SSC CGL level Solution Set 37, Geometry 5

37th SSC CGL level Solution Set, 5th on topic Geometry

SSC CGL level solution set 37 geometry-5

This is the 37th solution set of 10 practice problem exercise for SSC CGL exam and 5th on topic Geometry.

If you have not taken the corresponding test yet, you may refer to SSC CGL level Question Set 37, Geometry 5 and then continue with this solution.

Method for taking the test and get the best results from the test set:

  1. Before start, go through the tutorial on Geometry basic concepts part 1, Geometry basic concepts part 2, and Geometry basic and rich concepts part 3 on Circles or any other short but good material to refresh your concepts if you so require. This question set is in fact the set of exercise problems at the end of the first tutorial. Don't do the exercise as you are preparing for a hard competitive test. For you, taking the test will involve a different more stringent method.
  2. Answer the questions in an undisturbed environment with no interruption, full concentration and alarm set at 15 minutes.
  3. When the time limit of 15 minutes is over, mark up to which you have answered, but go on to complete the set.
  4. At the end, refer to the answers given at the end to mark your score at 15 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.
  5. Identify and analyze the problems that you couldn't do to learn how to solve those problems.
  6. Identify and analyze the problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledge improve it by referring to only that part of concept from the best source you get hold of. You might google it. If it is because of your method of answering, analyze and improve those aspects specifically.
  7. Identify and analyze the problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.
  8. Give a gap before you take a 10 problem practice test again.

Important: both practice tests and mock tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.


37th solution set- 10 problems for SSC CGL exam: 5th on Geometry - answering time 15 mins

Problem 1.

Two chords AB and CD of a circle with centre at O intersect each other at point P. If the two non-overlapping angles $\angle AOD=100^0$ and $\angle BOC=70^0$ then the value of $\angle APC$ is,

  1. $80^0$
  2. $75^0$
  3. $95^0$
  4. $85^0$

Solution 1.

The following figure describes the problem,

ssc cgl level solution set 37 geometry-5-1

Problem analysis

While scanning the problem description, the word "non-overlapping" caught our attention and so we sketched quickly the intersection of the two chords so that the two given angles don't overlap. In a separate discussion, Geometry basic and rich concepts part 4, proof of arc angle subtending concept, on the mechanism of how angle subtending takes place for chord or arcs in circles, we will take up this problem in more details.

To return to our problem at hand, the first conclusion we have is, whenever one or more than one chord with angle subtended at the centre is given, we will expect the use of a paired angle subtended by the chord at a point on the complementary arc (the other part of the circle). This is use of the arc angle subtending concept that is so frequently used in geometry problem solving especially for SSC CGL.

Problem solving

Searching for the angles subtended by the arcs BC and AD on the complementary major arcs we locate angles,

$\angle BAC=\frac{1}{2}\text{ of }70^0=35^0$ and

$\angle ACD=\frac{1}{2}\text{ of }100^0=50^0$.

Automatically the third angle in the $\triangle APC$ comes to be,

$\angle APC = 180^0 - (35^0 + 50^0)=95^0$.

It is that easy if you can use the given information properly and quickly.

Answer: c: $95^0$.

Key concepts used: Visualization of formation of intersection of the chords and placement of the given angles in the form of a quick sketch -- deductive reasoning helped conclude existence of pairs of half angles on the complementary arcs -- use of arc angle subtending concept -- locating these two angles for the two given angles held by the two chords at the centre -- use of triangle angle property.

Problem 2.

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the larger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at E. The length of line segment AD is,

  1. 17 cm
  2. 19 cm
  3. 18 cm
  4. 20 cm

Solution 2.

