## 5th SSC CGL Tier II level Solution Set, 2nd 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 (circles, tangents, inscribed rectangle, common tangents, parallelogram, quarter circle, medians and more).

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 this test you should take it by referring to the * SSC CGL Tier II level question set 5 on Geometry 2* before going through the solutions.

### 5th solution set - 10 problems for SSC CGL exam: 2nd on topic Geometry - answering time 15 mins

**Problem 1. **

In figure below, ABCD is a rectangle inscribed inside the circle with centre at O so that ratio of area of the rectangle to the circle is $\sqrt{3} : \pi$.

Line segment CP intersects AB at P. If $\angle OCD=\angle BCP$, ratio of $BP:BC$ equals,

- $1:\sqrt{2}$
- $1:\sqrt{3}$
- $1:2$
- $1:2\sqrt{3}$

#### Solution 1 - Problem analysis and solving first stage

The following figure describes the problem solution,

From the given ratio we can find the relation between the radius, say $r$ and the side lengths, say $AD=BC=a$ and $CD=AB=b$. Without going deeper into the problem this is what we can do easily,

$\displaystyle\frac{\text{Area of rectangle}}{\text{Area of circle}}=\frac{ab}{\pi{r^2}}=\frac{\sqrt{3}}{\pi}$,

Or, $r^2=\displaystyle\frac{ab}{\sqrt{3}}$.

#### Solution 1 - Problem solving second stage

To use the second important information, $\angle OCD=\angle BCP$, we identify $\triangle PBC$ and $\triangle OQC$ as similar, as in these two right triangles, another pair of angles of value $\alpha$ are equal which will make third pair of angles also equal.

Ratio of corresponding sides are then equal and we get,

$\displaystyle\frac{BP}{OQ}=\frac{BC}{CQ}$,

Or the desired ratio,

$\displaystyle\frac{BP}{BC}=\frac{OQ}{CQ}$.

Again in two similar triangles, $\triangle ACD$ and $\triangle OCQ$,

$\displaystyle\frac{OQ}{CQ}=\frac{AD}{CD}=\frac{a}{b}$,

So the desired ratio is,

$\displaystyle\frac{BP}{BC}=\frac{a}{b}$.

Now we we have to eliminate radius $r$ from ratio of areas and ratio of sides.

#### Solution 1 - Problem solving third stage

First result,

$r^2=\displaystyle\frac{ab}{\sqrt{3}}$.

But hypotenuse, $AC^2=AD^2+CD^2$, because in any parallelogram inscribed in a circle its diagonals must pass through the centre and are in fact the diameters of the circle.

So, $AC^2=4r^2=a^2+b^2$.

Thus,

$4r^2=\displaystyle\frac{4ab}{\sqrt{3}}=a^2+b^2$.

Radius $r$ is eliminated and now we need to evaluate $\displaystyle\frac{a}{b}$.

Rearranging the above equation in $a$ and $b$,

$a^2-\displaystyle\frac{4}{\sqrt{3}}ab +b^2=0$.

Dividing the equation by $b^2$ we convert it to a quadratic equation in compound variable, $p=\displaystyle\frac{a}{b}$. This is use of abstraction and substitution technique,

$\left(\displaystyle\frac{a}{b}\right)^2 -\displaystyle\frac{4}{\sqrt{3}}\displaystyle\frac{a}{b} +1=0$,

Or, $p^2-\displaystyle\frac{4}{\sqrt{3}}p +1=0$, where $p=\displaystyle\frac{a}{b}$.

This is a quadratic equation in single variable $p$ and its roots are given by the Sreedhar Acharya's relation. The roots are,

$\displaystyle\frac{\displaystyle\frac{4}{\sqrt{3}} \pm \sqrt{\displaystyle\frac{16}{3}-4}}{2}$

$=\displaystyle\frac{2}{\sqrt{3}} \pm \displaystyle\frac{1}{\sqrt{3}}$.

