- identifying the triangles as isosceles -- deducing the vertex angles of the two triangles from two equal base angles -- getting the target value of the angle as a difference of two vertex angles.

Ability to arrive at the target in stages using related geometric entities and concepts starting from the given values.

Chosen Problem 2.

Two chords $AB$ and $CD$ in a circle subtend angles $90^0$ and $60^0$ respectively at the centre $O$. If the length of the chord $AB = 4$cm then length of the chord $CD$ (in cm) is,

How to solve difficult SSC CGL geometry problems in a few steps 1-2

  1. $4\sqrt{2}$
  2. $2$
  3. $2\sqrt{2}$
  4. $\sqrt{2}$

Problem analysis

This is a two chord circle problem where two triangles with vertices at the centre of the circle are involved. The base of the two triangles are the chords.

We have to establish useful relationship between the two triangles so that the unknown chord length can be deduced from the known length of the chord of the first triangle with angle subtended as $90^0$ at the centre.

As we see it, the first triangle is a right triangle with two equal sides as radii and the given known value of hypotenuse which is the chord $AB$. We can then easily deduce the length of the radius by applying the most basic Pythagoras theorem.

We go over next to the second triangle with this known value of radius and then the final result is only a step away.

Problem solving execution

In right $\triangle AOB$,

$AO=OB=R$, say, and $AB=4$cm.

So by Pythagoras theorem,

$2R^2 = 4^2=16$,

Or, $R^2 = 8$,

Or, $R=2\sqrt{2}$cm.

In the second $\triangle COD$, $\angle COD=60^0$ and as $CO=OD=R$, the triangle is isosceles with equal base angles summing up to $180^0 - 60^0=120^0$. These two angles are also then $60^0$ and the $\triangle COD$ turns out to be an equilateral one. This is a rich geometric concept under the class equilateral triangle conditions.

Rich concept of Equilateral triangle conditions

The rich concept that we have used here says,

An isosceles triangle of equal base angles and vertex angle of $60^0$ is an equilateral triangle.

This concept can be confirmed easily,

$\text{Base angle } = \frac{1}{2}(180^0 - 60^0) = 60^0$.

Thus in the equilateral $\triangle COD$,

$CD = R = 2\sqrt{2}$cm.

Ans. Option c: $2\sqrt{2}$.

Key concepts used: Two equal sides of a right triangle immediately urges us to use Pythagoras theorem and determines the length of radius $R$ -- while on the second triangle, the vertex angle of $60^0$ in an isosceles triangle classifies the triangle as equilateral and hands us the solution without any more deduction.

A roundabout way to the solution

None can miss the first part of getting the value of radius $R=2\sqrt{2}$.

But while on the $\triangle COD$, with a little bit of imagination you can cook up the relation,

$CD = 2R\cos \text{(base angle)}$

$\hspace{8mm}=2R\cos 60^0 $

$\hspace{8mm}=R $

$\hspace{8mm}= 2\sqrt{2}$, as $\cos 60^0 = \frac{1}{2}$

This solution takes more steps as the rich concept of Equilateral triangle conditions is not used.

Nevertheless it shows different facets of the solution by the application of Many ways technique.


We have shown two elegant methods of solution each with its own spcialties; shape analysis and concept based deductive reasoning lying at the heart of the few step solutions.

In the first solution assessment of what is required with respect to what are given and use of the rich resource of equal radii as two sides of a triangle in a circle, quickly handed us the solution.

The second elegant solution saw the use of the most basic geometric concept of Pythagoras theorem to determine the first unknown of length of the radius and then by using the rich concept of Equilateral triangle conditions identification of the second triangle as an equilateral triangle immediately provided us the solution.

In the second problem by following Many ways technique we have shown a second method of solution that missed the rich geometric concept used in the elegant solution and consequently took more steps to the solution. Though this solution is inefficient, practice of Many ways technique enhances the power of your problem solving skill in general.