Selina Solutions Concise Maths Class 10 Chapter 17 Circles

A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point in the same plane always remains constant. In this chapter, students will learn the basic types and parts of a circle, cyclic properties and some important theorems and results. Students wanting to get a clear picture of concepts and problem solving techniques can access the Selina Solutions for Class 10 Mathematics prepared by expert faculty at BYJU’S. The solutions will also boost confidence among students to take up their Class 10 ICSE. The Selina Solutions for Class 10 Mathematics Chapter 17 Circles PDFs are available exercise-wise in the link below.

Selina Solutions Concise Maths Class 10 Chapter 17 Circles Download PDF

Exercises of Concise Selina Solutions Class 10 Maths Chapter 17 Circles

Exercise 17(A) Solutions

Exercise 17(B) Solutions

Exercise 17(C) Solutions

Access Selina Solutions Concise Maths Class 10 Chapter 17 Circles

Exercise 17(A) Page No: 257

1. In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30^o and 40^o respectively. Find ∠AOC Show your steps of working.

Solution:

Firstly, let’s join AC.

And, let ∠OAC = ∠OCA = x [Angles opposite to equal sides are equal]

So, ∠AOC = 180^o – 2x

Also,

∠BAC = 30^o + x

∠BCA = 40^o + x

Now, in ∆ABC

∠ABC = 180^o – ∠BAC – ∠BCA [Angles sum property of a triangle]

= 180^o – (30^o + x) – (40^o + x)

= 110^o – 2x

And, ∠AOC = 2∠ABC

180^o – 2x = 2(110^o – 2x)

2x = 40^o

x = 20^o

Thus, ∠AOC = 180^o – 2×20^o = 140^o

2. In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°

(i) Prove that AC is a diameter of the circle.

(ii) Find ∠ACB.

Solution:

(i) In ∆ABD,

∠DAB + ∠ABD + ∠ADB = 180^o

65^o + 70^o + ∠ADB = 180^o

135^o + ∠ADB = 180^o

∠ADB = 180^o – 135^o = 45^o

Now,

∠ADC = ∠ADB + ∠BDC = 45^o + 45^o = 90^o

As ∠ADC is the angle of semi-circle for AC as the diameter of the circle.

(ii) ∠ACB = ∠ADB [Angles in the same segment of a circle]

Hence, ∠ACB = 45^o

3. Given O is the centre of the circle and ∠AOB = 70^o.

Calculate the value of:

(i) ∠OCA,

(ii) ∠OAC.

Solution:

Here, ∠AOB = 2∠ACB

∠ACB = 70^o/ 2 = 35^o

Now, OC = OA [Radii of same circle]

Thus,

∠OCA = ∠OAC = 35^o

4. In each of the following figures, O is the centre of the circle. Find the values of a, b and c.

Solution:

(i) (ii)

(i) Here, b = ½ x 130^o

Thus, b = 65^o

Now,

a + b = 180^o [Opposite angles of a cyclic quadrilateral are supplementary]

a = 180^o – 65^o = 115^o

(ii) Here, c = ½ x Reflex (112^o)

Thus, c = ½ x (360^o – 112^o) = 124^o

5. In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.

Solution:

(i) Here, ∠BAD = 90^o [Angle in a semi-circle]

So, ∠BDA = 90^o – 35^o = 55^o

And,

a = ∠ACB = ∠BDA = 55^o

(ii) Here, ∠DAC = ∠CBD = 25^o

And, we have

120^o = b + 25^o

b = 95^o

(iii) ∠AOB = 2∠AOB = 2 x 50^o = 100^o

Also, OA = OB

∠OBA = ∠OAB = c

c = (180^o– 100^o)/ 2 = 40^o

(iv) We have, ∠APB = 90^o [Angle in a semicircle]

∠BAP = 90^o – 45^o = 45^o

Now, d = ∠BCP = ∠BAP = 45^o

6. In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O₁ and O₂ are the centers of two circles.

Solution:

It’s seen that,

∠DBA = ∠CBA = 90^o [Angle in a semi-circle is a right angle]

So, adding both

∠DBA + ∠CBA = 180^o

Thus, DBC is a straight line i.e. D, B and C form a straight line.

