Cyclic quadrilateral questions for SSC CGL and other geometry questions solved set 97
Cyclic quadrilateral questions for SSC CGL and other questions to be answered 10 in 15 minutes. Verify correctness from answers. Learn from solutions.
A number of these 10 problems are difficult.
The solved question set contains,
- Cyclic quadrilateral questions for SSC CGL and other geometry questions to be answered in 15 minutes (10 chosen questions)
- Answers to the questions, and,
- Detailed conceptual solutions to the questions.
The question set can very well be used as a unit of mock test or mini mock test.
For best results, take the test first and then learn to solve from solutions.
Cyclic quadrilateral questions for SSC CGL and other geometry questions set 97 - answer 10 questions in 15 mins
Q1. If $O$ be the circumcentre of a $\triangle PQR$ and $\angle QOR=110^0$, $\angle OPR=25^0$, then the measure of $\angle PRQ$ is,
- $65^0$
- $55^0$
- $50^0$
- $60^0$
Q2. Two poles of height 7 metres and 12 metres stand on a plane ground. If the distance between their feet is 12 metres, distance between their top will be,
- $13$ metres
- $19$ metres
- $17$ metres
- $15$ metres
Q3. A square is inscribed in a quarter circle in such a way that two of its adjacent vertices on the radius are equidistant from the centre and the other two vertices lie on the circumference. If the side of the square is $\sqrt{\displaystyle\frac{5}{2}}$ cm then what is the radius (in cm) of the circle?
- $5$
- $10$
- $2.5$
- $2$
Q4. In $\triangle PQR$, $PQ=PR=18$ cm, and AB and AC are parallel to lines PR and PQ respectively with B and C lying on sides PQ and PR respectively. If A is the mid-point of side QR, then what is the perimeter (in cm) of the quadrilateral ABPC?
- $36$
- $32$
- $28$
- $18$
Q5. The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at point P. Then it always true that,
- $BP.AP=CD.CP$
- $AP.CD=AB.CP$
- $AP.CP=BP.DP$
- $AP.BP=CP.DP$
Q6. If opposite sides of a quadrilateral and also its diagonals are equal, then each of the angles of the quadrilateral is,
- $60^0$
- $120^0$
- $100^0$
- $90^0$
Q7. In the circumcircle to equilateral $\triangle PQR$, PS, the bisector of $\angle P$, meets the circumference at S. If PR and QS extended meet at T, find value of the ratio $RT:RQ$.
- $1:2$
- $2:1$
- $1:1$
- $1:3$
Q8. P, Q and R are three points so that, $PR=3$ cm, $QR=5$ cm and $PQ=8$ cm. The number of circles passing through the points P, Q and R is,
- $0$
- $1$
- $2$
- $3$
Q9. $\triangle ABC$ is inscribed in a circle so that BC is diameter. The tangent at point C intersects BA when produced at a point D. If $\angle ABC=36^0$, then the value of $\angle ADC$ is,
- $36^0$
- $54^0$
- $44^0$
- $48^0$
Q10. $\triangle LMN$ is right angled at M. $\angle N$ is $45^0$. What is the length of MN (in cm) if $NL=9\sqrt{2}$ cm?
- $9$
- $9\sqrt{2}$
- $\displaystyle\frac{9}{\sqrt{2}}$
- $18$
Answers to the Cyclic quadrilateral questions for SSC CGL and other geometry questions
Q1. Answer: Option d: $60^0$.
Q2. Answer: Option a: $13$ metres.
Q3. Answer: Option c: $2.5$.
Q4. Answer: Option a: $36$.
Q5. Answer: Option c: $AP.CP=BP.DP$.
Q6. Answer: Option d: $90^0$.
Q7. Answer: Option c: $1:1$.
Q8. Answer: Option a: $0$.
Q9. Answer: Option b: $54^0$.
Q10. Answer: Option a: $9$.
Solutions 10 Geometry questions for SSC CGL including cyclic quadrilateral questions for SSC CGL set 97 - time to solve was 15 minutes
Q1. If $O$ be the circumcentre of a $\triangle PQR$ and $\angle QOR=110^0$, $\angle OPR=25^0$, then the measure of $\angle PRQ$ is,
- $65^0$
- $55^0$
- $50^0$
- $60^0$
Solution 1: Immediate solution by key pattern identification of isosceles triangles
The following figure describes the problem,
All three of $OP$, $OQ$ and $OR$ being the radii of the circumcircle, these are equal. This forms two isosceles triangles, $\triangle POR$ and $\triangle POQ$.
In $\triangle POR$, base angles are equal,
$\angle OPR=25^0=\angle PRO$.
In $\triangle OQR$, apex angle $\angle QOR=110^0$ and base angles are equal. So,
$2\angle ORQ=180^0-110^0=70^0$,
Or, $\angle ORQ=35^0$.