The following figure describes the problem.

ssc cgl level solution set 37 geometry-5-2

Problem analysis

After sketching the problem figure when we analyze the problem against the figure, we recognize presence of two important concepts,

  • Diameter subtending an angle of $90^0$ on the periphery. Here, diameter AB subtends $\angle AEB = 90^0$, and
  • Chord bisection by a perpendicular from the centre and coincidence of bisected chord with the tangent in case of concentric circles. The radius OD in the smaller circle is perpendicular to the tangent BE at D and at the same time, OD in the larger circle is the perpendicular to the chord BE at D thus bisecting it.
Problem solving

With these facts in place we have in right $\triangle BDO$,

$DE=BD=\sqrt{BO^2 - OD^2}=\sqrt{105}$.

Now we can see that if we know the length of segment AE, by applying Pythagoras theorem a second time now in $\triangle ADE$ we can easily find the desired value of length of segment AD.

Invoking the concept of Similar triangles

For two right triangles $\triangle AEB$ and $\triangle ODB$, the vertex angle is also common and so the third angles are equal and the triangles are similar. The similarity results in corresponding sides to be in equal ratio. So BE being double of BD, AE will also be double of corresponding side OD of the smaller triangle $\triangle ODB$.

Thus, $AE=16$ cm.

Finally then,

$AD = \sqrt{16^2 + 105}=\sqrt{361}=19$ cm.

Answer: b: 19 cm.

Key concepts used: Visualization -- concept of diameter subtending a right angle at periphery -- chord bisection property -- radius perpendicular to tangent concept -- coincidence of radius of smaller circle and chord bisecting perpendicular of larger circle for concentric circles concept -- Pyhthagoras theorem twice -- identifying similarity of triangles -- using equal ratio of corresponding sides concept for similar triangles.

Problem 3.

In a $\triangle ABC$, AD is perpendicular dropped on side BC and E is a point on AD so that AE : ED = 5 : 1. If $\angle BAD = 30^0$ and $\tan \angle ACB = 6\tan \angle DBE$, then $\angle ACB$ is,

  1. $15^0$
  2. $30^0$
  3. $45^0$
  4. $60^0$

Solution 3.

The problem is depicted in the figure below.

ssc cgl level solution set 37 geometry 5-3

Problem analysis

As $\tan$ of desired angle $\angle ACB$ is given in terms of $\tan$ of a second angle, we will first find the value of the second $\tan$. The reason for our approach lies in the fact that the angle $\angle DBE$ is in the region of all given values so that we should be able to find its value easily from the given information.

Problem solving

Using ratio concepts we may assume, $AE=5x$ and $ED=x$, so that $AD=6x$.

In right triangle $\triangle ABD$, $\angle BAD$ being equal to $30^0$, we have,

$\tan 30^0 = \displaystyle\frac{1}{\sqrt{3}} = \displaystyle\frac{BD}{AD}$,

Or, $BD = \displaystyle\frac{6x}{\sqrt{3}}$.

Thus in triangle $\triangle BDE$,

$\tan DBE = \displaystyle\frac{ED}{BD} = \frac{\sqrt{3}x}{6x}=\frac{\sqrt{3}}{6}$

Finally then from the given relation of $\tan$ functions,

$\tan ACB = 6\tan DBE = \sqrt{3}=\tan 60^0$,

Or, $\angle ACB = 60^0$.

Answer: d: $60^0$.

Key concepts used: Target driven requirement analysis -- End state analysis -- basic ratio concepts -- basic trigonometry concepts -- deductive reasoning.

In this solution we have used quite a bit of basic trigonometric concepts which is called for because of the given conditions of the problem.

Problem 4.

In $\triangle ABC$ when the points D and E on sides AB and AC are joined, the line DE turns out to be parallel to BC and also bisects the triangle into two equal areas. Under this condition, the ratio of DB : AB is,

  1. $1 : 2$
  2. $\sqrt{2} : 1$
  3. $(\sqrt{2} - 1) : \sqrt{2}$
  4. $1 : \sqrt{2}$

Solution 4.

The following figure depicts the given problem.

ssc cgl level solution set 37 geometry 5-4

Problem analysis

We recognize that the triangles $\triangle ADE$ and $\triangle ABC$ are similar as DE || BC. But because areas are involved we resort to using the heights of the triangles AQ and AP.