7. In the figure, given below, find:

(i) ∠BCD,

(ii) ∠ADC,

(iii) ∠ABC.

Show steps of your working.

Solution:

From the given fig, it’s seen that

In cyclic quadrilateral ABCD, DC || AB

And given, ∠DAB = 105^o

(i) So,

∠BCD = 180^o – 105^o = 75^o

(ii) Now,

∠ADC and ∠DAB are corresponding angles.

So,

∠ADC + ∠DAB = 180^o

∠ADC = 180^o – 105^o

Thus,

∠ADC = 75^o

(iii) We know that, the sum of angles in a quadrilateral is 360^o

So,

∠ADC + ∠DAB +∠BCD + ∠ABC = 360^o

75^o + 105^o + 75^o + ∠ABC = 360^o

∠ABC = 360^o – 255^o

Thus,

∠ABC = 105^o

8. In the figure, given below, O is the centre of the circle. If ∠AOB = 140^o and ∠OAC = 50^o;

find:

(i) ∠ACB,

(ii) ∠OBC,

(iii) ∠OAB,

(iv) ∠CBA.

Solution:

Given, ∠AOB = 140^o and ∠OAC = 50^o

(i) Now,

∠ACB = ½ Reflex (∠AOB) = ½ (360^o – 140^o) = 110^o

(ii) In quadrilateral OBCA,

∠OBC + ∠ACB + ∠OCA + ∠AOB = 360^o [Angle sum property of a quadrilateral]

∠OBC + 110^o + 50^o + 140^o = 360^o

Thus, ∠OBC = 360^o – 300^o = 60^o

(iii) In ∆AOB, we have

OA = OB (radii)

So, ∠OBA = ∠OAB

Hence, by angle sum property of a triangle

∠OBA + ∠OAB + ∠AOB = 180^o

2∠OBA + 140^o = 180^o

2∠OBA = 40^o

∠OBA = 20^o

(iv) We already found, ∠OBC = 60^o

And, ∠OBC = ∠CBA + ∠OBA

60^o = ∠CBA + 20^o

Therefore,

∠CBA = 40^o

9. Calculate:

(i) ∠CDB,

(ii) ∠ABC,

(iii) ∠ACB.

Solution:

Here, we have

∠CDB = ∠BAC = 49^o

∠ABC = ∠ADC = 43^o

Now, by angle sum property of a triangle we have

∠ACB = 180^o – 49^o – 43^o = 88^o

10. In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75^o; ∠ABD = 58^o and ∠ADC = 77^o.

Find:

(i) ∠BDC,

(ii) ∠BCD,

(iii) ∠BCA.

Solution:

(i) By angle sum property of triangle ABD,

∠ADB = 180^o – 75^o – 58^o = 47^o

Thus, ∠BDC = ∠ADC – ∠ADB = 77^o – 47^o = 30^o

(ii) ∠BAD + ∠BCD = 180^o

Thus, ∠BCD = 180^o – 75^o = 105^o

(iii) ∠BCA = ∠ADB = 47^o

11. In the figure given below, O is the centre of the circle and triangle ABC is equilateral.

Find:

(i) ∠ADB, (ii) ∠AEB

Solution:

(i) As, it’s seen that ∠ACB and ∠ADB are in the same segment,

So,

∠ADB = ∠ACB = 60^o

(ii) Now, join OA and OB.

And, we have

∠AEB = ½ Reflex (∠AOB) = ½ (360^o – 120^o) = 120^o

12. Given: ∠CAB = 75^o and ∠CBA = 50^o. Find the value of ∠DAB + ∠ABD.

Solution:

Given, ∠CAB = 75^o and ∠CBA = 50^o

In ∆ABC, by angle sum property we have

∠ACB = 180^o – (∠CBA + ∠CAB)

= 180^o – (50^o + 75^o) = 180^o – 125^o

= 55^o

And,

∠ADB = ∠ACB = 55^o

Now, taking ∆ABD

∠DAB + ∠ABD + ∠ADB = 180^o

∠DAB + ∠ABD + 55^o = 180^o

∠DAB + ∠ABD = 180^o – 55^o

∠DAB + ∠ABD = 125^o

13. ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130^o, find ∠BAC.

