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

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,

- $120^0$
- $90^0$
- $110^0$
- $100^0$

**Problem analysis and execution**

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 --

*and*

**target modification***-- use of*

**working backwards***--*

**basic geometric property of same arc subtends same angle***-- 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*

**Deductive reasoning***-- 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.*

**external angle equality property of a triangle**### 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,

- 18cm
- 30cm
- 12cm
- 24cm

**Problem analysis and solving**

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,

- $\displaystyle\frac{7}{2}$ cm.
- $\displaystyle\frac{5}{2}$ cm.
- $\displaystyle\frac{9}{2}$ cm.
- $\displaystyle\frac{11}{2}$ 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,

- $80^0$, $60^0$, $40^0$
- $80^0$, $55^0$, $45^0$
- $70^0$, $50^0$, $60^0$
- $80^0$, $65^0$, $35^0$

**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***and other staple*

**deductive reasoning***produced desired result. Being purely geometric, initial*

**problem solving strategies***played an important part for successful further analysis.*

**visualization of the geometric figure**and**shape analysis**In contrast, the last two problems were * primarily of algebraic nature* and could be elegantly solved using

*. Visualization played insignificant part in problem solving in these cases.*

**basic and rich algebraic concepts****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.**