## 4th SSC CGL Tier II level Solution Set, 1st on Geometry

Hope that you have already tried to solve the corresponding 10 SSC CGL Tier II level questions that had quite a variety and depth (triangles, circles, medians, parallelogram, interior bisectors and inscribed circle).

These detailed solutions of the questions should give you a clear idea on how to solve the questions easily and quickly.

If you have not yet taken the test you should take it by referring to the * SSC CGL Tier II level question set 4 on Geometry 1* before going through the solutions.

### 4th solution set - 10 problems for SSC CGL exam: 1st on topic Geometry - answering time 12 mins

**Problem 1. **

If P and Q are two points on sides AB and AD of a parallelogram ABCD respectively, and areas of $\triangle CPD=A_1$ and that of $\triangle BQC=A_2$, then,

- $2A_1=A_2$
- $A_1=A_2$
- $A_1=A_2$
- $2A_1=3A_2$

#### Solution 1 - Problem analysis and solving

The following figure describes the problem,

In finding the area of $\triangle BQC$ as $\frac{1}{2}BC\times{height}$, we note that when the vertex Q is moved along the line AD parallel to base BC, as BC||AD, the height of the triangle remains constant as is the base and consequently area. So for all positions of vertex Q on AD, the area of the $\triangle BQC$ remains unchanged. When Q reaches the corner point A, the side CQ becomes the diagonal CA that divides the whole parallelogram into two equal parts.

So area of $\triangle BQC =\displaystyle\frac{1}{2}\text{ of area of parallelogram } ABCD$.

Because of symmetry, the same applies for the area of $\triangle CPD$.

Thus,

$A_1=A_2$.

**Answer:** b: $A_1=A_2$.

**Key concepts used:** Area of a triangle -- parallelogram concepts -- area of a parallelogram -- key pattern identification of unchanging nature of the area of $\triangle BQC$ as Q is moved along the line AD parallel to base BC -- * geometric object movement technique*.

**Problem 2.**

If three non-colllinear points A, B and C lie on the periphery of a circle so that $AB=AC=BC=3$ cm, then the radius of the circle (in cm) is,

- $\sqrt{3}$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $\displaystyle\frac{\sqrt{3}}{2}$
- $\displaystyle\frac{2}{\sqrt{3}}$

#### Solution 2 - Problem analysis and solving

The following figure describes the problem.

By problem definition the $\triangle ABC$ effectively turns out to be an equilateral triangle of side length 3 cm inscribed within the circle. By the relationship between an equilateral triangle and its circumcircle, the centroid of the equilateral triangle coincides with the centre of the circle $O$.

The length of median of the equilateral triangle is,

$AE=\sqrt{(AB)^2 - (BE)^2}$

$\hspace{7mm}=\sqrt{3^2 - \left(\displaystyle\frac{3}{2}\right)^2}$, as the median AE bisects side BC of length 3 cm

$\hspace{7mm}=\sqrt{\displaystyle\frac{27}{4}}$

$\hspace{7mm}=\displaystyle\frac{3\sqrt{3}}{2}$.

By the * median section ratio at centroid* concept, vertex to centroid length AO being $\frac{2}{3}$rd of the median length,

$AO=\displaystyle\frac{2}{3}\times{\displaystyle\frac{3\sqrt{3}}{2}}=\sqrt{3}$.

As the centre of the circle and centroid are coincident, AO of length $\sqrt{3}$ cm is indeed the radius itself.

**Answer:** a: $\sqrt{3}$.

**Key concepts used:** Visualization -- concept of relationship between equilateral triangle and its circumcircle -- circumcircle concepts -- the centroid and the cente coincides -- concepts of median and centroid of an equilateral triangle -- concept of **median section ratio at centroid.**

**Problem 3.**

Two similar triangles have areas 96 cm$^2$ and 150 cm$^2$. If the largest side of the larger triangle is 20 cm, the largest side of the smaller triangle (in cm) is,

- 20
- 15
- 16
- 18

#### Solution 3 - Problem analysis

The problem is depicted in the figure below.

