SSC CGL level Solution Set 80, Geometry 8 | SureSolv

SSC CGL level Solution Set 80, Geometry 8

80th SSC CGL level Solution Set, 8th on topic Geometry

SSC CGL solution set 80 geometry 8

This is the 80th solution set of 10 practice problem exercise for SSC CGL exam and 8th on topic Geometry. Some of the problems need suitable approach for quick solution.

If you have not taken the corresponding test yet, you may refer to SSC CGL level Question Set 80, Geometry 8 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 tutorials on Geometry basic concepts part 1 on points lines triangles, Geometry basic concepts part 2 on Quadrilaterals Squares Rectangles, Geometry basic and rich concepts part 3 on Circles, Basic and rich Geometry concepts part 4 on proof of arc angle subtending concept or any other short but good material to refresh your concepts if you so require.
  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.


80th solution set- 10 problems for SSC CGL exam: 8th on Geometry - answering time 15 mins

Problem 1.

The internal bisectors of the $\angle B$ and the $\angle C$ of $\triangle ABC$ intersect at $O$. If $\angle A=100^0$, the $\angle BOC$ is,

  1. $110^0$
  2. $140^0$
  3. $120^0$
  4. $130^0$

Problem analysis and solving 1

The following figure describes the problem,

SSC-CGL-Solution-Set-80-geometry8-q1

Bisectors BE and CD of angles $\angle B$ and $\angle C$ meet at point O, and $\angle A=100^0$. We are to find value of $\angle BOC$.

First conclusion from angle values in $\triangle ABC$ is,

$\angle B+\angle C=180^0 - 100^0=80^0$,

Or, $\displaystyle\frac{1}{2}(\angle B + \angle C) = 40^0$.

We have used the given angle value to reach this first conclusion. Now we will use the second fact that BO and CO are angle bisectors.

Using this last resource, our second conclusion is,

$\angle OBC+\angle OCB=40^0$.

So value of the third angle in $\triangle OBC$ is,

$\angle BOC=180^0-40^0=140^0$.

Answer: b: $140^0$.

Key concepts used: Angle in triangles -- Angle bisectors -- Step by step use of given conditions or resources for quick solution of the problem wholly mentally -- Solving in mind.

Problem 2.

G is the centroid of $\triangle ABC$. The medians AD and BE intersect at right angles. If the lengths of AD and BE are 9 cm and 12 cm respectively, the length of AB (in cm) is,

  1. 10
  2. 10.5
  3. 9.5
  4. 11

Problem analysis and solving 2

The relevant figure is shown below.

SSC-CGL-Solution-Set-80-geometry8-q2

When a given angle in a triangle is $90^0$, the first method we try to apply is Pythagoras theorem. But to apply this most used theorem we need to know values of two sides in a right triangle. To apply Pythagoras theorem here and find the length of AB, first we need to find the lengths of AG and BG in the $\triangle ABG$.

To fulfil this objective, we use the second given fact that G is the centroid. By centroid property, it divides a median into two parts with a length ratio 2 : 1. Thus we conclude that the length of AG is two-thirds of AD, that is, 6 cm, and the length of BG is two-thirds of BE, that is, 8 cm. This is our first conclusion.

Finally we are in a position to apply Pythagoras theorem, that was our objective from the beginning,

$AB^2=AG^2 + BG^2=100$,

Or, $AB=10$ cm.

Answer: a: 10.

Key concepts used: Pythagoras theorem -- Centroid property of dividing a median into a length ratio of 2 : 1 -- Problem analysis and strategic problem solving -- Strategic rules -- Patterns and methods -- Solving in mind.

The problem could easily be solved in mind in a few tens of seconds.

Strategic rule applied:

When in a geometry problem involving triangles, an angle given is $90^0$ and desired goal is to find length of a side and a few side values are given, the main strategy is to use Pythagoras theorem. This falls under the broader category of finding patterns and using suitable methods.

From this main goal, other actions follow.

Try to find the length of BC.

Hint: you need to use property of median bisecting the opposite side. We didn't need to take this additional action in this problem.

Problem 3.

An interior angle of a regular polygon is 5 time its exterior angle. Then the number of sides of the polygon is,

  1. 16
  2. 18
  3. 12
  4. 14

Problem analysis and solving 3.

One of the very few formulas used is the value of total of interior angles in a regular polygon with $n$ sides as,

$(n-2)\pi$.

check this out on a triangle and a quadrilateral.