Problem solving

By the condition of DE halving the triangle area we have the relation of areas of the two triangles $\triangle ABC$ and $\triangle ADE$,

$\text{Area of }\triangle ABC = 2 \text{ Area of } \triangle ADE$,

Or, $\frac{1}{2}AQ\times{BC} = 2\times{\frac{1}{2}}\times{AP\times{DE}}$

Or, $\displaystyle\frac{BC}{DE}=2\displaystyle\frac{AP}{AQ}$

Now as DE is parallel to BC, the perpendicular APQ makes the two triangles $\triangle ADP$ and $\triangle ABQ$ similar so that,

$\displaystyle\frac{AP}{AQ} = \frac{AD}{AB} = \frac{DE}{BC}$, as triangles $\triangle ADE$ and $\triangle ABC$ are already similar.

Using this result we have then,

$\displaystyle\frac{BC}{DE}=2\displaystyle\frac{AP}{AQ}$,

Or, $\displaystyle\frac{AB}{AD}=2\displaystyle\frac{AD}{AB}$

Or, $\displaystyle\frac{AB^2}{AD^2} = 2$

Or, $\displaystyle\frac{AB}{AD} = \sqrt{2}$

We invert the expression as AB is in the denominator of the desired ratio, thus getting,

$\displaystyle\frac{AD}{AB}=\frac{1}{\sqrt2}$,

Now we subtract both sides from 1 to get the desired ratio,

$1 - \displaystyle\frac{AD}{AB}=1 - \frac{1}{\sqrt2}$,

Or, $\displaystyle\frac{DB}{AB}=\frac{(\sqrt{2} - 1)}{\sqrt{2}}$

Answer: c: $\displaystyle\frac{(\sqrt{2} - 1)}{\sqrt{2}}$

Key concepts used: Similarity of triangles rich concepts -- area of triangle concept -- ratio manipulation concept.

Problem 5.

The external bisectors of $\angle B$ and $\angle C$ of $\triangle ABC$ with AB and AC extended to E and F respectively, meet at point P. If $\angle BAC=100^0$, then $\angle BPC$ is,

  1. $40^0$
  2. $50^0$
  3. $80^0$
  4. $100^0$

Solution 5.

The figure depicting the problem is shown below.

ssc cgl level solution set 37 geometry 5-5

Problem analysis and solution

BP and CP are the external bisectors of the external angles $\angle CBE$ and $\angle BCF$.

Let's assume equal bisected angles,

$\angle EBP = \angle CBP =\angle x$,

$\angle FCP = \angle BCP =\angle y$, and internal angles,

$\angle ABC = \angle q$, and

$\angle ACB =\angle r$.

So,

$2\angle x + \angle q = 180^0$, and

$2\angle y + \angle r = 180^0$.

Adding the two we get,

$2(\angle x + \angle y) + (\angle q + \angle r) = 360^0$.

From $\triangle ABC$,

$\angle q + \angle r = 180^0 - 100^0 = 80^0$.

So,

$2(\angle x+\angle y) = 360^0 - 80^0 = 280^0$,

Or, $(\angle x+\angle y) = 140^0$,

Thus in $\triangle BPC$,

$\angle BPC = 180^0 - (\angle x+ \angle y) = 40^0$.

Answer: a: $40^0$.

Key concepts used: Visualization -- angle concepts.

Problem 6.

Angle between the two internal bisectors of angles $\angle B$ and $\angle C$ in $\triangle ABC$ is $120^0$. The $\angle A$ is,

  1. $20^0$
  2. $90^0$
  3. $30^0$
  4. $60^0$

Solution 6.

The following figure represents the problem.

ssc cgl level solution set 37 geometry 5-6

Problem analysis and solving

Approaching the problem in the simplest way, we work on the relationship between the base angles and the given angle of $120^0$.

In $\triangle BDC$,

$\angle DBC + \angle DCB = 180^0 - 120^0 = 60^0$.

In $\triangle ABC$,

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

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

$\hspace{8mm}=180^0 - 2\times{60^0}$

$\hspace{8mm}=60^0$.

