## 97th SSC CGL level Solved question set, 12th on Geometry

This is the 97th set of 10 solved questions for SSC CGL and 12th on Geometry. All problems are solved by basic and rich geometry concepts based on patterns and methods.

A number of these 10 problems may seem to be new and difficult.

The solved question set contains,

**Question set on Geometry for SSC CGL**to be answered in 15 minutes (10 chosen questions)**Answers to the questions,**and,**Detailed conceptual solutions**to the questions.

The **question set can very well be used as a unit of mock test **or mini mock test.

For maximum gains, the test should be taken first, and then the solutions are to be read.

The answers and detailed solutions follow the questions.

### 97th Question set - 10 problems for SSC CGL exam: 12th on topic Geometry - answering time 15 mins

**Q1.** If $O$ be the circumcentre of a $\triangle PQR$ and $\angle QOR=110^0$, $\angle OPR=25^0$, then the measure of $\angle PRQ$ is,

- $65^0$
- $55^0$
- $50^0$
- $60^0$

**Q2.** Two poles of height 7 metres and 12 metres stand on a plane ground. If the distance between their feet is 12 metres, distance between their top will be,

- $13$ metres
- $19$ metres
- $17$ metres
- $15$ metres

**Q3.** A square is inscribed in a quarter circle in such a way that two of its adjacent vertices on the radius are equidistant from the centre and the other two vertices lie on the circumference. If the side of the square is $\sqrt{\displaystyle\frac{5}{2}}$ cm then what is the radius (in cm) of the circle?

- $5$
- $10$
- $2.5$
- $2$

**Q4.** In $\triangle PQR$, $PQ=PR=18$ cm, and AB and AC are parallel to lines PR and PQ respectively with B and C lying on sides PQ and PR respectively. If A is the mid-point of side QR, then what is the perimeter (in cm) of the quadrilateral ABPC?

- $36$
- $32$
- $28$
- $18$

**Q5.** The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at point P. Then it always true that,

- $BP.AP=CD.CP$
- $AP.CD=AB.CP$
- $AP.CP=BP.DP$
- $AP.BP=CP.DP$

**Q6.** If opposite sides of a quadrilateral and also its diagonals are equal, then each of the angles of the quadrilateral is,

- $60^0$
- $120^0$
- $100^0$
- $90^0$

**Q7.** In the circumcircle to equilateral $\triangle PQR$, PS, the bisector of $\angle P$, meets the circumference at S. If PR and QS extended meet at T, find value of the ratio $RT:RQ$.

- $1:2$
- $2:1$
- $1:1$
- $1:3$

**Q8.** P, Q and R are three points so that, $PR=3$ cm, $QR=5$ cm and $PQ=8$ cm. The number of circles passing through the points P, Q and R is,

- $0$
- $1$
- $2$
- $3$

**Q9.** $\triangle ABC$ is inscribed in a circle so that BC is diameter. The tangent at point C intersects BA when produced at a point D. If $\angle ABC=36^0$, then the value of $\angle ADC$ is,

- $36^0$
- $54^0$
- $44^0$
- $48^0$

**Q10.** $\triangle LMN$ is right angled at M. $\angle N$ is $45^0$. What is the length of MN (in cm) if $NL=9\sqrt{2}$ cm?

- $9$
- $9\sqrt{2}$
- $\displaystyle\frac{9}{\sqrt{2}}$
- $18$

Answers to the questions

**Q1. Answer:** Option d: $60^0$.

**Q2. Answer:** Option a: $13$ metres.

**Q3. Answer:** Option c: $2.5$.

**Q4. Answer:** Option a: $36$.

**Q5. Answer:** Option c: $AP.CP=BP.DP$.

**Q6. Answer:** Option d: $90^0$.

**Q7. Answer:** Option c: $1:1$.

**Q8. Answer:** Option a: $0$.

**Q9. Answer:** Option b: $54^0$.

**Q10. Answer:** Option a: $9$.

### 97th solution set - 10 problems for SSC CGL exam: 12th on topic Geometry - answering time 15 mins

**Q1.** If $O$ be the circumcentre of a $\triangle PQR$ and $\angle QOR=110^0$, $\angle OPR=25^0$, then the measure of $\angle PRQ$ is,

- $65^0$
- $55^0$
- $50^0$
- $60^0$

#### Solution 1: Immediate solution by key pattern identification of isosceles triangles

The following figure describes the problem,

All three of $OP$, $OQ$ and $OR$ being the radii of the circumcircle, these are equal. This forms two isosceles triangles, $\triangle POR$ and $\triangle POQ$.

