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SSC CGL level Solution Set 39, Geometry 7

Learn to Solve difficult circle geometry problems: SSC CGL Set 39

Quick solutions to difficult circle geometry and other geometry problems

Learn how to solve quickly difficult circle geometry and other geometry problems in SSC CGL Set 39. Quick solutions use geometry problem solving techniques.

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SSC CGL level Question Set 39, Geometry 7.

Solutions to 10 difficult circle geometry and other geometry problems in SSC CGL Set 39 - answering time was 15 mins

Problem 1.

In a rhombus ABCD, AB is produced to F and BA is produced to E such that AB = AE = BF. Then,

  1. $ED \bot CF$
  2. $ED \gt CF$
  3. $ED^2 + CF^2=EF^2$
  4. $ED || CF$

Problem analysis and solving 1

The following figure describes the problem,

SSC CGL level Solution Set 39 Geometry 7-1

In $\triangle ABC$, as BC=BA,

$\angle \alpha = 180^0 - 2\angle x = 2\angle y$, as $\angle \alpha = \angle y + \angle y$, being the external angle in $\triangle CBF$,

Or, $2(\angle x + \angle y) = 180^0$,

Or, $(\angle x + \angle y) = \angle ACF = 90^0$. That is, $AC \bot CF$.

Next, as BC || AD, $\angle ABC = \angle EAD = \angle \alpha$; and the two pairs of sides that include this equal angle, AD=BC, and AE=AB. This is SAS rule of triangle congruency. Result is, $\triangle AED$ is congruent to the $\triangle ABC$ so that $\angle ADE=\angle BCA = \angle x=\angle CAD$, as BC || AD.

This makes $ED || AC$ and $ED \bot CF$.

Answer: a: $ED \bot CF$.

Key concepts used: Basic rhombus properties -- triangle external angle concept -- isosceles triangle property -- congruent triangle tests -- congruent triangle concepts -- intersected parallel line concept.

We have used problem breakdown technique in dividing the problem into two parts, first showing $AC \bot CF$ and then $ED || AC$ so that final result obtained is surely $ED \bot CF$.

Second logic

After showing triangle congruency we can just superimpose $\triangle EAD$ on $\triangle ABC$ matching vertices so that ED merges with AC and automatically becomes perpendicular to CF. By this logic you can bypass showing line parallelism, a small step though. This is use of Many ways technique.

Geometric object movement technique

This action of translation or moving of $\triangle EAD$ so that it is placed exactly on top of $\triangle ABC$ is a powerful technique of problem solving in Geometry. Depending on the situation you may discover useful result bearing such movement (which may be a rotation) of a line, an arc, a chord or a closed regular shape that quickly gives you the solution. Visualization skills, geometric conceptual maturity and intelligent practice is needed to be able to detect such possibilities. We name this technique as, Geometric object movement technique.

Note: This is a rich problem with many possibilities of solution paths. You should try out for more solution paths and evaluate the efficiency of different solutions to get a deeper insight into the concepts involved. Using rhombus diagonals properties is one promising alternative.

Lastly, the perpendicularity condition means that the angle between two lines in direct intersection, in extended form or in parallely moved form is $90^0$.

To actually see that $ED \bot CF$ just extend FC and ED to intersect each other at P. The following figure shows the situation. This is the extended form of perpendicularity.

SSC CGL level Solution Set 39 Geometry 7-1-2

Problem 2.

In a quadrilateral ABCD with unequal sides if the diagonals AC and BD intersect at right angles then,

  1. $AB^2 + BC^2 = 2(CD^2 + DA^2)$
  2. $AB^2 + BC^2 = CD^2 + DA^2$
  3. $AB^2 + CD^2 = BC^2 + DA^2$
  4. $AB^2 + AD^2 = BC^2 + CD^2$

Problem analysis and solving 2

The relevant figure is shown below.

SSC CGL level Solution Set 39 Geometry 7-2

In two right triangles $\triangle APB$ and $\triangle APD$,

$AP^2 = AB^2 - PB^2 = DA^2 - PD^2$.

Similarly in two triangles $\triangle CPB$ and $\triangle CPD$,

$CP^2 = BC^2 - PB^2 = CD^2 - PD^2$.

Subtracting the second from the first we eliminate PB and PD,

$AB^2 - BC^2 = DA^2 - CD^2$,

Or, $AB^2 + CD^2 = BC^2 + DA^2$.

Answer: c: $AB^2 + CD^2 = BC^2 + DA^2$.

