How to solve difficult SSC CGL Geometry problems in a few steps 2

Basic geometric concepts together with End state analysis and deductive reasoning for quick solution

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

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 second episode of Geometry problem solving in a few steps we will analyze solution process of four chosen geometry 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.

Recommendation:

Before referring to the solutions, try to solve each of the four problems yourself first.


Chosen Problem 1.

In a circle with centre at $O$ and four points $A$, $B$, $C$, $D$ on its periphery, the two lines $AC$ and $BD$ intersect at $E$ so that $\angle BEC = 130^0$ and $\angle ECD=20^0$. The $\angle BAC$ is then,

  1. $120^0$
  2. $90^0$
  3. $110^0$
  4. $100^0$

Problem analysis and execution

How to solve difficult ssc cgl geometry problems in a few steps 2-1

On analysis of the target angle, it follows immediately that, being subtended by the same arc $BC$,

$\angle BAC = \angle BDC$.

We are now working backwards.

As we modify the target to $\angle BDC$, through external angle equality to sum of two opposite internal angles property of a triangle, the relationship between the modified target angle and the two given angles becomes apparent. Let us see how.

The $\angle BEC$ being the external angle in the $\triangle EDC$,

$\angle BEC = \angle EDC + \angle ECD$,

Or, $130^0 = \angle BDC + 20^0$,

Or, $\angle BDC = \angle BAC = 130^0 - 20^0 = 110^0$.

We have used only two basic geometric concepts along with problem solving techniques and strategies for the elegant solution.

Answer: Option c: $110^0$.

Key concepts used: Visualization -- Shape analysis -- End state analysis revealed equality of the target angle with the angle subtended by the same arc -- target modification and working backwards -- use of basic geometric property of same arc subtends same angle -- Deductive reasoning -- with the new target we look into the two given angles and find that the new target angle and the two given angles are related through external angle equality property of a triangle -- two basic geometric concepts along with end state analysis, working backwards and deductive reasoning make it possible to reach the solution in just a few steps.

Chosen Problem 2.

If the length of radii of two concentric circles are 9cm and 15cm, length of a chord of the larger circle tangent to the smaller circle is,

  1. 18cm
  2. 30cm
  3. 12cm
  4. 24cm

Problem analysis and solving

How to solve difficult ssc cgl geometry problems in a few steps 2-2

This is a two concentric circle problem where the length of the tangent chord $AB$ of the larger circle is to be found out.

Being a tangent to the smaller circle at D, the radius of the smaller circle $OD$ is perpendicular to the chord at $D$. Again, by the property of a chord, a perpendicular $OD$ from co-located centre $O$ to the chord $AB$ of the larger circle bisects the chord.

Thus in the $\triangle AOD$, $OD$ is radius of the smaller circle 9cm long whereas $AO$ is the radius of the larger circle 15cm long.

By applying Pythagoras theorem in $\triangle AOD$ then,

$AD^2 = 15^2 - 9^2 = 144$,

Or, $AD = 12$,

Or, $AB = 24$cm.

Ans. Option d: 24cm.

Key concepts used: Visualization -- Shape analysis -- tangent perpendicular property -- chord bisection property -- Pythagoras theorem -- basic geometric concepts.

Chosen Problem 3.

In a $\triangle ABC$, $AB + BC = 12 $cm,  $BC + CA = 14 $cm and $CA + AB = 18 $cm. The radius of the circle with same perimeter as the triangle is,

  1. $\displaystyle\frac{7}{2}$ cm.
  2. $\displaystyle\frac{5}{2}$ cm.
  3. $\displaystyle\frac{9}{2}$ cm.
  4. $\displaystyle\frac{11}{2}$ cm.

Problem analysis and solving

If we add together the three given equations, we get twice the perimeter of the triangle. This is the key information discovery and it follows rich algebraic concept of principle of collection of friendly terms.

Adding togther the three given equations,

$2(AB + BC + CA) = 44$cm

So perimeter of the triangle is,

$(AB + BC + CA) = 22$cm.

Perimeter of the circle with radius $R$ is,

$2\pi{R} = 22$,

Or, $2\times{\frac{22}{7}}R = 22$,

Or, $R = \displaystyle\frac{7}{2}$ cm.

Ans. Option a: $\displaystyle\frac{7}{2}$ cm.

Key concepts used: Key information discovery that adding the three equations together will result in twice the perimeter of the triangle -- application of rich algebraic concept of principle of collection of friendly terms -- circle perimeter concept.

Chosen Problem 4.

$\angle A$, $\angle B$,  and $\angle C$ are three angles of triangle. If $\angle A - \angle B = 15^0$ and $\angle B - \angle C = 30^0$, then $\angle A$, $\angle B$, and $\angle C$ are,

  1. $80^0$, $60^0$, $40^0$
  2. $80^0$, $55^0$, $45^0$
  3. $70^0$, $50^0$, $60^0$
  4. $80^0$, $65^0$, $35^0$

Problem analysis

With the two given equations in three unknown angles of the triangle, when we consider in addition the equation arising out of the sum of three angles as $180^0$, we become sure of evaluating three unknown variables from three linear equations.  Basically the problem is transformed to a simple algebraic problem where we apply the linear equation solution procedure.

Problem solving execution

Subtracting first equation from second,

$ -\angle A - \angle C + 2\angle B = 15^0$

We chose this route for two reasons,

  • subtraction because, by only this route we will get a sum of two angles [$-(\angle A + \angle C)$] that can be substituted later by the third angle from sum of three angles = $180^0$ relation
  • specific subtraction so that we get a positive angle as a result. We could very well have subtracted second equation from the first. But in any case, addition won't have taken us nearer to the solution.

In a triangle we have the sum of three angles as $180^0$, that is,

$\angle A + \angle B + \angle C = 180^0$

Adding the two equations eliminates $\angle A $ and $\angle C$ leaving only $\angle B$,

$3\angle B = 195^0$,

Or, $\angle B = 65^0$.

From second given equation,

$\angle C = 65^0 - 30^0 = 35^0$, and

from first given equation,

$\angle A = 15^0 + 65^0 = 80^0$.

So the three angles are, $80^0$, $65^0$, $35^0$

Ans. Option d: $80^0$, $65^0$, $35^0$.

Key concepts used: Conversion of the geometric problem to an algebraic problem of solving for three unknown variables from three linear equations which we know from basic algebraic concepts, to be always solvable -- solving for three variables from three simple linear equation.

Summarization

We have shown elegant methods to reach solution in a few steps for all the four problems. The first two problems were purely geometric ones where basic geometric concepts along with end state analysis, deductive reasoning and other staple problem solving strategies produced desired result. Being purely geometric, initial visualization of the geometric figure and shape analysis played an important part for successful further analysis.

In contrast, the last two problems were primarily of algebraic nature and could be elegantly solved using basic and rich algebraic concepts. Visualization played insignificant part in problem solving in these cases.

It is good to be aware that Geometry and Algebra are strongly integrated.

Regarding use of geometric concepts, we have used only the basic concepts. Usually if we can use only basic concepts in solving a problem, we can achieve the most elegant solution.