In $\triangle POR$, base angles are equal,

$\angle OPR=25^0=\angle PRO$.

In $\triangle OQR$, apex angle $\angle QOR=110^0$ and base angles are equal. So,

$2\angle ORQ=180^0-110^0=70^0$,

Or, $\angle ORQ=35^0$.

Sum up the two components of the desired angle,

$\angle PRQ=25^0+35^0=60^0$.

**Answer:** Option d: $60^0$.

**Key concepts used:** **Key pattern identification of two isosceles triangles -- Total of angles in a triangle -- Solving in mind.**

**Q2.** Two poles of height 7 metres and 12 metres stand on a plane ground. If the distance between their feet is 12 metres, distance between their top will be,

- $13$ metres
- $19$ metres
- $17$ metres
- $15$ metres

#### Solution 2: Quick solution by parallel projection of shorter pole length on the longer one and by using Pythagoras theorem

The following figure shows the problem situation.

AB and CD are the 12 metre and 7 metre poles standing 12 metre apart on a plane ground.

The poles by usual nature of pole erection, are erected vertically on the plane ground and so are parallel.

Project CD on AB to get difference of length $AE=5$ metres.

Also being parallel distance apart 12 metres CE is the separation of E and C.

In Pythagoras theorem then, hypotenuse,

$AC=\sqrt{AE^2+CE^2}=\sqrt{25+144}=\sqrt{169}=13$ metres.

**Answer:** Option a: $13$ metres.

**Key concepts used:** **Real world reasonable assumption -- Parallel projection of smaller pole on longer pole -- Pythagoras theorem -- Solving in mind.**

**Q3.** A square is inscribed in a quarter circle in such a way that two of its adjacent vertices on the radius are equidistant from the centre and the other two vertices lie on the circumference. If the side of the square is $\sqrt{\displaystyle\frac{5}{2}}$ cm then what is the radius (in cm) of the circle?

- $5$
- $10$
- $2.5$
- $2$

#### Solution 3: Quick solution by proper problem representation in figure, isosceles triangle properties, square diagonal properties and new geometric element introduction

The first difficulty in this problem is to sketch the figure specific to the problem.

As the two adjacent vertices of the square lie on the two radii, the square must be positioned as tilted by $45^0$ as shown in following graphic.

The following figure shows the problem graphic.

As vertices B and C are equidistant from centre O, with $\angle O$ of the quarter circle a $90^0$, in the isosceles $\triangle BOC$, the base angles are each $45^0$.

So, in right $\triangle BOC$ as side $BC=\sqrt{\displaystyle\frac{5}{2}}$ cm,

$BO=OC=\displaystyle\frac{BC}{\sqrt{2}}=\frac{\sqrt{5}}{2}$ cm.

A second help is now provided by the new element diagonal AC introduced.

By the property of a square, a diagonal of a square bisects each corner angle of $90^0$.

So, $\angle ACO=\angle BCO+\angle ACB=90^0$.

That is, $\angle AOC$ is a right triangle with desired radius as hypotenuse AC.

We need to find only the diagonal length of the square.

Again by square property, a diagonal is $\sqrt{2}$ times its side length.

So, $AC=\sqrt{5}$ cm.

Finally by Pythagoras theorem in $\triangle AOC$,

$AO^2=OC^2+AC^2=\displaystyle\frac{5}{4}+5=\displaystyle\frac{25}{4}$,

Or, $\text{Radius }AO=\displaystyle\frac{5}{2}=2.5$ cm.

**Answer:** Option c: $2.5$.

**Key concepts used:** **Geometric visualization -- Isosceles angle property -- New geometric element introduction -- Diagonal length of a square -- Square properties -- Pythagoras theorem -- Solving in mind.**

Calculations being simple, if you can follow the conceptual steps quickly, the problem can easily be solved in mind. But Geometric visualization is important to start solving the problem.

**Q4.** In $\triangle PQR$, $PQ=PR=18$ cm, and AB and AC are parallel to lines PR and PQ respectively with B and C lying on sides PQ and PR respectively. If A is the mid-point of side QR, then what is the perimeter (in cm) of the quadrilateral ABPC?

- $36$
- $32$
- $28$
- $18$

#### Solution 4: Quick solution by key pattern identification of a pair of new mid-points using similar triangles property and identifying a crucial pair of isosceles triangles

The following figure describes the problem and helps explaining the solution.

We will take the simplest path to the solution.

As $AB||RP$ in $\triangle PQR$, the two triangles $\triangle BQA$ and $\triangle PQR$ are similar.