Key concepts used: Pythagoras theorem -- basic algebraic concepts of variable elimination.

Deductive reasoning:

The first thing that occurred to us is, if we translate point D down towards P along DP, the problem definition is not affected at all but adjacent sides AD and CD gets smaller and smaller with respect to the corresponding adjacent sides AB and BC and so, option 1 and 2 won't be valid in general.

Similarly when we translate point A down towards point P along AP adjacent sides AB and AD gets gradually smaller in comparison to BC and CD, making the option 4 invalid in general leaving only option 3 which sums up squares of opposite sides not adjacent sides.

This is purely conceptual reasoning using Geometric object movement technique and is the first step of deductive reasoning.

In the second step we look for the way to be sure of our conjecture.

The quadrilateral is not a special one such as cyclic quadrilateral leaving the only way to proceed is to use Pythagoras theorem in two pairs of right triangles. Being experienced in Algebra, visualizing the way to eliminate the common elements was easy.

Problem 3.

Two chords AC and BD of a circle with centre at O intersect at right angles at E. If $\angle OAB = 25^0$, then $\angle EBC$ is,

  1. $15^0$
  2. $20^0$
  3. $25^0$
  4. $30^0$

Problem analysis and solving 3.

The problem is depicted in the figure below.

SSC CGL level Solution Set 39 Geometry 7-3

The two chords AC and BD intersect perpendicularly at E so that $\angle OAB=25^0$. As radii OA=OB, in isosceles $\triangle OAB$,

$\angle OBA = \angle OAB = 25^0$.

And the third angle in the same triangle will be,

$\angle AOB = 180^0 - 2\times{25^0}=130^0$.

This is the angle held by the arc AB at the centre and so by arc angle subtending concept it will be twice the $\angle ACB$ at C on the complementary arc. So,

$\angle ACB = \angle ECB = \frac{1}{2}\text{ of } 130^0=65^0$.

Finally thus in the right $\triangle ECB$,

$\angle EBC = 90^0 - 65^0 = 25^0$.

Answer: c: $25^0$.

Key concepts used: Isosceles triangle properties -- arc angle subtending concept -- angles in triangle concept -- deductive reasoning.

Problem 4.

In a circle of radius 21 cm, an arc subtend an angle of $72^0$ at the centre. The length of the arc is,

  1. 26.4 cm
  2. 19.8 cm
  3. 21.6 cm
  4. 13.2 cm

Problem analysis and solving 4.

The following figure depicts the given problem.

SSC CGL level Solution Set 39 Geometry 7-4

Fact is, the whole of the circle circumference of length $2\pi{r}$, where $r$ is the radius, holds a total angle of $360^0$ at the centre. So what would be the proportion of the circumference or length of arc that would hold angle of $72^0$ at the centre? It is a simple case of using unitary method, but we detect that $72^0$ is one-fifth of $360^0$ and so the desired length of the arc would simply be,

$\displaystyle\frac{1}{5}\text{th of }2\pi{r} = \displaystyle\frac{2\times{22}\times{21}}{5\times{7}}$


$\hspace{20mm}=26.4$ cm

Answer: a: 26.4 cm.

Key concepts used: Proportionality of arc length and angle held at the centre -- total angle of $360^0$ held at the centre by the circumference of length $2\pi{r}$, where $r$ is the radius -- Utilizing the direct proportionality to get to the solution.

Problem 5.

A circle with centre at O touches two intersecting lines AX and BY. The two points of contact A and B subtend and angle of $65^0$ at any point C on the major arc of the circle. If P is the point of intersection of the two lines, then the measure of $\angle APO$ is,

  1. $65^0$
  2. $25^0$
  3. $90^0$
  4. $40^0$

Problem analysis and solving 5.

The figure depicting the problem is shown below.

SSC CGL level Solution Set 39 Geometry 7-5

In the figure above, $\angle ACB$ held by tangent points A and B at any point C on the major arc is $65^0$. This is actually the angle held by the minor arc AB on any point C on the complementary arc ACB. So the angle held by the minor arc AB at the centre will be double the angle held by it at the major arc, that is,

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

Now tangents AX and BY intersecting at P with tangent points A and B, the two right triangles $\triangle APO$ and $\triangle BPO$ are congruent. This is because, in addition to common side OP, also the radii OA = OB so that the third side is also equal by Pythagoras theorem in the right triangles (as radius to the tangent point is always perpendicular to the tangent). It means the line OP bisects the angles $\angle AOB$ and $\angle APB$.