Solution:

From the fig. its seem that,

∠ACB = 90^o [Angle in a semi-circle is 90^o]

Also,

∠ABC = 180^o – ∠ADC = 180^o – 130^o = 50^o

By angle sum property of the right triangle ACB, we have

∠BAC = 90^o – ∠ABC

= 90^o – 50^o

Thus, ∠BAC = 40^o

14. In the figure given alongside, AOB is a diameter of the circle and ∠AOC = 110^o, find ∠BDC.

Solution:

Let’s join AD first.

So, we have

∠ADC = ½ ∠AOC = ½ x 110^o = 55^o

Also, we know that

∠ADB = 90^o

Therefore,

∠BDC = 90^o – ∠ADC = 90^o – 55^o

∠BDC = 35^o

15. In the following figure, O is the centre of the circle; ∠AOB = 60^o and ∠BDC = 100^o, find ∠OBC.

Solution:

Form the figure, we have

∠ACB = ½ ∠AOB = ½ x 60^o = 30^o

Now, by applying angle sum property in ∆BDC,

∠DBC = 180^o – 100^o – 30^o = 50^o

Therefore,

∠OBC = 50^o

16. In ABCD is a cyclic quadrilateral in which ∠DAC = 27^o, ∠DBA = 50^o and ∠ADB = 33^o. Calculate (i) ∠DBC, (ii) ∠DCB, (iii) ∠CAB.

Solution:

(i) It’s seen that,

∠DBC = ∠DAC = 27^o

(ii) It’s seen that,

∠ACB = ∠ADB = 33^o

And,

∠ACD = ∠ABD = 50^o

Thus,

∠DCB = ∠ACD + ∠ACB = 50^o + 33^o = 83^o

(iii) In quad. ABCD,

∠DAB + ∠DCB = 180^o

27^o + ∠CAB + 83^o = 180^o

Thus,

∠CAB = 180^o – 110^o = 70^o

17. In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80^o and ∠CDE = 40^o. Find the number of degrees in: (i) ∠DCE; (ii) ∠ABC.

Solution:

(i) Form the fig. its seen that,

∠DCE = 90^o – ∠CDE = 90^o – 40^o = 50^o

Therefore,

∠DEC = ∠OCB = 50^o

(ii) In ∆BOC, we have

∠AOC = ∠OCB + ∠OBC [Exterior angle property of a triangle]

∠OBC = 80^o – 50^o = 30^o  [Given ∠AOC = 80^o]

Therefore, ∠ABC = 30^o

18. In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.

Solution:

Firstly, join OB.

Then, ∠OBA = 90^o [Angle in a semi-circle is a right angle]

That is, OB is perpendicular to AE.

Now, we know that the perpendicular draw from the centre to a chord bisects the chord.

Therefore,

AB = BE

19. (a) In the following figure,

(i) if ∠BAD = 96^o, find ∠BCD and ∠BFE.

(ii) Prove that AD is parallel to FE.

(b) ABCD is a parallelogram. A circle

Solution:

(i) ABCD is a cyclic quadrilateral

So, ∠BAD + ∠BCD = 180^o

∠BCD = 180^o – 96^o = 84^o

And, ∠BCE = 180^o – 84^o = 96^o [Linear pair of angles]

Similarly, BCEF is a cyclic quadrilateral

So, ∠BCE + ∠BFE = 180^o

∠BFE = 180^o – 96^o = 84^o

(ii) Now, ∠BAD + ∠BFE = 96^o + 84^o = 180^o

But these two are interior angles on the same side of a pair of lines AD and FE.

Therefore, AD || FE.

20. Prove that:

(i) the parallelogram, inscribed in a circle, is a rectangle.

(ii) the rhombus, inscribed in a circle, is a square.

Solution:

(i) Let’s assume that ABCD is a parallelogram which is inscribed in a circle.

So, we have

∠BAD = ∠BCD [Opposite angles of a parallelogram are equal]

And ∠BAD + ∠BCD = 180^o

So, 2∠BAD = 180^o

Thus, ∠BAD = ∠BCD = 90^o

Similarly, the remaining two angles are 90^o each and pair of opposite sides are equal.

