SSC CGL Tier II level Solution Set 16, Geometry 5 | SureSolv

SSC CGL Tier II level Solution Set 16, Geometry 5

ssc cgl tier 2 solutions 16 geometry 5

16th SSC CGL Tier II level Solution Set, 5th on Geometry

This is the 16th solution set of 10 practice problem exercise for SSC CGL Tier II exam and the 5th on topic Geometry.

A few of the problems look simple but deceptively so. In the same way, a few of the problems seem to be quite difficult, but can be solved quickly by identifying key geometric patterns and applying corresponding methods that are based on core concepts.

In this solution set we have even touched upon the issue of why and how to draw the right diagram for quick solution.

If you have not yet taken this test yet you may take it by referring to the SSC CGL Tier II level question set 16 on Geometry 5 before going through the solution.


16th solution set - 10 problems for SSC CGL exam: 5th on topic Geometry - answering time 15 mins

Problem 1.

If ABCD is a cyclic quadrilateral and AD is its diameter. If $\angle DAC=55^0$, then the value of $\angle ABC$ is,

  1. $35^0$
  2. $55^0$
  3. $125^0$
  4. $145^0$

Solution 1: Problem analysis and solving execution

The following figure describes the problem solution,

ssc cgl tier ii solutions 16 geometry 5 q1

Key pattern identified—diameter AD holds an angle of $90^0$ at the point C on the perimeter.

In $\triangle ACD$ then, with given $\angle DAC=55^0$, the third angle value is,

$\angle CDA=180^0-(55^0+90^0)=35^0$.

In the cyclic quadrilateral ABCD, as sum of opposite angles is $180^0$,

$\angle ABC=180^0-35^0=145^0$.

Answer: Option d: $145^0$.

Key concepts used: Diameter holding an angle of $90^0$ at the periphery -- Diameter subtending right angle  -- Opposite angles sum up to $180^0$ in a cyclic quadrilateral -- Cyclic quadrilateral property.

Problem 2.

AD is perpendicular to the internal bisector of $\angle ABC$ of $\triangle ABC$. DE is drawn through D and parallel to BC to meet AC at E. If the length of AC is 12 cm, then the length of AE (in cm) is,

  1. 3
  2. 4
  3. 6
  4. 8

Solution 2: Problem solving by mathematical reasoning based on educated guess

The following figure will help to solve the problem.

ssc cgl tier ii solutions 16 geometry 5 q2

The first action that you have to take is,

When in a problem involving a triangle, a line segment is parallel to one of the sides and length of a line segment is to evaluated, extend the parallel segment to intersect both the other sides thus creating the two similar triangles, a very important resource in solving triangle problems.

We consider this as an important geometry problem solving rule classified under strategic rules in geometry.

Thus we extend ED to intersect AB at F so that $\triangle AFE$ is similar to $\triangle ABC$.

Now we apply mathematical reasoning.

First, as only length of AC is given, and there is no visible way to use Pythagoras theorem, the only way length of AE can be evaluated is by using the equal ratio of the segments in the two similar triangles,

$\displaystyle\frac{AE}{AC}=\frac{AF}{AB}$.

Secondly, none of the values of AF or AB is known or derivable. By educated guess we decide with close to certainty that, AF must be equal to FB, which should be possible to prove using the last given information of angle equality followed by identification of isosceles triangles.

Answer should then be half of 12 cm, that is 6 cm.

Solution 2: Mathematical proof

With $\angle FBD=\angle DBC$ (as BD bisects $\angle ABC$) and $\angle DBC=\angle FDB$ (as FD || BC), we get in $\triangle FBD$,

$\angle FBD=\angle FDB$.

So in isosceles $\triangle FBD$, side length of FD is equal to FB.

Again in right triangle, $\triangle ABD$,

$\angle BAD=90^0-\angle ABD=90^0-\angle FDB$.

Lastly, in right angle, $\angle ADB$,

$\angle ADF=90^0-\angle FDB=\angle BAD$.

So in second isosceles $\triangle ADF$,

$AF=FD=FB$.

In the two similar triangles $\triangle AFE$ and $\triangle ABC$, by the property of equal ratio of corresponding sides,

$\displaystyle\frac{AE}{AC}=\frac{AF}{AB}=\frac{1}{2}$.

Thus length of AE is 6 cm as we have reasoned earlier.

Answer: Option c: 6.

Key concepts used: Geometry problem solving rule -- Strategic rules in Geometry -- Educated guess -- Mathematical reasoning -- Similar triangles property -- Deductive reasoning -- Key pattern identification -- Isosceles triangle identification.