Two similar triangles can always be superimposed on each other with a pair of similar vertices coinciding. In figure, the smaller triangle is $\triangle ADE$ of area 96 cm$^2$ and the larger triangle is $\triangle ABC$ of area 150 cm$^2$. This is **similar triangle superimposition technique.**

Additionally the heights of the triangles are AP and AQ and by * similar triangle rich concepts*, the two triangles $\triangle APD$ and $\triangle AQB$ are also simlar to each other.

#### Solution 3 - Problem solving execution

The two triangles $\triangle ADE$ and $\triangle ABC$ being similar, the ratios of corresponding sides are equal so that,

$\displaystyle\frac{DE}{BC}=\frac{AD}{AB}$.

Again the two triangles $\triangle APD$ and $\triangle AQB$ being similar we have by the same principle,

$\displaystyle\frac{AP}{AQ}=\frac{AD}{AB}=\frac{DE}{BC}$.

The ratio of areas of the two triangles $\triangle ADE$ and $\triangle ABC$ is,

$\displaystyle\frac{A_1}{A_2}=\frac{DE\times{AP}}{BC\times{AQ}}$

$=\left(\displaystyle\frac{DE}{BC}\right)^2$

$=\displaystyle\frac{96}{150}$

$=\displaystyle\frac{16}{25}$,

Or, $\displaystyle\frac{DE}{BC}=\frac{4}{5}$.

This is the ratio of corresponding sides of the smaller and larger triangles.

So if the largest side of the larger triangle is 20 cm long the length of the corresponding largest side of the smaller triangle will be,

$L=\displaystyle\frac{4}{5}\times{20}=16$ cm.

**Answer:** c: 16.

**Key concepts used:** * Similar tringle superimposition technique* --

*-- equality of ratio of corresponding sides of two similar triangles including their heights -- simplification -- unitary method.*

**similar triangle rich concepts****Problem 4. **

The interior bisectors of $\angle B$ and $\angle C$ of $\triangle ABC$ meet at point P. If $\angle A=70^0$, then the value of $\angle BPC$ is,

- $55^0$
- $125^0$
- $150^0$
- $135^0$

#### Solution 4 - Problem analysis and solving

The following figure depicts the given problem.

Approaching the problem in the simplest way, we work on the relationship between the base angles and the given apex angle of $70^0$.

In $\triangle ABC$,

$\angle A = 180^0 - (\angle B + \angle C)$

$\hspace{8mm}=180^0 - 2(\angle PBC + \angle PCB)$

$\hspace{8mm}=70^0$

So,

$2(\angle PBC + \angle PCB)=180^0-70^0=110^0$,

Or, $(\angle PBC + \angle PCB)=55^0$

Again in $\triangle BPC$,

$\angle PBC + \angle PCB = 180^0 - \angle BPC$.

So,

$\angle BPC=180^0-55^0=125^0$

**Answer:** b: $125^0$.

**Key concepts used:** Basic angle in triangle concepts -- angle bisection concept.

**Problem 5. **

In $\triangle ABC$ the medians BE and CF intersect at point G. If GD=3 cm, then the length of AD is,

- 12 cm
- 4.5 cm
- 6 cm
- 9 cm

#### Solution 5 - Problem analysis and solving

The figure depicting the problem is shown below.

This is relatively a simple problem with only the concept of * median section ratio at centroid* involved. As we know, a median is divided at the centroid in two parts of ratio of length 2:1, the longer section being towards the vertex.

Thus in this case as $GD=3$ cm, $AG=2\times{GD}=6$cm and the length of the median, $AD=9$ cm.