Therefore,

ABCD is a rectangle.

– Hence Proved

(ii) Let’s assume that ABCD is a rhombus which is inscribed in a circle.

So, we have

∠BAD = ∠BCD [Opposite angles of a rhombus are equal]

And ∠BAD + ∠BCD = 180^o

So, 2∠BAD = 180^o

Thus, ∠BAD = ∠BCD = 90^o

Similarly, the remaining two angles are 90^o each and all the sides are equal.

Therefore,

ABCD is a square.

– Hence Proved

21. In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.

Solution:

Give, AB = AC

So, ∠B = ∠C … (1)

And, DECB is a cyclic quadrilateral.

So, ∠B + ∠DEC = 180^o

∠C + ∠DEC = 180^o …. (Using 1)

But this is the sum of interior angles on one side of a transversal.

DE || BC.

But, ∠ADE = ∠B and ∠AED = ∠C [Corresponding angles]

Thus, ∠ADE = ∠AED

AD = AE

AB – AD = AC = AE [As AB = AC]

BD = CE

Hence, we have DE || BC and BD = CE

Therefore,

DECB is an isosceles trapezium.

22. Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.

Solution:

Let O and O’ be the centres of two intersecting circles, where points of the intersection are P and Q and PA and PB are their diameters respectively.

Join PQ, AQ and QB.

Thus, ∠AQP = 90^o and ∠BQP = 90^o

Now, adding both these angles we get

∠AQP + ∠BQP = 180^o

∠AQB = 180^o

Therefore, the points A, Q and B are collinear.

23. The figure given below, shows a circle with centre O. Given: ∠AOC = a and ∠ABC = b.

(i) Find the relationship between a and b

(ii) Find the measure of angle OAB, if OABC is a parallelogram.

Solution:

(i) It’s seen that,

∠ABC = ½ Reflex (∠COA)

So, b = ½ (360^o – a)

a + 2b = 180^o ….. (1)

(ii) As OABC is a parallelogram, the opposite angles are equal.

So, a = b

Now, using the above relationship in (1)

3a = 180^o

a = 60^o

Also, OC || BA

∠COA + ∠OAB = 180^o

60^o + ∠OAB = 180^o

Therefore,

∠OAB = 120^o

24. Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the center O is equal to twice the angle APC

Solution:

Required to prove: ∠AOC + ∠BOD = 2∠APC

OA, OB, OC and OD are joined.

Also, AD is joined.

Now, it’s seen that

∠AOC = 2∠ADC …. (1)

Similarly,

∠BOD = 2∠BAD …. (2)

Adding (1) and (2), we have

∠AOC + ∠BOD = 2∠ADC + 2∠BAD

= 2(∠ADC + ∠BAD) ….. (3)

And in ∆PAD,

Ext. ∠APC = ∠PAD + ∠ADC

= ∠BAD + ∠ADC …. (4)

So, from (3) and (4) we have

∠AOC + ∠BOD = 2∠APC

25. In the figure given RS is a diameter of the circle. NM is parallel to RS and MRS = 29^o

Calculate: (i) ∠RNM; (ii) ∠NRM.

Solution:

(i) Join RN and MS

∠RMS = 90^o [Angle in a semi-circle is a right angle]

So, by angle sum property of ∆RMS

∠RMS = 90^o – 29^o = 61^o

And,

∠RNM = 180^o – ∠RSM = 180^o – 61^o = 119^o

(ii) Now as RS || NM,

∠NMR = ∠MRS = 29^o [Alternate angles]

∠NMS = 90^o + 29^o = 119^o

Also, we know that

∠NRS + ∠NMS = 180^o

∠NRM + 29^o + 119^o = 180^o

∠NRM = 180^o – 148^o

Therefore,

∠NRM = 32^o

26. In the figure given alongside, AB || CD and O is the center of the circle. If ∠ADC = 25^o; find the angle AEB. Give reasons in support of your answer.

Solution:

Join AC and BD.