Similarity of triangles and identification of two isosceles triangles in sequence are the key patterns here.

Problem 3.

Among the angles $30^0$, $36^0$, $45^0$ and $50^0$ one angle cannot be an exterior angle of a regular polygon. The angle is,

  1. $36^0$
  2. $30^0$
  3. $50^0$
  4. $45^0$

Solution 3: Problem analysis and solving

Relation between an internal angle and number of sides of a regular polygon is,

$\theta=\displaystyle\frac{(n-2)\pi}{n}$, where $\theta$ is any internal angle of a polygon of $n$ number of sides.

And exterior angle of such a regular polygon of $n$ sides,

$\phi=180^0-\theta=\displaystyle\frac{2\pi}{n}$.

For $n=6$, $n=8$, $n=10$ and $n=12$, the values of the external angles are, $\phi=60^0$, $\phi=45^0$, $\phi=36^0$ and $\phi=30^0$ respectively reducing with increasing number of sides.

Exterior angle of $50^0$ falls between $45^0$ and $60^0$ so that it could have been the exterior angle for $n=7$ only, which actually gives an exterior angle value with a fraction.

So the angle $50^0$ cannot be the exterior angle of a regular polygon.

Answer: Option c: $50^0$.

Key concepts used: Properties of a regular polygon -- Exterior angle of a regular polygon -- Polygon side angle relation.

Problem 4.

The distance between the centres of two circles having radii 8 cm and 3 cm is 13 cm. The length (in cm) of the direct common tangent of the two circles is,

  1. 15
  2. 12
  3. 16
  4. 18

Solution 4: Problem analysis and solving

With two radii dropping to the direct common tangent at two right angles, and radii lengths and centre distance given, chances are—application of Pythagoras relation would give the solution. This is application of a Strategic geometric problem solving rule

So after drawing the two circles, connecting the centres and dropping the two perpendicular radii on the common tangent, we go on to form a rectangle by extending the perpendicular radius of the smaller circle and drawing the fourth side of the rectangle parallel to the common tangent. The following figure depicts the situation.

ssc cgl tier ii solutions 16 geometry 5 q4

Extending radius QB to R and completing the rectangle APQR by joining AR are the two key actions for quick solution. This is application of Geometric element introduction technique.

Let us explain the mathematical reasoning for the solution.

In the rectangle APQR, length of PQ will be equal to length of its opposite parallel side AR.

Again, AR is the third side of right $\triangle ABR$ in which length of diagonal AB is 13 cm. We need only to know the length of BR to evaluate the length of the third side AR and hence length of PQ.

How do we get the length of line segment BR? The answer is easy—QR is the opposite parallel side of AP in the rectangle and hence is of length 8 cm (AP being the diameter of the larger circle). Again this QR length comprises of two lengths, length of QB, that is 3 cm, plus length of BR.

So,

$\text{BR}=8-3=5$ cm.

Now applying Pythagoras theorem on the $\triangle ABR$, we get the length of AR and hence length of PQ as,

$\text{PQ}^2=\text{AR}^2=13^2-5^2=12^2$,

Or, $\text{PQ}=12$ cm.

Answer: Option b: 12.

Key concepts used: Problem analysis -- Strategic rules in Geometry -- Mathematical reasoning -- Pythagoras theorem -- Geometric element introduction technique -- Rectangle properties -- Direct common tangent -- Tangent concepts.

Problem 5.

AB is the diameter of a circle with centre at O. The tangent at C meets AB produced at Q. If $\angle CAB=34^0$, then the value of $\angle CBA$ is,

  1. $34^0$
  2. $56^0$
  3. $124^0$
  4. $68^0$

Solution 5: Problem analysis and solving

The figure depicting the problem is shown below.

ssc cgl tier ii solutions 16 geometry 5 q5

This is a problem that contains some information not needed for the solution, but was put in for diverting your attention from the simple pattern that would solve the problem in a few seconds.

Solution 5: Reasoning chain to identify key pattern for quick solution disregarding tangent information altogether

Value of one angle $\angle CAB$ of $\triangle ACB$ is given and value of a second angle of the same triangle, $\angle CBA$ is to be found out. We are focusing our attention on the heart of the problem.

Naturally then we would examine the $\triangle ACB $ more closely and identify that its base AB is a diameter of the circle and so the third angle $\angle ACB=90^0$. This is the key pattern that we identify.

The value of $\angle CBA$ follows immediately,

$\angle CBA=90^0-34^0=56^0$.

Though we had drawn the figure, we never needed to use any information on the tangent and its intersection.

Answer: Option b: $56^0$.