So,

$p=\displaystyle\frac{1}{\sqrt{3}}$, or,

$p=\sqrt{3}$.

As only the first value appears in the choice values, the desired ratio,

$\displaystyle\frac{BP}{BC}=\frac{a}{b}=p=\displaystyle\frac{1}{\sqrt{3}}$.

**Answer:** b: $1:\sqrt{3}$.

#### Mathematical reasoning

If we move point P along the side AB towards point A, the angle equality condition holds till P merges with A. At this point the rectangle is transformed to a square and ratio of its area to the circle area increases to the maximum value $2:\pi$, and the desired ratio of $BP:BC$ reaches terminal maximum value $1:1$. We cannot move point P further on the right without violating the angle equality condition. Thus, with the given angle equality condition, even disregarding the area ratio value, BP cannot be larger than BC.

So the valid ratio will be $1:\sqrt{3}$ and not $\sqrt{3}:1$ which will be invalid.

**Key concepts used:** Area of circle -- area of parallelogram -- Ratio simplification -- triangle similarity concepts -- * inscribed parallelogram properties *-- Pythagorean theorem -- basic algebraic concepts -- variable elimination technique --

*-- substitution technique --*

**roots of quadratic equation concepts***-- free resource use of the choice values -- mathematical reasoning --*

**Sreedhar Acharya's formula***.*

**geometric object movement technique****Note:** we feel this problem to be a bit too involved to attempt in SSC CGL Tier II test. But concept-wise this problem imparts a rich learning.

**Problem 2.**

In the given figure $O$ is the centre of the circle and A, B, C, and D are four points on the circumference of the circle. Line segments AD and BC intersect at Q so that $\angle AQB=100^0$, while segments CA and DB extended meet at a point P outside the circle and $\angle CPD=60^0$.

The angle $\angle AOB$ is then,

- $60^0$
- $40^0$
- $50^0$
- $55^0$

#### Solution 2 - Problem analysis and solving

Looking at the figure analytically for a few seconds while trying to detect useful geometric concepts forming parts of the figure, the very first useful structure we notice is the two similar triangles $\triangle AQC$ and $\triangle BQD$.

Here $\angle C=\angle D=\angle x$ (we assume) by * arc angle subtending concept*, and $\angle AQC = \angle BQD$, being the two opposite angles in an intersection of two straight lines.

So the two triangles are similar and the third pair of angles,

$\angle CAQ = \angle DBQ=\angle y$, let us assume.

Furthermore as $\angle AQB$ is an external angle to the triangle $\triangle BQD$,

$\angle AQB=100^0=\angle x+\angle y$,

Or, $\angle y=100^0-\angle x$.

Again in $\triangle PCB$ the $\angle y$ is the external angle. So,

$\angle y=\angle x+\angle CPB=\angle x+60^0=100^0-\angle x$.

Or, $2\angle x=100^0-60^0=40^0=\angle AOB$.

As $\angle AOB$ is the angle held at centre by the same arc that holds $\angle x$ at the circumference, it is twice of $\angle x$.

**Answer:** b: $40^0$.

**Key concepts used:** Visualization -- * key pattern identification *of two similar triangles formed by the same arc AB --

*--*

**arc angle subtending concept***-- use of*

**intersecting pair of lines concept***in moving towards the target in two stages for finding $\angle x$ and hence twice its value $\angle AQB$ -- target driven information use.*

**triangle external angle concept****Problem 3.**

CD is a common tangent to two circles which intersect each other at points A and B. Then, $\angle CAD+\angle CBD$ is,

- $120^0$
- $360^0$
- $90^0$
- $180^0$

#### Solution 3 - Problem analysis and solving

The problem is depicted in the figure below.

By arc angle subtending concept, In circle with centre P,

$\angle CPB=2\angle CAB$, as arc CB holds an angle at centre double the angle it holds at a point on complementary arc.

By same concept in circle with centre Q,

$\angle BQD=2\angle BAD$, as arc BD holds an angle at centre double the angle it holds at a point on complementary arc.