So, we have

∠CAD = 90^o and ∠CBD = 90^o

And, AB || CD

So, ∠BAD = ∠ADC = 25^o [Alternate angles]

∠BAC = ∠BAD + ∠CAD = 25^o + 90^o = 115^o

Thus,

∠ADB = 180^o – 25^o – ∠BAC = 180^o – 25^o – 115^o = 40^o

Finally,

∠AEB = ∠ADB = 40^o

27. Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.

Solution:

Let’s join AC, PQ and BD.

As ACQP is a cyclic quadrilateral

∠CAP + ∠PQC = 180^o ……. (i)

Similarly, as PQDB is a cyclic quadrilateral

∠PQD + ∠DBP = 180^o ……. (ii)

Again, ∠PQC + ∠PQD = 180^o …… (iii) [Linear pair of angles]

Using (i), (ii) and (iii) we have

∠CAP + ∠DBP = 180^o

Or ∠CAB + ∠DBA = 180^o

We know that, if the sum of interior angles between two lines when intersected by a transversal are supplementary.

Then, AC || BD.

28. ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.

Solution:

Let’s assume that ABCD be the given cyclic quadrilateral.

Also, PA = PD [Given]

So, ∠PAD = ∠PDA …… (1)

And,

∠BAD = 180^o – ∠PAD [Linear pair of angles]

Similarly,

∠CDA = 180^o – ∠PDA = 180^o – ∠PAD [From (1)]

As the opposite angles of a cyclic quadrilateral are supplementary,

∠ABC = 180^o – ∠CDA = 180^o – (180^o – ∠PAD) = ∠PAD

And, ∠DCB = 180^o – ∠BAD = 180^o – (180^o – ∠PAD) = ∠PAD

Thus,

∠ABC = ∠DCB = ∠PAD = ∠PDA

Which is only possible when AD || BC.

Exercise 17(B) Page No: 265

1. In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.

Prove it.

Solution:

Let ABCD be the cyclic trapezium in which AB || DC, AC and BD are the diagonals.

Required to prove:

(i) AD = BC

(ii) AC = BD

Proof:

It’s seen that chord AD subtends ∠ABD and chord BC subtends ∠BDC at the circumference of the circle.

But, ∠ABD = ∠BDC [Alternate angles, as AB || DC with BD as the transversal]

So, Chord AD must be equal to chord BC

AD = BC

Now, in ∆ADC and ∆BCD

DC = DC [Common]

∠CAD = ∠CBD [Angles in the same segment are equal]

AD = BC [Proved above]

Hence, by SAS criterion of congruence

∆ADC ≅ ∆BCD

Therefore, by CPCT

AC = BD

2. In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110^o, calculate:

(i) ∠AFE, (ii) ∠FAB.

Solution:

Join AE, OB and OC.

(i) As AOD is the diameter

∠AED = 90^o [Angle in a semi-circle is a right angle]

But, given ∠DEF = 110^o

So,

∠AEF = ∠DEF – ∠AED = 110^o – 90^o = 20^o

(ii) Also given, Chord AB = Chord BC = Chord CD

So,

∠AOB = ∠BOC = ∠COD [Equal chords subtends equal angles at the centre]

But,

∠AOB + ∠BOC + ∠COD = 180^o [Since, AOD is a straight line]

Thus,

∠AOB = ∠BOC = ∠COD = 60^o

Now, in ∆OAB we have

OA = OB [Radii of same circle]

So, ∠OAB = ∠OBA [Angles opposite to equal sides]

But, by angle sum property of ∆OAB

∠OAB + ∠OBA = 180^o – ∠AOB

= 180^o – 60^o

= 120^o

Therefore, ∠OAB = ∠OBA = 60^o

Now, in cyclic quadrilateral ADEF

∠DEF + ∠DAF = 180^o

∠DAF = 180^o – ∠DEF

= 180^o – 110^o

= 70^o

Thus,

∠FAB = ∠DAF + ∠OAB

= 70^o + 60^o = 130^o

3. If two sides of a cycli-quadrilateral are parallel; prove that:

(i) its other two sides are equal.

(ii) its diagonals are equal.

Solution:

Let ABCD is a cyclic quadrilateral in which AB || DC. AC and BD are its diagonals.