In this problem, interior angle is 5 times an exterior angle. As these two angles sum up to $180^0$, 6 times an exterior angle equals $180^0$, that is, the value of an exterior angle is $30^0$ and interior angle is, $150^0$.

So,

$150n=(n-2)\times{180}$, as the polygon is regular, all its internal angles are of equal value,

Or, $30n=360$,

Or, $n=12$.

Answer: c: 12.

Key concepts used: Relation between Interior angle and exterior angle of a polygon -- Total angle of a polygon -- Regular polygon property of all angles same -- Solving in mind.

Problem 4.

ABCD is a cyclic trapezium with AB || DC and AB as the diameter. If $\angle CAB=30^0$, then $\angle ADC$ is,

  1. $120^0$
  2. $30^0$
  3. $150^0$
  4. $60^0$

Problem analysis and solving 4.

The following figure depicts the given problem.

SSC-CGL-Solution-Set-80-geometry8-q4

Just introducing the non-existing CO makes the new $\triangle AOC$ isosceles with the two radii, AO = CO so that $\angle AOC$ is $120^0$.

Again, in isosceles $\triangle COB$, $\angle OCB=\angle OBC$. As $\angle AOC$ is the external angle, the sum of these two internal angles in the triangle is $120^0$.

Individually each of the angles is then $60^0$.

Now we will use the property of a cyclic quadrilateral that sum of its opposite angles is $180^0$.

By this property then, the value of $\angle ADC$ is,

$180^0-60^0=120^0$.

Answer: a: $120^0$

Key concepts used: Technique of introduction of new element of a common side of two isosceles triangles -- Angle property of an isosceles triangle -- External angle of a triangle is equal to sum of two opposite internal angles -- Property of a cyclic quadrilateral.

We have not used the property of a trapezium here.

Second alternate solution

Instead of using the external angle of a triangle property, we observe that the diameter AB holds an angle of $90^0$ at any point in the periphery.

So, $\angle ACB = 90^0$.

In $\triangle ACB$ then the third $\angle ABC=60^0$.

By cyclic quadrilateral property then, opposite $\angle ADC=180^0-60^0=120^0$.

Neither did we introduce a new side BO, nor used the property of a trapezium in this solution.

This solution is also independent of the nature of the quadrilateral ABCD.

This is the cleanest and fastest solution, but this also is independent of whether the quadrilateral is a trapezium or not.

We have used here many ways technique to find many ways to the solution. Practice of this technique improves ability to discover new possibilities.

First question: Can you find any solution using property of trapezium?

Second question: Can you say whether $\angle ADC$ will remain unchanged if we move D along the periphery?

Problem 5.

Two circles with centers P and Q intersect at B and C. A and D are points on the first and second circles respectively, such that A, C and D are collinear. If $\angle APB=130^0$, the value of $\angle BQD$ is,

  1. $65^0$
  2. $135^0$
  3. $130^0$
  4. $195^0$

Problem analysis and solving 5.

The figure depicting the problem is shown below.

SSC-CGL-Solution-Set-80-geometry8-q5

In a problem involving two circles intersecting, it is strongly recommended that, unless circle radii are given, draw the two circles of similar sizes. This keeps drawing the full diagram simpler. Then,

  • First action that you must take is to connect the two intersection points BC. We have seen this line segment BC never remaining unused in solving such a problem.
  • And then draw the rest of the line segments required to describe the problem, putting a question mark on the angle or the length of line segment to be evaluated.

In this problem, when we finished drawing of the problem description, the solution came quickly in a few steps.

Solution Problem 5: Step 1: Relating the angle desired with the angle given: Working backwards and deductive reasoning

Thinking backwards, the first conclusion we draw is,

to evaluate $\angle BQD$ we need to find the value of $\angle BCD$ which will be half of $\angle BQD$ by arc angle subtending concept.

Going one more step backwards, we make the second conclusion,

the second action we must take is to evaluate $\angle BCA$, as this angle and $\angle BCD$ will sum up to $180^0$ by angles at point of incidence concept (when a line is incident on a second line at point P, the two angles formed will always sum up to $180^0$).

At this point we have come out of the second circle on the right and entered the first circle in which value of an angle is given.

This is an essential strategic objective in solving such geometric problems,

where the element to be evaluated and the element with given value lie in two different components of the whole structure. 

How to relate then $\angle BCA$ with $\angle APB$? Both now belong to the same circle.

Solution Problem 5: Step 2: Introducing a new angle $\angle ARB$ creating a cyclic quadrilateral

Now the solution path is clearly visible. We need to create a new $\angle ARB$ so that in the resulting cyclic quadrilateral, opposite pair of angles sum up to $180^0$,

$\angle BCA+\angle ARB=180^0$.