Summing up the two,

$\angle CPB +\angle BQD=2(\angle CAB +\angle BAD)=2\angle CAD$.

Again two radii in $\triangle CPB$ being equal its base angles are equal, and so,

$\angle CPB=180^0-2\angle PCB=2(90^0-\angle PCB)=2\angle BCD$.

Similarly in $\triangle BQD$ radii being equal the base angles are equal and so,

$\angle BQD=2\angle BDC$.

Summing up these last two results and using earlier result,

$\angle CPB +\angle BQD=2\angle CAD=2(\angle BDC+\angle BCD)$,

Or, $\angle CAD = \angle BDC+\angle BCD = 180^0 - \angle CBD$,

Or, $\angle CAD + \angle CBD=180^0$.

**Answer:** d: $180^0$.

**Key concepts used:** * Tangent concepts* --

*-- useful patter identification -- isosceles triangle concept -- deductive reasoning -- angles in a triangle concept --*

**arc angle subtending concept***--*

**efficient simplification***.*

**end state analysis****Problem 4. **

If two equal circles are such that the centre of one lies on the periphery of the other, the ratio of the common chord to the radius of any of the circles is,

- $\sqrt{3}:2$
- $\sqrt{3}:1$
- $\sqrt{5}:1$
- $1:\sqrt{3}$

#### Solution 4 - Problem analysis and solving

The following figure depicts the given problem.

As the two circles are equal and as centre joining line AB is perpendicular to the common chord CE, the line AB is in fact the perpendicular bisector of CE.

In right triangle $\triangle ADC$,

$AC^2 = AD^2 + CD^2$,

Or, $r^2=\left(\frac{1}{2}c\right)^2+\left(\frac{1}{2}r\right)^2$, radius is $r=AC=AB$ and chord is $c=CE$

So,

$\frac{3}{4}r^2=\frac{1}{4}c^2$,

Or, $\sqrt{3}r=c$,

Or, $c:r=\sqrt{3}:1$.

**Answer:** b: $\sqrt{3}:1$.

**Key concepts used:** Chord bisection -- Pythagorean theorem.

**Problem 5. **

In the figure below an arc ABC of a circle subtends an angle of $100^0$ at the centre $O$.

If AB is extended to a point D outside the circle, the $\angle CBD$ is,

- $40^0$
- $140^0$
- $50^0$
- $130^0$

#### Solution 5 - Problem analysis and solving

The figure depicting the problem is shown below.

Drawing two new elements line segments AP and CP where P is a point on complementary arc of primary arc AC, we use the * arc angle subtending concept* to deduce,

$\angle AOC=100^0=2\angle APC$.

So,

$\angle APC = 50^0$.

Turning our attention now to cyclic quadrilateral APCB, as sum of opposite angles equal $180^0$,

$\angle ABC = 180^0 - \angle APC=180^0 - 50^0=130^0$.

Finally then at the point of incidence B of straight line CB on line ABD, as sum of two angles at the point of incidence equals $180^0$,

$\angle CBD=180^0-\angle ABC=180^0-130^0=50^0$.

**Answer:** c: $50^0$.

**Key concepts used:** Visualization -- new geometric element introduction -- arc angle subtending concept -- cyclic quadrilateral properties -- angles at point of incidence of two lines concept.

**Problem 6.**

In figure below two lines RP and SP touch a circle with centre at O at points A and B respectively and meet at point P.

If the line CD also touches the circle at point Q, then,

- $BP=DP+PC+CD$
- $3BP=DP+PC+CD$
- $4BP=DP+PC+CD$
- $2BP=DP+PC+CD$

#### Solution 6 - Problem analysis

The following figure represents the problem solution.

By the property of * two tangents to a circle from a common point* outside a circle, both the tangent segment lengths from the common point to the tangent point are equal. In our problem then, $BP=AP$.

Let us see how this happens. If you know this concept you may skip the follwing section.