Required to prove:

(i) AD = BC

(ii) AC = BD

Proof:

(i) As AB || DC (given)

∠DCA = ∠CAB [Alternate angles]

Now, chord AD subtends ∠DCA and chord BC subtends ∠CAB at the circumference of the circle.

So,

∠DCA = ∠CAB

Hence, chord AD = chord BC or AD = BC.

(ii) Now, in ∆ABC and ∆ADB

AB = AB [Common]

∠ACB = ∠ADB [Angles in the same segment are equal]

BC = AD [Proved above]

Hence, by SAS criterion of congruence

∆ACB ≅ ∆ADB

Therefore, by CPCT

AC = BD

4. The given figure show a circle with centre O. Also, PQ = QR = RS and ∠PTS = 75°.

Calculate:

(i) ∠POS,

(ii) ∠QOR,

(iii) ∠PQR.

Solution:

Join OP, OQ, OR and OS.

Given, PQ = QR = RS

So, ∠POQ = ∠QOR = ∠ROS [Equal chords subtends equal angles at the centre]

Arc PQRS subtends ∠POS at the centre and ∠PTS at the remaining part of the circle.

Thus,

∠POS = 2 x ∠PTS = 2 x 75^o = 150^o

∠POQ + ∠QOR + ∠ROS = 150^o

∠POQ = ∠QOR = ∠ROS = 150^o/ 3 = 50^o

In ∆OPQ we have,

OP = OQ [Radii of the same circle]

So, ∠OPQ = ∠OQP [Angles opposite to equal sides are equal]

But, by angle sum property of ∆OPQ

∠OPQ + ∠OQP + ∠POQ = 180^o

∠OPQ + ∠OQP + 50^o = 180^o

∠OPQ + ∠OQP = 130^o

2 ∠OPQ = 130^o

∠OPQ = ∠OPQ = 130^o/ 2 = 65^o

Similarly, we can prove that

In ∆OQR,

∠OQR = ∠ORQ = 65^o

And in ∆ORS,

∠ORS = ∠OSR = 65^o

Hence,

(i) ∠POS = 150^o

(ii) ∠QOR = 50^o and

(iii) ∠PQR = ∠PQO + ∠OQR = 65^o + 65^o = 130^o

5. In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:

(i) ∠AOB,

(ii) ∠ACB,

(iii) ∠ABC.

Solution:

(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∠ACB = ½ ∠AOB

And as AB is the side of a regular hexagon, we have

∠AOB = 60^o

(ii) Now,

∠ACB = ½ (60^o) = 30^o

(iii) Since AC is the side of a regular octagon,

∠AOC = 360^o/ 8 = 45^o

Again, arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.

∠ABC = ½ ∠AOC

∠ABC = 45^o/ 2 = 22.5^o

Exercise 17(C) Page No: 265

1. In the given circle with diameter AB, find the value of x.

Solution:

Now,

∠ABD = ∠ACD = 30^o [Angles in the same segment]

In ∆ADB, by angle sum property we have

∠BAD + ∠ADB + ∠ABD = 180^o

But, we know that angle in a semi-circle is 90^o

∠ADB = 90^o

So,

x + 90^o + 30^o = 180^o

x = 180^o – 120^o

Hence, x = 60^o

2. In the given figure, ABC is a triangle in which ∠BAC = 30^o. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose center is O.

Solution:

Firstly, join OB and OC.

Proof:

∠BOC = 2∠BAC = 2 x 30^o = 60^o

Now, in ∆OBC

OB = OC [Radii of same circle]

So, ∠OBC = ∠OCB [Angles opposite to equal sides]

And in ∆OBC, by angle sum property we have

∠OBC + ∠OCB + ∠BOC = 180^o

∠OBC + ∠OBC + 60^o = 180^o

2 ∠OBC = 180^o – 60^o = 120^o

∠OBC = 120^o/ 2 = 60^o

So, ∠OBC = ∠OCB = ∠BOC = 60^o

Thus, ∆OBC is an equilateral triangle.

So,

BC = OB = OC

But, OB and OC are the radii of the circum-circle.

Therefore, BC is also the radius of the circum-circle.

3. Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Solution:

Let’s consider ∆ABC, AB = AC and circle with AB as diameter is drawn which intersects the side BC and D.