Moreover, as A is the mid-point of side QR, B must also be the mid-point of side PQ from side ratio equality property of two similar triangles.

Similarly, C is also the mid-point of side PR.

As a result, $BQ=PB=9$ cm, and also,

$RC=PC=9$ cm.

In the final stage, we'll identify two crucial isosceles triangles.

With $PQ=QR$ in $\triangle PQR$,

$\angle Q=\angle R$.

As $AB||PR$ and QR intersects this pair of parallel lines,

$\angle BAQ=\angle R=\angle Q$.

$\triangle BQA$ is isosceles, and,

$AB=BQ=9$ cm.

Same way, $AC=RC=9$ cm.

Perimeter of quadrangle ABPC is,

$4\times{9}=36$ cm.

**Answer:** Option a: $36$.

**Key concepts used:** **Key pattern identification of similar triangles -- Side ratio equality property of similar triangles --**** Identification of two crucial isosceles triangles -- Angles formed by a line intersecting a pair of parallel lines -- ***Intersected parallel lines concepts -- ***Solving in mind.**

**Q5.** The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at point P. Then it is always true that,

- $BP.AP=CD.CP$
- $AP.CD=AB.CP$
- $AP.CP=BP.DP$
- $AP.BP=CP.DP$

#### Solution 5: Immediate solution key pattern identification of similar triangles by arc angle subtending property and side ratio equality of a pair of similar triangles

The following figure will be used for explaining the solution,

As each of the choices is an equality of product of two pairs of sides, it is clear that our target is to find a pair of similar triangles and spot the suitable two pairs of sides for ratio equality.

With this clear objective, the key pattern of two pairs of equal angles could be identified by using **arc angle subtending property,**

$\angle BAC=\angle BAP=\angle BDC=\angle PDC$ on same arc BC (blue colored arc and angles), and,

$\angle ABD=\angle ABP=\angle ACD=\angle PCD$, on same arc AD (red colored arc and angles).

In triangles $\triangle ABP$ and $\triangle CPD$ two pairs of angles being equal, the third pair of angles also are equal, and the triangles are similar by A-A-A criterion of similarity.

By side ratio equality property in these two similar triangles then,

$\displaystyle\frac{AP}{DP}=\frac{BP}{CP}$, first ratio are for sides opposite red colored angles and the second opposite blue colored angles,

Or, $AP.CP=BP.DP$.

**Answer:** Option c: $AP.CP=BP.DP$.

**Key concepts used:** **Key pattern identification -- ***Cyclic quadrilateral -- ***A-A-A similar triangle criterion -- Arc angle subtending concept -- Side ratio equality of a pair of similar triangles -- Solving in mind.**

**Q6.** If opposite sides of a quadrilateral and also its diagonals are equal, then each of the angles of the quadrilateral is,

- $60^0$
- $120^0$
- $100^0$
- $90^0$

#### Solution 6: Instant solution using choice values and total angles of a quadrilateral with subsequent conceptual explanation

**Instant solution:**

Total of four internal angles of a quadrilateral is $360^0$. So each of the four angles to be equal, the equal value must be one-fourth of $360^0$, that is, value must be of Option d: $90^0$.

First we will show why with two pairs of intersecting parallel lines, a quadrangle must be a parallelogram.

#### With two pairs intersecting parallel lines, the quadrilateral must be a parallelogram, Proof

The following figure is used for proof,

In the figure ABCD is a quadrilateral with pairs of opposite sides equal.

$AD=BC$, and $AB=CD$.

In two triangles $\triangle ABC$ and $\triangle ACD$, with side AC common, all three corresponding pairs of sides are equal in length.

So by S-S-S criterion, the two triangles are congruent and opposite internal angles made by the diagonal are equal,

$\angle CAD =\angle ACB$,

$\angle BAC = \angle ACD$.

So $AD||BC$, and $AB||CD$ and ABCD is a parallelogram.

#### With two diagonals of equal length, a parallelogram must be a rectangle, Proof

The following figure will be used for explanation,

BCD is a parallelogram with diagonals, $AC=BD$.

In a parallelogram, the diagonals bisect each other and as the diagonals are equal to each other, all four sections of the bisected diagonals are equal to each other.

So, $AX=DX=CX=BX$.

As a result, two pairs of isosceles triangles are formed.

And we get two sets of four equal angles,

$\angle CAD=\angle ACB=\angle DBC=\angle BDA$, and similarly,

$\angle CAB=\angle ACD=\angle BDC=\angle DBA$.

Sum up adjacent pairs,

$\angle CAD+\angle CAB=\angle A=\angle DBC+\angle DBA=\angle B=\angle C$, as opposite angles of a parallelogram are equal.