**Answer:** d: 9 cm.

**Key concepts used:** Visualization -- **median section ratio at centroid.**

**Problem 6.**

D is the mid-point of side AB of a right angled $\triangle ABC$ with right angle at B. If AD subtends an angle $\alpha$ at C and BC is $n$ times of AB, then $\tan \alpha$ is,

- $\displaystyle\frac{n}{2n^2+1}$
- $\displaystyle\frac{n}{n^2+1}$
- $\displaystyle\frac{n}{n^2-1}$
- $\displaystyle\frac{n^2-1}{n^2+1}$

#### Solution 6 - Problem analysis

The following figure represents the problem.

The requirement that value of $\tan \alpha$ is needed as well as the notion that from the given information, value of $\tan \beta$ can easily be found out urges us to use the rich trigonometry concept of compound angle relationship for $\tan (\alpha +\beta)$.

#### Solution 6 - Problem solving execution

In $\triangle BDC$,

$\tan \beta=\displaystyle\frac{BD}{BC}$

$=\displaystyle\frac{AB}{2BC}$, as D is the mid-point of AB

$=\displaystyle\frac{1}{2n}$, as $BC=nAB$.

By the compound angle relationship for $\tan$ function,

$\tan (\alpha +\beta)=\displaystyle\frac{\tan \alpha + \tan \beta}{1-\tan \alpha.{\tan \beta}}$,

Or, $\displaystyle\frac{AB}{BC}=\frac{\tan \alpha + \displaystyle\frac{1}{2n}}{1-\displaystyle\frac{\tan \alpha}{2n}}$,

Or, $1-\displaystyle\frac{\tan \alpha}{2n}=n\times{\left(\tan \alpha + \displaystyle\frac{1}{2n}\right)}$, as $\displaystyle\frac{AB}{BC}=\frac{1}{n}$

Or, $\tan \alpha\left(n+\displaystyle\frac{1}{2n}\right)=1-\displaystyle\frac{1}{2}$,

Or, $\tan \alpha=\displaystyle\frac{n}{2n^2+1}$

**Answer:** a: $\displaystyle\frac{n}{2n^2+1}$.

**Key concepts used:** Basic trigonometry concepts -- rich trigonometry concepts -- algebraic simplification -- **compound angle rich concept.**

** Problem 7.**

If A, B and C are three points lying on the same plane and $AB=5$ cm and $BC=10$ cm, then the possible length of AC (in cm) is,

- 15
- 5
- 3
- 6

#### Solution 7 - Problem analysis and solution

The following figure represents the problem situation.

This problem calls for use of the * basic triangle formation condition.* By definition,

A triangle is a bounded region in a plane with three sides and three vertices so that length of no single side exceeds or equals the sum of length of the other two sides.

In the above figure formation of three different triangles have been shown with the length of two given sides. In each of these three cases the basic triangle formation condition is satisfied.

Testing this condition against the choice values we find only the length 6 cm of the third side satisfies the basic triangle formation condition.

**Answer:** d: 6.

**Key concepts used:** * Basic triangle formation condition* -- choice value testing against condition.

** Problem 8.**

In the following figure, a square ABCD is formed with its vertices as the mid-points of a larger square PQRS. A circle is inscribed in square ABCD and the $\triangle EFG$ is an equilateral triangle inscribed in the circle. If length of side of square PQRS is $a$, the area of the $\triangle EFG$ is,

- $\displaystyle\frac{\sqrt{3}a^2}{16}$
- $\displaystyle\frac{3\sqrt{3}a^2}{32}$
- $\displaystyle\frac{5\sqrt{3}a^2}{32}$
- $\displaystyle\frac{5\sqrt{3}a^2}{64}$

#### Solution 8 - Problem analysis and solution

The following is a depiction of the problem graphically.

As side length of the larger square is $a$, side length of the smaller square is,

$\sqrt{\left(\displaystyle\frac{a}{2}\right)^2 + \left(\displaystyle\frac{a}{2}\right)^2}=\displaystyle\frac{a}{\sqrt{2}}$.

This is also the diameter of the inscribed circle.

So the radius of the inscribed circles is,

$EO=\displaystyle\frac{a}{2\sqrt{2}}$.