Sum up the two components of the desired angle,
$\angle PRQ=25^0+35^0=60^0$.
Answer: Option d: $60^0$.
Key concepts used: Key pattern identification of two isosceles triangles -- Total of angles in a triangle -- Solving in mind.
Q2. Two poles of height 7 metres and 12 metres stand on a plane ground. If the distance between their feet is 12 metres, distance between their top will be,
- $13$ metres
- $19$ metres
- $17$ metres
- $15$ metres
Solution 2: Quick solution by parallel projection of shorter pole length on the longer one and by using Pythagoras theorem
The following figure shows the problem situation.
AB and CD are the 12 metre and 7 metre poles standing 12 metre apart on a plane ground.
The poles by usual nature of pole erection, are erected vertically on the plane ground and so are parallel.
Project CD on AB to get difference of length $AE=5$ metres.
Also being parallel distance apart 12 metres CE is the separation of E and C.
In Pythagoras theorem then, hypotenuse,
$AC=\sqrt{AE^2+CE^2}=\sqrt{25+144}=\sqrt{169}=13$ metres.
Answer: Option a: $13$ metres.
Key concepts used: Real world reasonable assumption -- Parallel projection of smaller pole on longer pole -- Pythagoras theorem -- Solving in mind.
Q3. A square is inscribed in a quarter circle in such a way that two of its adjacent vertices on the radius are equidistant from the centre and the other two vertices lie on the circumference. If the side of the square is $\sqrt{\displaystyle\frac{5}{2}}$ cm then what is the radius (in cm) of the circle?
- $5$
- $10$
- $2.5$
- $2$
Solution 3: Quick solution by proper problem representation in figure, isosceles triangle properties, square diagonal properties and new geometric element introduction
The first difficulty in this problem is to sketch the figure specific to the problem.
As the two adjacent vertices of the square lie on the two radii, the square must be positioned as tilted by $45^0$ as shown in following graphic.
The following figure shows the problem graphic.
As vertices B and C are equidistant from centre O, with $\angle O$ of the quarter circle a $90^0$, in the isosceles $\triangle BOC$, the base angles are each $45^0$.
So, in right $\triangle BOC$ as side $BC=\sqrt{\displaystyle\frac{5}{2}}$ cm,
$BO=OC=\displaystyle\frac{BC}{\sqrt{2}}=\frac{\sqrt{5}}{2}$ cm.
A second help is now provided by the new element diagonal AC introduced.
By the property of a square, a diagonal of a square bisects each corner angle of $90^0$.
So, $\angle ACO=\angle BCO+\angle ACB=90^0$.
That is, $\angle AOC$ is a right triangle with desired radius as hypotenuse AC.
We need to find only the diagonal length of the square.
Again by square property, a diagonal is $\sqrt{2}$ times its side length.
So, $AC=\sqrt{5}$ cm.
Finally by Pythagoras theorem in $\triangle AOC$,
$AO^2=OC^2+AC^2=\displaystyle\frac{5}{4}+5=\displaystyle\frac{25}{4}$,
Or, $\text{Radius }AO=\displaystyle\frac{5}{2}=2.5$ cm.
Answer: Option c: $2.5$.
Key concepts used: Geometric visualization -- Isosceles angle property -- New geometric element introduction -- Diagonal length of a square -- Square properties -- Pythagoras theorem -- Solving in mind.
Calculations being simple, if you can follow the conceptual steps quickly, the problem can easily be solved in mind. But Geometric visualization is important to start solving the problem.
Q4. In $\triangle PQR$, $PQ=PR=18$ cm, and AB and AC are parallel to lines PR and PQ respectively with B and C lying on sides PQ and PR respectively. If A is the mid-point of side QR, then what is the perimeter (in cm) of the quadrilateral ABPC?
- $36$
- $32$
- $28$
- $18$
Solution 4: Quick solution by key pattern identification of a pair of new mid-points using similar triangles property and identifying a crucial pair of isosceles triangles
The following figure describes the problem and helps explaining the solution.
We will take the simplest path to the solution.
As $AB||RP$ in $\triangle PQR$, the two triangles $\triangle BQA$ and $\triangle PQR$ are similar.
Moreover, as A is the mid-point of side QR, B must also be the mid-point of side PQ from side ratio equality property of two similar triangles.
Similarly, C is also the mid-point of side PR.
As a result, $BQ=PB=9$ cm, and also,
$RC=PC=9$ cm.
In the final stage, we'll identify two crucial isosceles triangles.
With $PQ=QR$ in $\triangle PQR$,
$\angle Q=\angle R$.
As $AB||PR$ and QR intersects this pair of parallel lines,
$\angle BAQ=\angle R=\angle Q$.