#### Tangents to a circle from a common point

As a tangent is perpendicular to the radius of the circle at the point of tangency, we have in the above figure the two right triangles, $\triangle OAP$ and $\triangle OBP$ congruent. This is so because in the two right triangles, radii $OA=OB$ and hypotenuse $OP$ common, so that by Pythagorean theorem, the third pair of sides are also equal, $BP=AP$.

#### Solution 6 - Problem solving execution

As CQ and CA are two tangents from a common point C,

$QC=AC$.

Similarly as DB and DQ are two tangents from the same common poind D, we have,

$QD=BD$.

Using these results then summing up the three sides of the $\triangle DPC$,

$DP+PC+CD=DP+PC+CQ+QD$

$=(DP+DB)+(PC+AC)$

$=2BP$.

**Note:** At whichever point Q between A and B on the arc AB the third line CD is tangent to the circle, the above relationship holds.

**Answer:** d: $2BP=DP+PC+CD$.

**Key concepts used:** Tangents from a common point rich concept -- triangle congruency concept -- Pythagorean theorem -- basic tangent concepts.

** Problem 7.**

In figure below ARBD is a quarter circle of radius 1cm and a second circle is inscribed within the quarter circle touching it at three points.

The radius of the inscribed circle (in cm) is,

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

#### Solution 7 - Problem analysis and solution

The following figure represents the problem situation.

In the quarter circle, $AD$ is perpendicular to $BD$ with each radius of unit length and tangent to the insribed circle with centre at $C$ at points P and Q.

Again as $CP$ and $CQ$ are perpendicular to $AD$ and $BD$ respectively by tangent concepts, and $CQ=CP$ being radii of inscribed circle, $PCQD$ forms a square.

Thus distance between two radii, the hypotenuse $CD$ in isosceles right $\triangle CQD$ is,

$CD=\sqrt{2}CQ=\sqrt{2}r$, where $r$ is the radius of the inscribed smaller circle.

Finally then, as $DR$ is the radius of the larger circle and is of unit length,

$DR=1=CD+r=\sqrt{2}r+r$,

Or, $r=\displaystyle\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$, rationalizing the surd fraction by multiplying the numerator and denominator by $(\sqrt{2}-1)$.

**Answer:** d: $\sqrt{2}-1$.

**Key concepts used:** * Visualization* -- Tangent concepts -- useful pattern identification of the square formation -- Pythagorean theorem -- deductive reasoning -- linear equation --

**geometric simplification -- surd rationalization.****Note: **By * geometric simplification* we mean the identification of sum of centre distance and smaller circle radius as larger circle radius and deduction of smaller circle radius from the resulting simple linear equation in one variable.

** Problem 8.**

In the figure below P and Q are the mid-points of two sides CD and AD of a rectangle ABCD respectively.

The ratio of areas of $\triangle BPQ$ and rectangle ABCD is,

- $4:9$
- $5:8$
- $8:13$
- $3:8$

#### Solution 8 - Problem analysis

On cursory analysis we find that the length of the sides of the $\triangle BPQ$ cannot easily be evaluated. But then we recognize that the actual evaluation of the area of the triangle is not called for, only the ratio between its area and the area of the rectangle ABCD is needed.

Thus we look at area of the $\triangle BPQ$ as the difference between area of the whole rectangle, $A=AB\times{BC}$ and the sum of three leftover areas of three triangles $\triangle DQP$, $\triangle CBP$ and $\triangle AQB$. On further examination we find, P and Q being mid-points of two adjacent sides, all these three leftover triangles will have areas in terms of proportions of product of two sides of the rectangle. In other words the ratio of areas of the $\triangle BPQ$ and the rectangle will have a numerical ratio value, product of two sides canceling out.

#### Solution 8- Problem solving execution

By the above reasoning we have,

Area of triangle BPQ$=$(Area of rectangle ABCD)$-$(Sum of areas of triangles, DQP, CBP and AQB).