Continuing summing up,

$\angle ACB+\angle ACD=\angle C=\angle BDA+\angle BDC=\angle D$.

So, $\angle A=\angle B=\angle C=\angle D=90^0$, as total of four angles in a quadrilateral is $360^0$.

**Brief reasoning is:**

With two pairs of parallel equal sides, a quadrilateral must be a parallelogram by the property of a parallelogram, and with the pair of diagonals equal, a parallelogram must at least be a rectangle by the property of a rectangle.

In any case, you don't have to prove any concept. All explanations here are for concept clearance only.

**Answer:** Option d: $90^0$.

**Key concepts used:** **Property of a parallelogram -- Property of a rectangle -- Solving in mind.**

**Q7.** In the circumcircle to equilateral $\triangle PQR$, PS, the bisector of $\angle P$, meets the circumference at S. If PR and QS extended meet at T, find value of the ratio $RT:RQ$.

- $1:2$
- $2:1$
- $1:1$
- $1:3$

#### Solution 7: Quick solution by target driven analysis, arc angle subtending property, equilateral triangle angle properties, external angle of a triangle property

The following figure will be used for explaining the solution,

It is apparent that the target ratio can be evaluated either by proving similarity and congruence of two triangles $\triangle QTR$ and $\triangle PST$ or by proving the $\triangle QTR$ isosceles.

In both cases, all data points to the ratio $1:1$.

Let us use the easiest second path.

PS being the bisector of $\angle P$, $\angle RPS=\angle QPS=30^0$, as in an equilateral triangle all three angles are $60^0$.

Next by arc angle subtending property, arc RS subtends two equal angles at points P and Q on alternate segment of circumference,

$\angle RPS=30^0=\angle RQS=\angle RQT$.

Finally $\angle PRQ$, the external angle to $\triangle QTR$ which is $60^0$ is equal to the sum of two opposite internal angles of the triangle,

$\angle PRQ=60^0=\angle RQT+\angle QTS$,

Or, $\angle QTS=60^0-\angle RQT=60^0-30^0=30^0$.

So $\triangle QTR$ is isosceles, and $RT=RQ$.

Desired ratio is, $1:1$.

**Answer:** Option c: $1:1$.

**Key concepts used:** **Mathematical reasoning -- Circumcircle -- **** Equilateral triangle angle properties -- Arc angle subtending property -- External angle of a triangle property -- Solving in mind.**

**Q8.** P, Q and R are three points so that, $PR=3$ cm, $QR=5$ cm and $PQ=8$ cm. The number of circles passing through the points P, Q and R is,

- $0$
- $1$
- $2$
- $3$

#### Solution 8: Instant solution by very basic constraint of a triangle formation

Though it is a matter of circles, the crucial key pattern stares you in the eye,

$PQ=PR+QR$.

All three points must lie on the same straight line.

No triangle can be formed with these three line segments as sides and no circle can pass through the three points P, Q and R as well.

Most basic condition for triangle formation by three line segments, AB, BC and CA is,

Sum of lengths of any two sides must exceed the length of the third side.

**Answer:** Option a: $0$.

**Key concepts used:** *Key pattern identification -- Most basic condition of triangle formation -- Solving in mind.*

**Q9.** $\triangle ABC$ is inscribed in a circle so that BC is diameter. The tangent at point C intersects BA when produced at a point D. If $\angle ABC=36^0$, then the value of $\angle ADC$ is,

- $36^0$
- $54^0$
- $44^0$
- $48^0$

#### Solution 9: Immediate solution by the concept that tangent is perpendicular to the radius at the tangent point

The following figure describes the problem and helps explaining the solution.

The tangent DC is perpendicular to diameter BC at tangent point C. So,

$\angle BCD=90^0$.

This makes third angle in $\triangle BCD$ as,

$\angle BDC=90^0-36^0=54^0$.

**Answer:** Option b: $54^0$.

**Key concepts used:** **Tangent perpendicular to diameter at tangent point -- Solving in mind.**

**Note:** You should always try to chose the shortest path to the solution.

**Q10.** $\triangle LMN$ is right angled at M. $\angle N$ is $45^0$. What is the length of MN (in cm) if $NL=9\sqrt{2}$ cm?

- $9$
- $9\sqrt{2}$
- $\displaystyle\frac{9}{\sqrt{2}}$
- $18$

#### Solution 10: Immediate solution by key pattern identification of $\triangle LMN$ as isosceles right-angled and well-known side to hypotenuse relation of such a triangle

The following figure describes the problem,

With $\angle N=45^0$, the third angle in the right-angled triangle $\angle L$ is also $45^0$.