Answer: d: $60^0$.

Key concepts used: Basic angle in triangle concepts.

Problem 7.

Two chords AB and CD of a circle with centre O intersect at point P. If $\angle ADP=23^0$ and $\angle APC=70^0$ then $\angle BCD$ is,

  1. $47^0$
  2. $45^0$
  3. $67^0$
  4. $57^0$

Solution 7.

The following figure represents the problem.

ssc cgl level solution set 37 geometry 5-7

Problem analysis and solution

First we apply concept of angles at the intersection of two straight lines.

At the intersection P one of four angles is given where these four angles form two pairs of equal angles.

So,

$\angle BPC = \frac{1}{2} \text{ of } (360^0 - 2\times{70^0})$

$\hspace{15mm}=\frac{1}{2} \text { of } 220^0$

$\hspace{15mm}=110^0$.

Again the angle $\angle ADC$ subtended by the arc AC equals the angle $\angle ABC$ subtended by the same arc AC on the complementary major arc.

So,

$\angle ABC=\angle ADC = 23^0$.

Finally then in triangle $\triangle BPC$,

$\angle BCP = \angle BCD $

$\hspace{15mm}= 180^0 - (110^0 + 23^0) = 47^0$.

Answer: a: $47^0$.

Key concepts used: Concept of angles in an intersection of two straight lines -- arc angle subtending concept -- total angle in a triangle concept.

Problem 8.

In $\triangle ABC$ AD is the internal bisector of the $\angle A$ meeting the side BC at D. If BD=5 cm and BC=7.5 cm, the ratio of AB : AC is,

  1. 1 : 2
  2. 2 : 1
  3. 4 : 5
  4. 3 : 5

Solution 8.

The following is a depiction of the problem graphically.

ssc cgl level solution set 37 geometry 5-8

As stated in the problem, AD is the angle bisector of $\angle BAC$ meeting side BC at D so that $\angle BAD = \angle DAC$. The ratio between the two sides containing the bisected angle is to found out.

In this case the rich concept of relation between the sides containing the bisected angle with the segments of the third side is,

$\displaystyle\frac{AB}{AC}=\frac{BD}{CD}$.

In the following section we will see how it happens. If you know the reasons behind the relation you may skip the explanation.


Rich concept of relation between sides of a triangle in case of bisection of the containing angle - the mechanism, how it happens

To understand the relationship clearly we need to extend the side AB and also draw a line through C parallel to AB to meet extended AB at E as shown below.

ssc cgl level solution set 37 geometry 5-8-1

The line AD being parallel to CE, the two triangles $\triangle ABD$ and $\triangle BCE$ are similar.

As a result the ratio of corresponding sides being equal we have,

$\displaystyle\frac{AB}{AE} = \frac{BD}{DC}$

But,

$\angle BAD = \angle DAC = \angle ACE = \angle BEC$.

This makes the $\triangle AEC$ isosceles with AE=AC.

So the equality ratio is transformed to,

$\displaystyle\frac{AB}{AC} = \frac{BD}{DC}$.


Coming back to our problem when we use this result we have,

$\displaystyle\frac{AB}{AC} = \frac{BD}{DC}$.

As BD=5 cm and BC=7.5 cm, DC=2.5 cm and so we have our desired ratio as,

$AB : AC = 5 : 2.5 = 2 : 1$.

Answer: b: 2 : 1.

Key concepts used: Rich concept of angle bisection property in a triangle -- simplification.

Problem 9.

P and Q are two points on a circle with centre at O. R is a point between P and Q on the minor arc formed by P and Q. If tangents to the circle at P and Q meet at point S with $\angle PSQ = 20^0$, then $\angle PRQ$ is,

  1. $80^0$
  2. $100^0$
  3. $160^0$
  4. $200^0$

Solution 9.

The following figure represents the problem description.

ssc cgl level solution set 37 geometry 5-9

Problem analysis

As R is any point on the minor arc, the value of the desired angle must be same over the minor arc. Thus the solution will come clearly through the use of arc angle subtending concept.

