Selina Solutions Concise Maths Class 10 Chapter 17 Circles Exercise 17(A)

Fundamentals of circles and relation between its parts like segment, arc, chord are discussed in these exercise problems. The cyclic properties are also covered in this exercise. The Selina Solutions for Class 10 Maths is one valuable resource a student can have when solving problems of the book. It helps clear doubts instantly and improves problem-solving skills of the students. The Concise Selina Solutions for Class 10 Maths Chapter 17 Circles Exercise 17(A) PDF, is provided in the link given below.

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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.