Only value we need to know is the value of $\angle ARB$.

Again by arc angle subtending concept, as minor arc AB subtends $\angle ARB$ at the periphery and $\angle APB=130^0$ at the center, the angle at the periphery will be half of angle held by the minor arc at the center, that is, 

$\angle ARB=65^0$.

Solution Problem 5: Final solution: last step of evaluation when deductive reasoning is complete

With all the reasoning of how we will relate given $\angle APB$ with desired $\angle AQB$ completed, now we work forwards to get the value of $\angle BCA$ as,

$\angle BCA = 180^0-\angle ARB=115^0$.

Then the second angle at the point of incidence, $\angle BCD$ is,

$\angle BCD =180^0-115^0=65^0$.

Lastly, the $\angle BQD$ being the angle held by the minor arc BD at the center, it will be double the $\angle BCD$ held by the same minor arc at pont C on the major arc, that is, it will be,

$\angle BQD=2\times{65^0}=130^0$.

Answer: Option c: $130^0$.

Key concepts used:  Visualization -- Working backwards reasoning -- Deductive reasoning --  arc angle subtending concept -- angles at point of incidence concept -- Cyclic quadrilateral property -- Strategic problem solving.

If you need to know more about the powerful arc angle subtending concept with all its variations, you should refer to our in-depth tutorial,

Basic and rich Geometry concepts part 4, proof of Arc angle subtending concept.

Problem 6.

AB and AC are tangents to a circle. PQ, another tangent of the circle at the point D on the circle, intersects AB and AC at P and Q respectively. If AB is 15 cm and QA is 9 cm, then DQ is,

  1. 4.5 cm
  2. 3 cm
  3. 7.5 cm
  4. 6 cm

Problem analysis and solving 6.

The following figure represents the problem. AB and AC are the tangents and O is the centre of the circle.

SSC-CGL-Solution-Set-80-geometry8-q6

As AB and AC are two tangents from a single point A, the tangent intercepts will be equal,

$AC=AB=15$ cm.

QA being 9 cm,

$QC=15-9=6$ cm.

Again we find QC and DQ are two tangents from same point Q so that,

$DQ=QC=6$ cm.

The difficulty in solving the problem lies in drawing the correct figure representing the problem. Some amount of visualization is needed to do it. Once you draw the figure, applying equal tangent intercepts concept twice, the problem can be solved in a few tens of seconds.

Answer: d: 6 cm.

Key concepts used: Visualization -- Equal tangent intercepts from single point concept.

Problem 7.

If two circles of radii 5 cm and 3 cm touch externally, the ratio in which the direct common tangent to the circles divide externally the line joining the centers of the circles is,

  1. 2.5 : 1.5
  2. 1.5 : 2.5
  3. 3 : 5
  4. 5 : 3

Problem analysis and solving 7.

The following figure describes the problem.

SSC-CGL-Solution-Set-80-geometry8-q7

Only the common tangent at the point of touching C can be the tangent described in the problem. Then, the required ratio is just the ratio of the radii, that is, 5: 3.

Answer: d: 5: 3.

Key concepts used: Common tangent -- Two circles touching.

Problem 8.

AC is the transverse common tangent to two circles with centers at P and Q and radii 6 cm and 3 cm at the points A and C respecively. If AC cuts PQ at the point B and AB is 8 cm, then the length of PQ is,

  1. 13 cm
  2. 15 cm
  3. 10 cm
  4. 12 cm

Problem analysis and solving 8.

The following is a depiction of the problem pictorially.

SSC-CGL-Solution-Set-80-geometry8-q8

As radius PA is perpendicular to the tangent BA at A, $\angle PAB=90^0$, and in right $\triangle PAB$, by Pythagoras theorem,

$PB^2=PA^2+BA^2=100$.

So, $PB=10$ cm.

At the second step we identify the two triangles, $\triangle PAB$ and $\triangle BCQ$ to be similar because of two right angles and as $\angle CBQ=\angle ABP$.

Applying the property of equal ratios of corresponding sides of two similar triangles, as radius 6 cm, one side of the larger triangle, is twice the radius 3 cm, corresponding side of the smaller triangle, PB will be twice the length of BQ.

So, 

$BQ=\displaystyle\frac{1}{2}\text{ of } 10=5$cm.

Thus length of PQ is,

$PQ=10+5=15$ cm.

Answer: b: 15 cm.