Thus in right triangle $\triangle AOP$, $\angle AOP = 65^0$ so that,

$\angle APO = 180^0- (90^0 + 65^0) = 25^0$.

Answer: Option b: $25^0$

Key concepts used:  Visualization -- arc angle subtending concept -- tangent concept -- two tangent intersection concept -- Pythagoras theorem -- Congruent triangle concept -- angles in triangle concept.

Problem 6.

Chords AB and CD of a circle intersect at E and are perpendicular to each other. Segments AE, EB and ED are of lengths 2 cm, 6 cm and 3 cm respectively. Then the length of the diameter is,

  1. $\sqrt{65}$ cm
  2. $65$ cm
  3. $\displaystyle\frac{65}{2}$ cm
  4. $\frac{1}{2}\sqrt{65}$ cm

Problem analysis and solving 6.

The following figure represents the problem. AB and CD are the chords and O is the centre of the circle.

SSC CGL level Solution Set 39 Geometry 7-6

We have drawn the diameter parallel to the chord CD for convenience of getting many perpendiculars so that section lengths can be easily calculated in rectangles formed.

Using problem breakdown technique in the first step we will find out the length of unknown fourth part, that is, length of EC.

We know that the intersected chords form two similar triangles, $\triangle ACE$ and $\triangle BDE$, as in addition to the right angle, the angles held by the arc AD are also same. As a result ratio of corresponding sides of similar triangles being same we have,


Or, $CE = \displaystyle\frac{2}{3}\times{6}=4$ cm.

At the second stage now we will use the concept of chord bisection by perpendicular from centre O. Thus half length of the chords,

$AP=\displaystyle\frac{6+2}{2}=4$ cm and

$CQ=\displaystyle\frac{3+4}{2}=\frac{7}{2}$ cm.

Also as $AE=2$ cm, $EP=OQ=4-2=2$ cm, because $AB \bot CD$ as also $OQ \bot CD$ and OP parallel to QE forming two opposite sides of a rectangle.

So finally in the right $\triangle CQO$, length of the radius is given by,

$OC^2 = CQ^2 + OQ^2$

Or, $OC = \sqrt{\left(\displaystyle\frac{7}{2}\right)^2 + 2^2} = \frac{1}{2}\sqrt{65}$.

Thus the diameter is, $\sqrt{65}$ cm.

Answer: a: $\sqrt{65}$ cm.

Key concepts used: Problem breakdown technique -- Arc angle subtending concept -- similar triangle property -- chord bisection property -- deductive reasoning -- rectangle concept -- Pythagoras theorem.

Problem 7.

I and O are the incentre and circumcentre of $\triangle ABC$ respectively. The line AI produced intersects the circumcircle at point D. If $\angle ABC=x^0$, $\angle BID = y^0$ and $\angle BOD = z^0$, then $\displaystyle\frac{z+x}{y}$ is,

  1. 1
  2. 2
  3. 3
  4. 4

Problem analysis and solving 7.

The following figure describes the problem.

SSC CGL level Solution Set 39 Geometry 7-7

The very first thing we notice is, $\angle BAD$ being the angle held by the arc BD at the point A on the complementary arc, it will be equal to half of the $\angle BOD$ held by the same arc at the centre which is $\angle z$. By this observation then,

$\angle BAD = \angle BAI = \displaystyle\frac{z}{2}$.

Also BI being the bisector of the $\angle B$ we have,

$\angle ABI = \displaystyle\frac{x}{2}$.

Now it remains to relate the third angle by observing that it is the external angle to the $\triangle BIA$ with half of the first two angles as opposite internal angles. So,

$y = \displaystyle\frac{z+x}{2}$,

Or, $\displaystyle\frac{z+x}{y}=2$

Answer: b: 2.

Key concepts used: Deductive reasoning regarding how to bring all three angles in the same platform so that we can sum them up -- the result of analysis is to use arc angle subtending concept to transform $\angle z$ to an angle on the periphery of the larger circle and simultaneously as an angle of a triangle in which $\angle x$ already is present in halved form -- use of incentre concept -- the last step is to detect that the $\angle y$ is the external angle to the triangle in which we have already positioned the other two half angles as opposite angles -- use of external angle of triangle property.

Problem 8.

The radius of two concentric circles are 17 cm and 10 cm. A straight line ABCD intersects the larger circle at A and D and the smaller circle B and C. If BC = 12 cm, then the length of AD is,

  1. 24 cm
  2. 34 cm
  3. 20 cm
  4. 30 cm

Problem analysis and solving 8.