$\triangle BQA$ is isosceles, and,
$AB=BQ=9$ cm.
Same way, $AC=RC=9$ cm.
Perimeter of quadrangle ABPC is,
$4\times{9}=36$ cm.
Answer: Option a: $36$.
Key concepts used: Key pattern identification of similar triangles -- Side ratio equality property of similar triangles -- Identification of two crucial isosceles triangles -- Angles formed by a line intersecting a pair of parallel lines -- Intersected parallel lines concepts -- Solving in mind.
Q5. The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at point P. Then it is always true that,
- $BP.AP=CD.CP$
- $AP.CD=AB.CP$
- $AP.CP=BP.DP$
- $AP.BP=CP.DP$
Solution 5: Immediate solution key pattern identification of similar triangles by arc angle subtending property and side ratio equality of a pair of similar triangles
The following figure will be used for explaining the solution,
As each of the choices is an equality of product of two pairs of sides, it is clear that our target is to find a pair of similar triangles and spot the suitable two pairs of sides for ratio equality.
With this clear objective, the key pattern of two pairs of equal angles could be identified by using arc angle subtending property,
$\angle BAC=\angle BAP=\angle BDC=\angle PDC$ on same arc BC (blue colored arc and angles), and,
$\angle ABD=\angle ABP=\angle ACD=\angle PCD$, on same arc AD (red colored arc and angles).
In triangles $\triangle ABP$ and $\triangle CPD$ two pairs of angles being equal, the third pair of angles also are equal, and the triangles are similar by A-A-A criterion of similarity.
By side ratio equality property in these two similar triangles then,
$\displaystyle\frac{AP}{DP}=\frac{BP}{CP}$, first ratio are for sides opposite red colored angles and the second opposite blue colored angles,
Or, $AP.CP=BP.DP$.
Answer: Option c: $AP.CP=BP.DP$.
Key concepts used: Key pattern identification -- Cyclic quadrilateral -- A-A-A similar triangle criterion -- Arc angle subtending concept -- Side ratio equality of a pair of similar triangles -- Solving in mind.
Q6. If opposite sides of a quadrilateral and also its diagonals are equal, then each of the angles of the quadrilateral is,
- $60^0$
- $120^0$
- $100^0$
- $90^0$
Solution 6: Instant solution using choice values and total angles of a quadrilateral with subsequent conceptual explanation
Instant solution:
Total of four internal angles of a quadrilateral is $360^0$. So each of the four angles to be equal, the equal value must be one-fourth of $360^0$, that is, value must be of Option d: $90^0$.
First we will show why with two pairs of intersecting parallel lines, a quadrangle must be a parallelogram.
With two pairs intersecting parallel lines, the quadrilateral must be a parallelogram, Proof
The following figure is used for proof,
In the figure ABCD is a quadrilateral with pairs of opposite sides equal.
$AD=BC$, and $AB=CD$.
In two triangles $\triangle ABC$ and $\triangle ACD$, with side AC common, all three corresponding pairs of sides are equal in length.
So by S-S-S criterion, the two triangles are congruent and opposite internal angles made by the diagonal are equal,
$\angle CAD =\angle ACB$,
$\angle BAC = \angle ACD$.
So $AD||BC$, and $AB||CD$ and ABCD is a parallelogram.
With two diagonals of equal length, a parallelogram must be a rectangle, Proof
The following figure will be used for explanation,
BCD is a parallelogram with diagonals, $AC=BD$.
In a parallelogram, the diagonals bisect each other and as the diagonals are equal to each other, all four sections of the bisected diagonals are equal to each other.
So, $AX=DX=CX=BX$.
As a result, two pairs of isosceles triangles are formed.
And we get two sets of four equal angles,
$\angle CAD=\angle ACB=\angle DBC=\angle BDA$, and similarly,
$\angle CAB=\angle ACD=\angle BDC=\angle DBA$.
Sum up adjacent pairs,
$\angle CAD+\angle CAB=\angle A=\angle DBC+\angle DBA=\angle B=\angle C$, as opposite angles of a parallelogram are equal.
Continuing summing up,
$\angle ACB+\angle ACD=\angle C=\angle BDA+\angle BDC=\angle D$.
So, $\angle A=\angle B=\angle C=\angle D=90^0$, as total of four angles in a quadrilateral is $360^0$.
Brief reasoning is:
With two pairs of parallel equal sides, a quadrilateral must be a parallelogram by the property of a parallelogram, and with the pair of diagonals equal, a parallelogram must at least be a rectangle by the property of a rectangle.
In any case, you don't have to prove any concept. All explanations here are for concept clearance only.
Answer: Option d: $90^0$.
Key concepts used: Property of a parallelogram -- Property of a rectangle -- Solving in mind.
