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.

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

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

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

**Solution: **

Firstly, letâ€™s join AC.

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

So, âˆ AOC = 180^{o} â€“ 2x

Also,

âˆ BAC = 30^{o} + x

âˆ BCA = 40^{o} + x

Now, in âˆ†ABC

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

= 180^{o} â€“ (30^{o} + x) â€“ (40^{o} + x)

= 110^{o}_{ }â€“ 2x

And, âˆ AOC = 2âˆ ABC

[Angle at the center is double the angle at the circumference subtend by the same chord]180^{o} â€“ 2x = 2(110^{o} â€“ 2x)

2x = 40^{o}

x = 20^{o}

Thus, âˆ AOC = 180^{o} â€“ 2×20^{o} = 140^{o}

**2. In the given figure,Â âˆ BAD = 65Â°,Â âˆ ABD = 70Â°,Â âˆ BDC = 45Â°**

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

**(ii) FindÂ âˆ ACB.**

**Solution:**

(i) In âˆ†ABD,

âˆ DAB + âˆ ABD + âˆ ADB = 180^{o}

65^{o} + 70^{o} + âˆ ADB = 180^{o}

135^{o} + âˆ ADB = 180^{o}

âˆ ADB = 180^{o} – 135^{o} = 45^{o}

Now,

âˆ ADC = âˆ ADB + âˆ BDC = 45^{o} + 45^{o} = 90^{o}

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

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

Hence, âˆ ACB = 45^{o}

**3. Given O is the centre of the circle andÂ âˆ AOB = 70 ^{o}. **

**Calculate the value of:**

**(i)Â âˆ OCA,**

**(ii)Â âˆ OAC.**

**Solution:**

Here, âˆ AOB = 2âˆ ACB

** **[Angle at the center is double the angle at the circumference subtend by the same chord]

âