## Visualization, shape analysis together with basic and rich geometric concepts delivered quick solution

Basic Geometry concepts with general problem solving strategies and techniques produce elegant solutions in difficult Geometry problems just as in Algebra or Profit and loss. In this third session of Geometry problem solving in a few steps we will analyze solution process of three 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 three problems yourself first.

### Problem 1.

$P$ and $Q$ are centres of two circles of radii 9cm and 2cm respectively. $PQ$ is 17cm and $R$ is the centre of a third circle that touches the other two circles externally. If $\angle PRQ = 90^0$, then the radius of the third circle is,

- 7cm
- 4cm
- 8cm
- 6cm

**Problem analysis and execution**

As the third circle touches the other two circles, the lengths of two lines $PR$ and $QR$ are each composed of the sum of a pair of radii of the circles touching, that is,

$PR = (9 + x)$ cm, and

$QR = (2 + x)$ cm

where we assume $x$ cm as the unknown length of the third radius.

This is true because at each touching point the two radii are perpendicular to the common tangent passing through the single point and **thus form two parts of a single centre to centre straight line.**

Now we will apply the Pythagoras theorem on $\triangle PRQ$ with $\angle PRQ = 90^0$,

$PQ^2 = PR^2 + QR^2$,

Or, $17^2 = (x + 9)^2 + (x + 2)^2$

Or, $289 = 2x^2 + 22x + 85$,

Or, $x^2 + 11x - 102 = 0$.

If we split the mid-term coefficient we get,

$17x - 6x = 11x$, and $17\times{6}=102$.

So the expression is factored into,

$(x - 6)(x + 17) = 0$,

As $x$ must be positive, finally the radius of the third circle is, $x=6$ cm.

**Answer:** Option d: 6 cm.

**Key concepts used: ***Visualization* --* Shape analysis* -- *Circles touching concept* -- *Pythagoras theorem* -- *solution of quadratic equation* -- *basic geometric concepts*-- basic algebraic concepts.

### Problem 2.

In the $\triangle ABC$ two medians $BE$ and $CF$ intersect at $G$. If $EF$ cuts $AG$ at $O$ and length of $OG=2$ cm, the length of $AO$ is

- 4cm
- 2cm
- 6cm
- 8cm

**Problem analysis and solving**

$CF$ and $BE$ being two medians, $E$ and $F$ are the mid-points of the two sides $AC$ and $AB$. Thus when we join these mid-points by the line $FE$ it becomes parallel to the base $BC$. This is a standard * rich geometric concept* that you may refer to in tutorial

**Basic geometry concepts - points, lines and triangles 1.**Under these conditions, the line $FE$ divides any line from the vertex to the base in two equal parts. This follows first from parallelism of $EF$ and $BC$ and secondly from the similarity of pairs of triangles such as $\triangle AFO$ and $\triangle ABD$, where $AF=FB$.

So all this geometric reasoning results in,

$AO = OD = GD + 2$,

Or, $AO + OG = GD + 4$, as $OG=2$,

Or, $AG = GD + 4$.

By the property of median and centroid (the centroid divides a median in a ratio of 2 : 1 in favor of the section originating from the vertex), $AG = 2GD$ and so,

$2GD = GD + 4$,

Or, $GD = 4$ cm and so,

$AG = 2GD = 8$cm,

Or, $AO = 8-2=6$ cm.

**Ans.** Option c: 6cm.

**Key concepts used: **Visualization -- Shape analysis -- rich concept of formation of similar triangles by a line joining mid-points of two sides and parallel to the third side -- deductive reasoning leads to bisection of the median $AG$ at $O$ -- concept of median section ratio of 2 : 1 at centroid of a triangle.

### Problem 3.

In a circle with centre at $O$ and radius 5 cm, the length of a chord $AB$ is 8 cm. If two tangents at $A$ and $B$ meet at $C$ length of tangent section $AC$ is,

- $\displaystyle\frac{20}{3}$ cm.
- $\displaystyle\frac{15}{4}$ cm.
- $\displaystyle\frac{10}{3}$ cm.
- $\displaystyle\frac{21}{4}$ cm.

**Problem analysis and partial solution**

The very first action that we can take is to find the length of section $OD$ drawn perpendicular from centre to the chord $AB$ thus bisecting it (* basic chord concepts*). We apply Pythagoras theorem in $\triangle OAD$,

$OA^2 = OD^2 + AD^2$,

Or, $5^2 = OD^2 + 4^2$, as $OD$ bisects $AB=8$ cm,

Or, $OD = 3$ cm.

The values given were so inviting we couldn't resist finding this value. But it is not the ease only, while looking at the shapes we expected to get a subsequent useful result out of this simple action.

#### Elegant path to the solution

Now we put our analytical mind to work seriously and notice that the target section of unknown length $AC$ is a part of the $\triangle OAC$ whereas the known section $OD$ is a part of a second $\triangle OAD$ where the side $OA$ is common between the triangles. When we examine this prospective pair of triangles more closely we find, each having a right angle and a common angle in $\angle AOD = \angle AOC$. This is a very encouraging situation of the * two triangles becoming similar with equal ratios of corresponding sides* (opposite to equal angles).

By this similarity we have the desired equal ratio,

$\displaystyle\frac{OD}{5} = \frac{4}{AC}$,

Or, $AC = \displaystyle\frac{20}{3}$ cm.

**Ans.** Option a: $\displaystyle\frac{20}{3}$ cm.

**Key concepts used:** Visualization -- shape analysis -- Use of Pythagoras theorem to find the length of the perpendicular to the chord -- while doing this it was not absolutely clear how it will be used. But this step was taken anyway as the problem practically demanded this step to be taken and this has been the most feasible step that could be taken with absolute ease.

Now, on serious analysis, the Key information discovered was the fact that the two similar triangles will deliver the final result in a single step.

*We were expecting this because of,*

the presence of two perpendiculars (or right angles) in two attached triangles (because of which one angle became common to the two triangles, this is nothing but useful pattern recognition in shapes) and the knowledge thatequal ratio of sides in two similar triangles produces the desired solution in a single stepvery often.

**Properties of similar triangles are a rich resource for solving problems involving triangles.**

This solution though very short, is not the usual. We will show you a longer solution applying **Many ways technique.**

#### Longer solution

*In this case instead of using the similarity of two suitable triangles we will use Pythagoras theorem in two right triangles suitably.*

Let $CD = y$ so that in right $\triangle OAC$,

$(y + 3)^2 = 5^2 + AC^2$,

Or, $AC^2 = (y + 3)^2 - 5^2$

Similarly in right $\triangle ADC$,

$y^2 + 4^2 = AC^2$.

Linking the two,

$AC^2 = (y + 3)^2 - 5^2 = y^2 + 4^2$,

Or, $6y = 25 + 16 - 9 = 32$,

Or, $y=\displaystyle\frac{16}{3}$, and

$AC^2 = \displaystyle\frac{256}{9} + 16 = \displaystyle\frac{400}{9}$,

Or, $AC = \displaystyle\frac{20}{3}$.

A truth,

Unless you know more than one path to the goal you won't be able to appreciate the value of the path you have followed to the goal.

**Summarization**

For all the three problems, hopefully we have been able to present to you with an elegant, and conceptually justified solution. It is not enough to present a quick solution without fully justifying it.

The primary objective here is not to know the elegant solution * but to understand the use of the reasoning and concepts behind each elegant solution*. Later you should be able to own up the concepts and methods to solve new problems elegantly yourself.

All the three problems have been largely geometric and in each visualization and shape analysis were needed to find the shortest route to the solution.

*The third problem saw clear possibility of a longer solution though you should evaluate more than one path to the solution for every problem we present. It is exercising Many ways technique that improves your problem solving skill.*

Regarding use of geometric concepts, we have used both basic and rich geometric concepts. Additionally we have used basic algebraic concepts.

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