Key concepts used: Key pattern identification -- Ignoring extra information -- Deductive reasoning -- Diameter holding an angle of $90^0$ at any point on the periphery of a circle -- Diameter subtending right angle.

Problem 6.

Two circles touch each other externally at A. PQ is a direct common tangent to the two circles with P and Q as points of contact. If $\angle APQ=35^0$ then $\angle PQA$ is,

  1. $75^0$
  2. $65^0$
  3. $35^0$
  4. $55^0$

Solution 6: Problem analysis

The following figure depicts the problem statement.

ssc cgl tier ii solutions 16 geometry 5 q6

With two isosceles triangles and two right angles, the figure has many resources to reach the solution. It is up to us which property we use to solve the problem quickly.

Solution 6: First approach to the solution using angle values of quadrilateral CPQD

CP being perpendicular to tangent PQ at P, $\angle CPQ=90^0$.

So,

$\angle CPA=90^0-35^0=55^0$.

With two radii CA and CP equal, the $\triangle CPA$ is isosceles and,

$\angle CAP=\angle CPA=55^0$.

So the third angle in $\triangle CPA$,

$\angle PCA=180^0-2\times{55^0}=70^0$.

Thus the fourth angle in the quadrilateral CPQD,

$\angle CDQ=\angle ADQ=360^0-(70^0+2\times{90^0})=110^0$.

From the first larger circle we have thus moved into the second smaller circle.

In this smaller circle we have the $\triangle AQD$ isosceles and so, 

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

Or, $\angle AQD=35^0$, and

$\angle PQA=90^0-35^0=55^0$.

Answer: Option d: $55^0$.

Key concepts used: Direct common tangent to two circles -- Tangent concepts -- Quadrilateral properties -- Isosceles triangle properties.

There is at least one more way to reach the solution. Try to find it.

Problem 7.

If $O$ is the circumcentre of a $\triangle ABC$ lying inside the triangle, then $\angle OBC+\angle BAC$ is equal to,

  1. $90^0$
  2. $110^0$
  3. $120^0$
  4. $60^0$

Solution 7: Problem analysis, key pattern identification and solution by strategic approach

The following figure depicts the problem situation.

ssc cgl tier ii solutions 16 geometry 5 q7

In problems involving circumcentre of a triangle, the circle lies outside the triangle and the vertices of the triangle lies on the periphery of the circle. This gives us an additional advantage of using each side of the triangle as chord and consequently use the rich set of chord properties. This falls under strategic rules in Geometry.

Thinking in this line of strategic approach, the first and most important key pattern we identify is the relation between two angles, $\angle BAC$ and $\angle BOC$. Both are subtended by the same chord BC and its corresponding minor arc BC, but the first angle is subtended on the periphery whereas the second angle is subtended at the centre.

So by arc angle subtending property,

$\angle BOC=2\angle BAC$.

We have applied the strategy of using the triangle sides as chords of the circle.

Now we will use the chord bisection property in a circle,

A perpendicular line dropped from the centre on any chord bisects the chord as well as the angle held by the chord at the centre.

$O$ being the circumcentre of the $\triangle ABC$, OE is perpendicular to the chord BC and bisects it, and also bisects the apex angle, $\angle BOC$.

So,

$\angle BAC=\displaystyle\frac{1}{2}\angle BOC=\angle BOE$.

It simply means, in triangle $\triangle OBE$,

$\angle OBC +\angle BAC=180^0-\angle OEB=90^0$.

To reach the solution fast, we have taken up the strategy of using the sides of the triangle as chords of the circle and that contributed two rich chord properties for the breakthrough—first the arc angle subtending concept, and then the second chord bisection property.

To know more, you may refer to our turtorial, Basic and rich concepts on arc angle subtending property.

Answer: Option a: $90^0$.

Key concepts used: Strategic rules in Geometry -- Use of triangle sides as chords of circumcircle of the triangle -- Circumcircle concepts -- Arc angle subtending concept -- Key pattern identification -- Chord bisection property.

Problem 8.

Three circles of radius 6 cm each touch one another externally. Then the shortest distance from the centre of one circle to the line joining the centres of the other two circles is equal to,

  1. $6\sqrt{7}$ cm
  2. $6\sqrt{2}$ cm
  3. $6\sqrt{5}$ cm
  4. $6\sqrt{3}$ cm

Solution 8: Problem analysis and solving

The following figure will aid solution of the problem.

ssc cgl tier ii solutions 16 geometry 5 q8

AB, BC and CA are the lines joining the centres of the three externally touching equal sized circles of radii 6 cm each. The centres of the circles thus formed an equilateral triangle of side length twice the radius length, that is, 12 cm.