Area of triangle DQP$=\frac{1}{2}DP\times{DQ}=\frac{1}{2}\times{\frac{1}{2}CD}\times{\frac{1}{2}}AD=\frac{1}{8}A$,

Area of triangle CBP$=\frac{1}{2}CP\times{BC}=\frac{1}{2}\times{\frac{1}{2}CD}\times{BC}=\frac{1}{4}A$, and

Area of triangle AQB$=\frac{1}{2}AB\times{AQ}=\frac{1}{2}\times{AB}\times{\frac{1}{2}AD}=\frac{1}{4}A$

So,

Sum of three triangles DQP, CPB and AQB$=A\left(\displaystyle\frac{1}{8}+\displaystyle\frac{1}{4}+\displaystyle\frac{1}{4}\right)=\displaystyle\frac{5}{8}A$.

Thus the area of the $\triangle BPQ=A-\displaystyle\frac{5}{8}A=\displaystyle\frac{3}{8}A$, where $A$ is the area of the rectangle.

And the desired ratio, $3:8$.

**Answer:** d: $3:8$.

**Key concepts used:** Visualization -- deductive reasoning -- area of rectangle -- area of triangle -- key pattern identification -- breakdown into parts technique -- Other way round principle -- basic set theoretic concepts -- fraction concepts.

**Problem 9.**

In a $\triangle ABC$, two medians AD and CF intersect at G. The ratio of areas of $\triangle DFG$ and the $\triangle ABC$ will then be,

- $1:12$
- $1:9$
- $1:6$
- $1:16$

#### Solution 9 - Problem analysis and solution

The following figure will aid the problem solution.

Line segment DF connecting the mid-points of two adjacent sides BC and AB, it is parallel to base BC.

QGP is perpendicular to both DF and AC passing through G so that QG and GP are the heights of the triangles $\triangle DFG$ and $\triangle AGC$. As G cuts CF into a ratio, $CG :GF=2:1$, in similar triangles $\triangle GQF$ and $\triangle GPC$, $GP:GQ=2:1$.

In other words, height of $\triangle DGF$, $GQ=\frac{1}{3}QP$.

Furthermore, we have dropped a perpendicular from B to AC through S on FD. As FD is parallel to AC and QP and SR are both perpendicular to AC, $QP=SR=3GQ$.

Finally, as again FD is parallel to AC joining the mid-points of the adjacent sides AB and BC, the two triangles $\triangle BFD$ and $\triangle ABC$ are similar with $AC=2FD$ and $BR=2BS=2SR=6GQ$ (as $BS=SR$).

Then between two triangles $\triangle DFG$ and $\triangle ABC$, the base AC of $\triangle ABC$ is 2 times the base FD of $\triangle DFG$ and the height BR of $\triangle ABC$ is 6 times the height GQ of $\triangle DFG$.

Thus the area of $\triangle ABC$ is $6\times{2}=12$ times the area of $\triangle DFG$ and the desired ratio is $1:12$.

**Answer:** a: $1:12$.

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

**-- deductive reasoning -- triangle area -- ratio concepts -- target driven information use.**

*triangle similarity rich concepts -- median section ratio at centroid*** Problem 10.**

TP is the common tangent to two circles, the smaller with centre at $O_1$ touching the larger circle it with centre at $O_2$ at point P. The smaller circle is inside the larger one. If TQ is a second tangent to the larger circle at Q and TR is a second tangent to the smaller circle at R, then the ratio $TQ:TR$ is,

- $8:7$
- $7:8$
- $1:1$
- $5:4$

#### Solution 10 - Problem analysis and solution

The following is the visualization relevant to the problem.

TR and TP being tangents from common point T to the same circle with centre at $O_1$, tangent sections are equal, that is, $TP=TR$.

Similarly TP and TQ are also the tangents from the same point T to the larger circle with centre at $O_2$ so that $TP=TQ$.

Thus, $TR=TQ$, and $TQ:TR=1:1$.

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

**Key concepts used:** Visualization -- tangents from common point concept.

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