And, join AD

Proof:

It’s seen that,

∠ADB = 90^o [Angle in a semi-circle]

And,

∠ADC + ∠ADB = 180^o [Linear pair]

Thus, ∠ADC = 90^o

Now, in right ∆ABD and ∆ACD

AB = AC [Given]

AD = AD [Common]

∠ADB = ∠ADC = 90^o

Hence, by R.H.S criterion of congruence.

∆ABD ≅ ∆ACD

Now, by CPCT

BD = DC

Therefore, D is the mid-point of BC.

4. In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65^o, calculate ∠DEC.

Solution:

Join OE.

Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.

∠EOC = 2∠EBC = 2 x 65^o = 130^o

Now, in ∆OEC

OE = OC [Radii of the same circle]

So, ∠OEC = ∠OCE

But, in ∆EOC by angle sum property

∠OEC + ∠OCE + ∠EOC = 180^o [Angles of a triangle]

∠OCE + ∠OCE + ∠EOC = 180^o

2 ∠OCE + 130^o = 180^o

2 ∠OCE = 180^o – 130^o

∠OCE = 50^o/ 2 = 25^o

And, AC || ED [Given]

∠DEC = ∠OCE [Alternate angles]

Thus,

∠DEC = 25^o

5. The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

Solution:

Let ABCD be a cyclic quadrilateral and PQRS be the quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.

Required to prove: PQRS is a cyclic quadrilateral.

Proof:

By angle sum property of a triangle

In ∆APD,

∠PAD + ∠ADP + ∠APD = 180^o …. (i)

And, in ∆BQC

∠QBC + ∠BCQ + ∠BQC = 180^o …. (ii)

Adding (i) and (ii), we get

∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 180^o + 180^o = 360^o …… (iii)

But,

∠PAD + ∠ADP + ∠QBC + ∠BCQ = ½ [∠A + ∠B + ∠C + ∠D]

= ½ x 360^o = 180^o

Therefore,

∠APD + ∠BQC = 360^o – 180^o = 180^o [From (iii)]

But, these are the sum of opposite angles of quadrilateral PRQS.

Therefore,

Quadrilateral PQRS is also a cyclic quadrilateral.

6. In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:

(i) ∠BDC

(ii) ∠BEC

(iii) ∠BAC

Solution:

(i) Given that BD is a diameter of the circle.

And, the angle in a semicircle is a right angle.

So, ∠BCD = 90°

Also given that,

∠DBC = 58°

In ∆BDC,

∠DBC + ∠BCD + ∠BDC = 180^o

58° + 90° + ∠BDC = 180^o

148^o + ∠BDC = 180^o

∠BDC = 180^o – 148^o

Thus, ∠BDC = 32^o

(ii) We know that, the opposite angles of a cyclic quadrilateral are supplementary.

So, in cyclic quadrilateral BECD

∠BEC + ∠BDC = 180^o

∠BEC + 32^o = 180^o

∠BEC = 148^o

(iii) In cyclic quadrilateral ABEC,

∠BAC + ∠BEC = 180^o [Opposite angles of a cyclic quadrilateral are supplementary]

∠BAC + 148^o = 180^o

∠BAC = 180^o – 148^o

Thus, ∠BAC = 32^o

7. D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.

Solution:

Given,

∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE.

And, DE is joined.

Required to prove: Points B, C, E and D are concyclic

Proof:

In ∆ABC,

AB = AC [Given]

So, ∠B = ∠C [Angles opposite to equal sides]

Similarly,

In ∆ADE,

AD = AE [Given]

So, ∠ADE = ∠AED [Angles opposite to equal sides]

Now, in ∆ABC we have

AD/AB = AE/AC

Hence, DE || BC [Converse of BPT]

So,

∠ADE = ∠B [Corresponding angles]

(180^o – ∠EDB) = ∠B

∠B + ∠EDB = 180^o

But, it’s proved above that

∠B = ∠C

So,

∠C + ∠EDB = 180^o

Thus, opposite angles are supplementary.

Similarly,

∠B + ∠CED = 180^o

Hence, B, C, E and D are concyclic.

8. In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92^o, ∠FAE = 20^o; determine ∠BCD. Given reason in support of your answer.