Problem solving

PS and QS being tangents at P and Q the radii OP and OQ are perpendiculars to PS and QS. So,

$\angle OPS = \angle OQS = 90^0$.

In quadrilateral OPSQ then,

$\angle QOP = 360^0 - (90^0 + 90^0 + 20^0) = 160^0$.

Thus the angle held by the major arc PAQ at the centre is,

$\angle POQ = 360^0 - 160^0 = 200^0$

Consequently, the angle held by the same major arc on its complementary minor arc should be half of it, that is,

$\angle PRQ=\frac{1}{2}\text{ of }200^0=100^0$.

Answer: b: $100^0$.

Key concepts used: Tangent concepts -- total angle in a quadrilateral concept -- arc angle subtending concept for major arc.

Alternate solution

By the reasoning of the previous solution, angle held by the minor arc at the centre,

$\angle QOP = 160^0$ and so angle held by the same arc on its complementary major arc should be half of it, that is,

$\angle QAP = 80^0$.

Now in cyclic quadrilateral AQRP, as the sum of two opposite angles is $180^0$, the desired angle,

$\angle PRQ=180^0 - 80^0 =100^0$.

Both these solutions are more or less equivalent to each other in terms of efficiency. Choice will be yours according to your comfort level. But the important point that we have highlighted here is the application of the powerful Many ways technique, practice of which improves your ability of problem solving, no less.

Problem 10.

Two circles intersect each other at points A and B. A straight line parallel to AB intersects the two circles consecutively at C, D, E and F. If CD=4.5 cm the length of line segment EF is,

  1. 4.5 cm
  2. 9 cm
  3. 1.5 cm
  4. 2.25 cm

Solution 10.

The following is the figure relevant to the problem.

ssc cgl level solution set 37 geometry 5-10

Problem analysis and solution

Only one value is given, but examining the choice values we find all to be multiples or factors of the given value. Thus we resort to geometric analysis.

First we observe the presence of the intersecting circle concepts.

Intersecting circles concepts among other things states,

When two circles intersect, the line joining two centres is the perpendicular bisector of the common chord.

This is a rich geometry concept on intersecting circles. You may skip the explanation of how this actually happens and go forward with the problem solving.


Concept of intersecting circles, the mechanism

When two circles intersect, the line joining the two centres will always be the perpendicular bisector of the common chord.

Let us see how it happens.

In the figure above, as $AQ=BQ$ in $\triangle AQB$, $\angle BAQ = \angle QBA$.

Similarly in $\triangle PAB$ as $AP=BP$, $\angle BAP = \angle PBA$.

Thus in two triangles, $\triangle PAQ$ and $\triangle PBQ$, $\angle PAQ=\angle PBQ$, $PA=PB$ and $QA=QB$.

This SAS (side-angle-side) condition makes these two triangles congruent and so,

$\angle AQP=\angle BQP$, and

$\angle APQ = \angle BPQ$.

So in two triangles $\triangle AHQ$ and $\triangle BHQ$, other two angles being equal, the third pair of angles must also be equal. So,

$\angle AHQ = \angle BHQ = 90^0$.

The same holds good in the other circle with,

$\angle AHP = \angle BHP = 90^0$.

Thus the centre joining line is in fact the perpendicular bisector of the common chord.


Continuing with our problem solving, as the centre joining line PQ is perpendicular to the common chord AB, it is perpendicular to the line CDEF parallel to AB also.

Now PQ being perpendicular to the chord DE in the larger circle it bisects it and so,

$DG=GE$. 

Similarly PQ being perpendicular to the chord CF in smaller circle, it bisects the chord and,

$CG=GF$,

Or, $CD + DG = GE + EF$.

So, $EF=CD=4.5$ cm.

Answer: a: 4.5 cm.

Key concepts used: Intersecting circles concepts -- line joining two centres of intersecting circles is the perpendicular bisector of the common chord of two intersecting circles -- laying out the equalities quickly gives us the solution.


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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