Key concepts used: Tangent property -- Pythagoras theorem -- Similar triangles condition -- Similar triangles property of equal ratio of corresponding sides.

Problem 9.

Length of a chord PQ in a circle is 18 cm. AB is the perpendicular bisector of PQ at M and intersects the circle at A and B. If MB = 3 cm, the length of AB is,

  1. 25 cm
  2. 27 cm
  3. 28 cm
  4. 30 cm

Problem analysis and solving 9.

The following figure represents the problem description.

SSC-CGL-Solution-Set-80-geometry8-q9

As PQ is bisected at M,

$QM = 9$ cm.

Also given, $BM=3$ cm, and so length of two sides of the $\triangle BMQ$ are known.

As we have mentioned the first strategic rule earlier,

in a Geometry problem, if side values are known, and triangles involved are right triangles, first attempt should be to use Pythagoras theorem.

Just a brief trial showed no promise in that approach.

Problem solution 9: Key pattern identification

Searching for an alternative path we identified AB to be a diameter holding an angle of $90^0$ at the point Q on the periphery. This has been the key pattern identification. Using this information, the problem could be solved just in a few steps.

Problem solution 9: Final solution by similarity of triangles—second strategic rule when triangle side values are given

The second strategic rule in solving a geometry problem where values of sides are known is,

When in a Geometry problem, a few side values of a triangle are given, the second approach should be to search for similarity of triangles. In similar triangles, equality of ratios of corresponding sides often help to solve such problems.

Knowing this second effective rule, and attempt to use Pythagoras theorem failing, we would identify a second triangle similar to the $\triangle BQM$ in which we know values of two sides. Which would be this second triangle? Naturally, it must be a triangle with AB as its one side. This is deductive reasoning.

We have now,

$\angle BQM=90^0-x=\angle QAM$.

Each with another angle as a right angle then, the two right triangles, $\triangle QAM$ and $\triangle BQM$ are similar.

In similar triangles, we know the ratios of corresponding sides to be equal. So,

$\displaystyle\frac{AM}{QM}=\frac{QM}{MB}=3$.

Thus diameter,

$AB=AM+MB=3\times{9}+3=30$ cm.

Answer: d: 30 cm.

Key concepts used: Key pattern identification -- Diameter holding a right angle on periphery -- Similar triangle identification -- Similar triangle property of equal ratio of corresponding sides -- Strategic rules -- Patterns and methods -- Strategic problem solving.

Note: By strategic methodical approach we briefly looked for an opportunity to apply Pythagoras theorem, the first rule of solving triangle problems with values of sides given. Failing, we resorted to the second rule of using similarity of triangles. In this quest, we identified the key pattern of diameter holding a right angle at a suitable point on the periphery. Solution was clearly visible from that point on.

Problem 10.

Two parallel chords are drawn in a circle of diameter 30 cm. The length of one chord is 24 cm and the distance between the chords is 21 cm. The length of the other chord is then,

  1. 10 cm
  2. 12 cm
  3. 18 cm
  4. 15 cm

Problem analysis and solving 10.

The following is the figure relevant to the problem.

SSC-CGL-Solution-Set-80-geometry8-q10

AB and CD are the two parallel chords in the circle with center at O.

Once we place the second chord CD on the opposite side of semicircle in which AB is, the rest is simple. The reason why CD cannot be placed on the same side as AB is, the perpendicular distance between AB and CD is 21 cm that is larger than the radius of 15 cm.

Following our first rule of trying for Pythagoras theorem when side values are given, immediately we get the value of OP in right triangle $\triangle AOP$ as,

$OP^2=AO^2-AP^2=225-144=81$.

So,

$OP=9$ cm.

At the next step we move into the second pair of triangles formed by the second chord and get the value of OQ first,

$OQ=PQ-OP=21-9=12$ cm.

Again in right triangle $\triangle OQD$, radius OD is 15 cm.

So, by Pythagoras theorem, $DQ=9$ cm, and $CD=18$cm as OQ is the perpendicular bisector of the chord.

Answer: c: 18 cm.

Key concepts used: Deductive reasoning -- Strategic problem solving -- Pythagoras theorem -- Chord bisection property.

Recommendation: In Geometry problem solving, first represent the problem figure correctly, analyze the problem to identify the useful patterns and apply suitable methods. Follow a strategic deductive reasoning approach starting with what element value is required to be known for the final answer, and then work backwards. If necessary, introduce new elements such as an angle or a side.

Keep using the basic concepts and use the advanced concepts or rules when you require it.


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