Solution:

Given,

In cyclic quad. ABCD

AF || CB and DA is produced to E such that ∠ADC = 92^o and ∠FAE = 20^o

So,

∠B + ∠D = 180^o

∠B + 92^o = 180^o

∠B = 88^o

As AF || CB, ∠FAB = ∠B = 88^o

But, ∠FAD = 20^o [Given]

Ext. ∠BAE = ∠BAF + ∠FAE

= 88^o + 22^o = 108^o

But, Ext. ∠BAE = ∠BCD

Therefore,

∠BCD = 108^o

9. If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66^o and ∠ABC = 80^o. Calculate:

(i) ∠DBC,

(ii) ∠IBC,

(iii) ∠BIC

Solution:

Join DB and DC, IB and IC.

Given, if ∠BAC = 66^o and ∠ABC = 80^o, I is the incentre of the ∆ABC.

(i) As it’s seen that ∠DBC and ∠DAC are in the same segment,

So, ∠DBC = ∠DAC

But, ∠DAC = ½ ∠BAC = ½ x 66^o = 33^o

Thus, ∠DBC = 33^o

(ii) And, as I is the incentre of ∆ABC, IB bisects ∠ABC.

Therefore,

∠IBC = ½ ∠ABC = ½ x 80^o = 40^o

(iii) In ∆ABC, by angle sum property

∠ACB = 180^o – (∠ABC + ∠BAC)

∠ACB = 180^o – (80^o + 66^o)

∠ACB = 180^o – 156^o

∠ACB = 34^o

And since, IC bisects ∠C

Thus, ∠ICB = ½ ∠C = ½ x 34^o = 17^o

Now, in ∆IBC

∠IBC + ∠ICB + ∠BIC = 180^o

40^o + 17^o + ∠BIC = 180^o

57^o + ∠BIC = 180^o

∠BIC = 180^o – 57^o

Therefore, ∠BIC = 123^o

10. In the given figure, AB = AD = DC = PB and ∠DBC = x^o. Determine, in terms of x:

(i) ∠ABD, (ii) ∠APB.

Hence or otherwise, prove that AP is parallel to DB.

Solution:

Given, AB = AD = DC = PB and ∠DBC = x^o

Join AC and BD.

Proof:

∠DAC = ∠DBC = x^o [Angles in the same segment]

And, ∠DCA = ∠DAC = x^o [As AD = DC]

Also, we have

∠ABD = ∠DAC [Angles in the same segment]

And, in ∆ABP

Ext. ∠ABD = ∠BAP + ∠APB

But, ∠BAP = ∠APB [Since, AB = BP]

2 x^o = ∠APB + ∠APB = 2∠APB

2∠APB = 2x^o

So, ∠APB = x^o

Thus, ∠APB = ∠DBC = x^o

But these are corresponding angles,

Therefore, AP || DB.

11. In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.

Solution:

Join EB.

Then, in cyclic quad.ABEP

∠APE + ∠ABE = 180^o ….. (i) [Opposite angles of a cyclic quad. are supplementary]

Similarly, in cyclic quad.BCQE

∠CQE + ∠CBE = 180^o ….. (ii) [Opposite angles of a cyclic quad. are supplementary]

Adding (i) and (ii), we have

∠APE + ∠ABE + ∠CQE + ∠CBE = 180^o + 180^o = 360^o

∠APE + ∠ABE + ∠CQE + ∠CBE = 360^o

But, ∠ABE + ∠CBE = 180^o [Linear pair]

∠APE + ∠CQE + 180^o = 360^o

∠APE + ∠CQE = 180^o

Therefore, ∠APE and ∠CQE are supplementary.

12. In the given, AB is the diameter of the circle with centre O.

If ∠ADC = 32^o, find angle BOC.

Solution:

Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle.

Thus, ∠AOC = 2∠ADC

∠AOC = 2 x 32^o = 64^o

As ∠AOC and ∠BOC are linear pair, we have

∠AOC + ∠BOC = 180^o

64^o + ∠BOC = 180^o

∠BOC = 180^o – 64^o

Therefore, ∠BOC = 116^o

The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2023-24 will be updated soon.