By the * properties of an inscribed equilateral triangle*, centroid of the triangle and the centre of the circle are coincident at $O$ and EOH is the median.

As the * centroid divides the median by a ratio of 2:1* with larger section towards the vertex, the median length is,

$EH=\displaystyle\frac{3}{2}\times{EO}$

$=\displaystyle\frac{3a}{4\sqrt{2}}$.

We know, if $x$ is the side length of an equilateral triangle, its median length is,

$M=\sqrt{x^2-\displaystyle\frac{x^2}{4}}=\displaystyle\frac{\sqrt{3}x}{2}$,

So the side length of our equilateral triangle is,

$x=\displaystyle\frac{2}{\sqrt{3}}\times{\displaystyle\frac{3a}{4\sqrt{2}}}=\frac{\sqrt{3}a}{2\sqrt{2}}$.

Area of the equilateral triangle is then,

$A=\displaystyle\frac{\sqrt{3}}{4}\times{\left(\displaystyle\frac{\sqrt{3}a}{2\sqrt{2}}\right)^2}$

$=\displaystyle\frac{3\sqrt{3}a^2}{32}$.

**Answer:** b: $\displaystyle\frac{3\sqrt{3}a^2}{32}$.

**Key concepts used:** Visualization -- deductive reasoning -- * square in a square concept* --

*-- median to side of equilateral triangle -- area of equilateral triangle by side.*

**median section ratio at centroid****Note**: This problem is not an easy one as it requires good visualization skills and a host of rich concepts alongwith careful simplification skills.

**Problem 9.**

The radii of two non-intersecting circles are $r_1$ and $r_2$ with $r_1$ being the larger one and the smaller circle inscribed within the larger circle. If the least distance between their circumference be $S$, the distance between their centres is,

- $r_1-r_2+S$
- $r_1-r_2$
- $r_1+r_2-S$
- $r_1-r_2-S$

#### Solution 9 - Problem analysis and solution

The following figure represents the problem description.

The least distance between the circumferences of the larger and the smaller circle with centres at P and Q is the lesser of AC and BD which is BD. As per given information, $BD=S$.

So the distance between the centres of the two circles,

$PQ=PB-BD-QD=r_1-r_2-S$.

**Answer:** d: $r_1-r_2-S$.

**Key concepts used:** * Visualization *--

**.**

*Problem understanding***Note:** This is a simple problem if one can visualize and understand the problem from the problem description. Visualization and problem understanding are the key skills for solving this problem.

** Problem 10.**

If $\triangle ABC$, $AD$, $BE$ and $CF$ are the altitudes and $AD$ and $BE$ intersect at $G$ with $BE+EG=BG$, then,

- $FC+CG=FG$
- $CF+FG=CG$
- $CG+GF=CF$
- $CF-FG=CG$

#### Solution 10 - Problem analysis

The following is the first visualization relevant to the problem taking the orthocentre inside the triangle as is usual.

Given,

$BE+EG=BG$,

Or, $BG+2EG=BG$,

Or, $EG=0$.

It siginifies coincidence of G, E, F and A. In other words, BA is perpendicular to CA. As a result, $GF$ is also of zero length and the answer maps to multiple choices. The assumption of an acute angled triangle is then a wrong one.

**Orthocentre Concept:**

The intersection of altitudes of a triangle is at a single point called orthocentre which will lie inside the triangle for an acute angled triangle but will lie outside the triangle in case of an obtuse angled triangle. Our problem involves this second class of triangles.

Revised visualization is then as follows with obtuse angle at $\angle C$,

In this case, the altitudes AD and BE extended intersect at G outside the triangle and given information, $BE+EG=BG$ satisfies the visualization perfectly. As G is the orthocentre where all three altitudes meet,

$FC+CG=FG$.

**Answer:** a: $FC+CG=FG$.

**Key concepts used:** Visualization -- Deductive reasoning -- Orthocentre concept -- alternative testing.

**Note:** Visualization and clear orthocentre concept are the keys to the solution to this problem.

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