AD, CE and BF are the perpendicular bisectors of the opposite sides as well as are the tangents passing through one centre to the touching point of the other two circles.

Thus AD being the perpendicular bisector of side BC of the equilateral triangle $\triangle ABC$ with side length 12 cm, the length of AD, shortest distance from one centre to the line joining the other two centres is given by,

$\text{AD}^2=\text{AB}^2-\text{BD}^2=12^2-6^2=108$.

So,

$\text{AD}=6\sqrt{3}$ cm.

Answer: Option d: $6\sqrt{3}$ cm.

Key concepts used: Three externally touching equal circles -- Tangent concepts -- Equilateral triangle -- Pythagoras theorem.

Problem 9.

The point of intersection of the diagonals AC and BD of the cyclic quadrilateral ABCD is P. If $\angle APB=64^0$ and $\angle CBD=28^0$, the value of $\angle ADB$ is,

  1. $36^0$
  2. $28^0$
  3. $32^0$
  4. $56^0$

Solution 9: Problem analysis, pattern identification and strategy decision

The following figure aid the problem solving.

ssc cgl tier ii solutions 16 geometry 5 q9

$\angle APB$ being the external angle to the $\triangle APD$, it equals sum of two oppposite internal angles,

$\angle ADP+\angle PAD=\angle APB=64^0$. 

We have focused our attention on the final desired angle.

To evaluate then $\angle ADP=\angle ADB$, we just need to know the third angle $\angle PAD$ of $\triangle APD$.

Looking for a pattern we identify the two angles, $\angle DAC$ and $\angle DBC=28^0$ to be held by the same chord and minor arc CD, and so the two angles are equal according to arc angle subtending property.

Finally then,

$\angle ADB=\angle ADP=64^0-\angle DAC=64^0-28^0=36^0$.

Answer: Option a: $36^0$.

Key concepts used: Identifying the key pattern of equality of two angles held by same minor arc -- Arc angle subtending concept -- Cyclic quadrilateral -- Deductive reasoning -- Triangle external angle property -- External angle in a triangle.

Problem 10.

A square is inscribed in a quarter circle in such a manner that two of its adjacent vertices lie on two radii at an equal distance from the centre, while the other two vertices lie on the circular arc. If the square has sides of length $x$, then the radius of the circle is,

  1. $\displaystyle\frac{\sqrt{5}x}{\sqrt{2}}$
  2. $\sqrt{2}x$
  3. $\displaystyle\frac{16x}{\pi + 4}$
  4. $\displaystyle\frac{2x}{\sqrt{\pi}}$

Solution 10: Problem analysis

The following is the visualization relevant to the problem.

ssc cgl tier ii solutions 16 geometry 5 q10

As the two adjacent vertices of the square A and B are at same distance from the centre $O$, $AO=BO$ in the right triangle $\triangle ABO$. Length of side AB being $x$, 

$AO=BO=\displaystyle\frac{x}{\sqrt{2}}$, applying Pythagoras theorem in isosceles right $\triangle ABO$.

What else can we find in terms of $x$? Naturally the diagonal AC of the square will have length in terms of side length $x$ as,

$AC=\sqrt{2}x$, applying Pythagoras theorem in $\triangle ABC$.

Now we join O to C completing the right $\triangle OCA$ with OC as its hypotenuse as well as the radius of the circle.

$\angle OAC=90^0$, because first, $\angle BAC=45^0$ in the square ABCD (diagonal bisecting a right angle) and second, $\angle BAO=45^0$ in the isosceles right $\triangle BOA$.

Thus applying Pythagoras theorem for the third time now in $\triangle OCA$ we get the length of radius of the circle as,

$r^2=\text{AO}^2+\text{AC}^2=\displaystyle\frac{x^2}{2}+2x^2=\displaystyle\frac{5x^2}{2}$,

Or, $r=\displaystyle\frac{\sqrt{5}x}{\sqrt{2}}$.

Answer: Option a: $\displaystyle\frac{\sqrt{5}x}{\sqrt{2}}$.

All that can be converted in terms of $x$ applying Pythagoras theorem yielded the solution quickly.

Key concepts used: Visualization -- Key pattern identification in two stages -- Pythagoras theorem -- Properties of right angled isosceles triangles -- Length of diagonal of a square -- Quarter circle problem.

Note: Though this problem looks simple, it needed a bit of extra attention.


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 section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but generally for any hard MCQ test.

Concept tutorials for SSC CGL and other competitive exams on Geometry

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

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