By Pythagoras theorem, if $a$ is side length and $h$ is hypotenuse length in an isosceles right-angled triangle,

$h^2=2a^2$,

Or, $h=\sqrt{2}a$.

As hypotenuse in this case is $9\sqrt{2}$ cm, the side length $MN=LM=9$ cm.

**Answer:** Option a: $9$.

**Key concepts used:** *Key pattern identification -- Hypotenuse to side length relation of a right-angled isosceles triangle -- Pythagoras theorem -- Solving in mind.*

### End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of steps using basic and rich geometry concept based key patterns and methods in each case.

This is the hallmark of quick problem solving:

**Concept based pattern and method formation**, and,**Identification of the key pattern and use of the method**associated with it. Every special pattern has its own method, and not many such patterns are there.

Important is the *concept based pattern identification and use of quick problem solving method.*

### Guided help on Suresolv Geometry

All of Suresolv Geometry articles are listed with links at the end, but it is an *unguided* list and **may not be up-to-date.**

To use this *extensive range of articles on geometry* problem solving **with best results**, *follow the guide,*

**5 step Suresolv Geometry Reading and Practice Guide for SSC CGL Tier II and Other competitive exams.**

*Basically, it is how to read and practice Suresolv Geometry guide.*

**It contains high school math articles on Geometry and even list of puzzles on Geometry.**

**The guide list of articles will be always UPTODATE.**

Wish you all the sure success.

**Related resources that should be useful for you**

**You may refer to:**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

**Concept tutorials for SSC CGL and other competitive exams on Geometry**

**Basic and Rich Geometry concepts part 9, Segment Relation for a Secant of a Circle**

**Basic and rich Geometry concepts part 7, Laws of sines and cosines**

**Basic and rich Geometry concepts part 6, proof of triangle area from medians**

**Basic and rich Geometry concepts part 5, proof of median relations**

**Basic and rich Geometry concepts part 4, proof of arc angle subtending concept**

**Geometry, basic and rich concepts part 3, Circles**

**Geometry, basic concepts part 2, Quadrilaterals polygons and squares**

**Geometry, basic concepts part 1, points lines and triangles**

**How to solve difficult Geometry problems quickly in a few steps**

**How to solve intriguing SSC CGL level Geometry problem in a few steps 4**

**How to solve difficult SSC CGL Geometry problems in a few steps 3**

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

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

**SSC CGL Tier II level question sets and solution sets on Geometry**

**SSC CGL Tier II level Solution Set 16, Geometry 5**

**SSC CGL Tier II level Question Set 16, Geometry 5**

**SSC CGL Tier II level Solution Set 15, Geometry 4**

**SSC CGL Tier II level Question Set 15, Geometry 4 **

**SSC CGL Tier II level Solution Set 6, Geometry 3**

**SSC CGL Tier II level Question Set 6, Geometry 3**

**SSC CGL Tier II level Solution Set 5, Geometry 2**

**SSC CGL Tier II level Question Set 5, Geometry 2**

**SSC CGL Tier II level Solution Set 4, Geometry 1**

**SSC CGL Tier II level Question Set 4, Geometry 1**

#### SSC CGL level question sets and solution sets on Geometry

**SSC CGL lervel solved question set 97 Geometry 12**

**SSC CGL level Solved Question set 96 Geometry 11**

**SSC CGL level Solved Question set 95 Geometry 10**

**SSC CGL level Solved Question set 94 Geometry 9**

**SSC CGL level Solution Set 80, Geometry 8**

**SSC CGL level Question Set 80, Geometry 8**

**SSC CGL level Solution Set 39, Geometry 7**

**SSC CGL level Question Set 39, Geometry 7**

**SSC CGL level Solution Set 38, Geometry 6**

**SSC CGL level Question Set 38, Geometry 6**

**SSC CGL level Solution Set 37, Geometry 5**

**SSC CGL level Question Set 37, Geometry 5**

**SSC CGL level Solution Set 36, Geometry 4**

**SSC CGL level Question Set 36, Geometry 4**

**SSC CGL level Solution Set 21, Geometry 3**

**SSC CGL Level Question Set 21, Geometry 3**

**SSC CGL level Solution Set 20, Geometry 2**

**SSC CGL level Question Set 20, Geometry 2**

**SSC CGL level Solution Set 18, Geometry 1**

**SSC CGL Level Question Set 18, Geometry 1**

If you like,you mayto get latest content on competitive exams.subscribe