Q7. In the circumcircle to equilateral $\triangle PQR$, PS, the bisector of $\angle P$, meets the circumference at S. If PR and QS extended meet at T, find value of the ratio $RT:RQ$.
- $1:2$
- $2:1$
- $1:1$
- $1:3$
Solution 7: Quick solution by target driven analysis, arc angle subtending property, equilateral triangle angle properties, external angle of a triangle property
The following figure will be used for explaining the solution,
It is apparent that the target ratio can be evaluated either by proving similarity and congruence of two triangles $\triangle QTR$ and $\triangle PST$ or by proving the $\triangle QTR$ isosceles.
In both cases, all data points to the ratio $1:1$.
Let us use the easiest second path.
PS being the bisector of $\angle P$, $\angle RPS=\angle QPS=30^0$, as in an equilateral triangle all three angles are $60^0$.
Next by arc angle subtending property, arc RS subtends two equal angles at points P and Q on alternate segment of circumference,
$\angle RPS=30^0=\angle RQS=\angle RQT$.
Finally $\angle PRQ$, the external angle to $\triangle QTR$ which is $60^0$ is equal to the sum of two opposite internal angles of the triangle,
$\angle PRQ=60^0=\angle RQT+\angle QTS$,
Or, $\angle QTS=60^0-\angle RQT=60^0-30^0=30^0$.
So $\triangle QTR$ is isosceles, and $RT=RQ$.
Desired ratio is, $1:1$.
Answer: Option c: $1:1$.
Key concepts used: Mathematical reasoning -- Circumcircle -- Equilateral triangle angle properties -- Arc angle subtending property -- External angle of a triangle property -- Solving in mind.
Q8. P, Q and R are three points so that, $PR=3$ cm, $QR=5$ cm and $PQ=8$ cm. The number of circles passing through the points P, Q and R is,
- $0$
- $1$
- $2$
- $3$
Solution 8: Instant solution by very basic constraint of a triangle formation
Though it is a matter of circles, the crucial key pattern stares you in the eye,
$PQ=PR+QR$.
All three points must lie on the same straight line.
No triangle can be formed with these three line segments as sides and no circle can pass through the three points P, Q and R as well.
Most basic condition for triangle formation by three line segments, AB, BC and CA is,
Sum of lengths of any two sides must exceed the length of the third side.
Answer: Option a: $0$.
Key concepts used: Key pattern identification -- Most basic condition of triangle formation -- Solving in mind.
Q9. $\triangle ABC$ is inscribed in a circle so that BC is diameter. The tangent at point C intersects BA when produced at a point D. If $\angle ABC=36^0$, then the value of $\angle ADC$ is,
- $36^0$
- $54^0$
- $44^0$
- $48^0$
Solution 9: Immediate solution by the concept that tangent is perpendicular to the radius at the tangent point
The following figure describes the problem and helps explaining the solution.
The tangent DC is perpendicular to diameter BC at tangent point C. So,
$\angle BCD=90^0$.
This makes third angle in $\triangle BCD$ as,
$\angle BDC=90^0-36^0=54^0$.
Answer: Option b: $54^0$.
Key concepts used: Tangent perpendicular to diameter at tangent point -- Solving in mind.
Note: You should always try to chose the shortest path to the solution.
Q10. $\triangle LMN$ is right angled at M. $\angle N$ is $45^0$. What is the length of MN (in cm) if $NL=9\sqrt{2}$ cm?
- $9$
- $9\sqrt{2}$
- $\displaystyle\frac{9}{\sqrt{2}}$
- $18$
Solution 10: Immediate solution by key pattern identification of $\triangle LMN$ as isosceles right-angled and well-known side to hypotenuse relation of such a triangle
The following figure describes the problem,
With $\angle N=45^0$, the third angle in the right-angled triangle $\angle L$ is also $45^0$.
By Pythagoras theorem, if $a$ is side length and $h$ is hypotenuse length in an isosceles right-angled triangle,
$h^2=2a^2$,
Or, $h=\sqrt{2}a$.
As hypotenuse in this case is $9\sqrt{2}$ cm, the side length $MN=LM=9$ cm.
Answer: Option a: $9$.
Key concepts used: Key pattern identification -- Hypotenuse to side length relation of a right-angled isosceles triangle -- Pythagoras theorem -- Solving in mind.
End note
Observe that, each of the problems could be quickly and cleanly solved in minimum number of steps using basic and rich geometry concept based key patterns and methods in each case.
This is the hallmark of quick problem solving:
- Concept based pattern and method formation, and,
- Identification of the key pattern and use of the method associated with it. Every special pattern has its own method, and not many such patterns are there.
Important is the concept based pattern identification and use of quick problem solving method.
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