The following is a depiction of the problem graphically.

SSC CGL level Solution Set 39 Geometry 7-8

As the two circles are concentric, the common perpendicular from the centre bisects the two chords AD and BC.

IT is given BC=12 cm. So half of BC,

$BP=6$ cm.

In right $\triangle BPO$ then,

$OP^2 = OB^2 - BP^2 = 10^2 - 6^2 = 64$,

Or, $OP=8$ cm

Finally then in right $\triangle APO$,

$AP^2 = AO^2 - OP^2 = 17^2 - 8^2 = 225$,

Or, half of chord AD, $AP = 15$ cm

So chord $AD=30$ cm.

Answer: d: 30 cm.

Key concepts used: Concentric circles concept -- Chord bisection property --  Pythagoras theorem.

Problem 9.

Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation between PT and PQ is,

  1. $PT \gt PQ$
  2. $PT=PQ$
  3. $PT \lt PQ$
  4. $PT=2PQ$

Problem analysis and solving 9.

The following figure represents the problem description.

SSC CGL level Solution Set 39 Geometry 7-9

The technique of seeing the key formation that will lead you to the solution in geometry problems is to draw only the elements that are just necessary. Don't draw anything extra in the beginning. This is minimal geometry drawing technique. In uncluttered formation it is usually easier to see through the problem. 

Then try to figure out the path to the problem solution from the given portion of the figure. At this point try to visualize which concept will relate the given information to the end state of the solution. It may just be in one step or in a number of steps. Now only you draw the elements that were absent but that were crucial to solving the problem by showing you clearly the key relationship.

Here as we looked at the bare bones figure above, immediately it reminded us of the secant of a tangent concept that we have discussed in Basic and rich concepts Geometry 3 Circles. This is a powerful rich concept that relates tangent intercept from an external point to segments of a line cutting through the circle.

More specifically using secant of a tangent concept we have for the smaller circle,

$PT^2 = PA\times{PB}$, and in the same way,

$PQ^2 = PA\times{PB}$.

So $PT=PQ$.

Answer: b: $PT=PQ$.

Key concepts used: Key concept discovery -- use of secant of tangent concept -- deductive reasoning.

Note: By deductive reasoning though we were sure of the answer in the very beginning without being aware of the reasons behind the solution. The logic of this conviction is, the relation must be independent of the relative size of the circles. Only equality condition satisfied that condition.

You may refer to the secant of a tangent concept in brief below or in a bit more details in our tutorial session Basic and rich concepts Geometry 3 Circles.

Secant of a tangent intercept relationship proof

SSC CGL level Solution set 39 geometry 7-9-2 secant of a circle

In two triangles $\triangle APB$ and $\triangle APC$, $\angle A$ is common and $\angle APB=\angle BCP=\angle ACP$. This again is another rich concept derived from VERY IMPORTANT arc angle subtending concept and explained in tutorial Basic and rich concepts Geometry 3.

Thus two angles being equal, third pair of angles are also equal and the two triangles are similar. In similar triangles corresponding side ratios are equal so that,

$\displaystyle\frac{AP}{AC} = \frac{AB}{AP}$,

Or, $AP^2 = AB\times{AC}$.

Problem 10.

O is the circumcentre of triangle ABC. If $\angle BAC = 85^0$ and $\angle BCA = 55^0$, then $\angle OAC$ is,

  1. $40^0$
  2. $60^0$
  3. $50^0$
  4. $80^0$

Problem analysis and solving 10.

The following is the figure relevant to the problem.

SSC CGL level Solution Set 39 Geometry 7-10

This is an example of analyzing the given bare bones figure to discover how to reach the solution end state from the given initial state, identifying the steps connecting the two stages and drawing a single missing element radius OC.

In $\triangle ABC$ the third angle,

$\angle ABC = 180^0 - (85^0+55^0)= 40^0$.

It is the angle held by the arc AC at a point on its complementary arc and so the angle held by the same arc at the centre will be double this angle. Thus,

$\angle AOC = 80^0$.

Again in $\triangle AOC$, radii AO=OC and the triangle is isosceles so that two base angles are equal. Thus,

$\angle OAC = \frac{1}{2} \text { of } (180^0 - 80^0) = 50^0$.

Answer: c: $50^0$.

Key concepts used: deductive reasoning -- angles in triangle concept -- arc angle subtending concept -- isosceles triangle concept.

Recommendation: First try to use the most basic concepts. If unable to get the solution then only use rich concepts. Basic concepts usually result in quick solution.

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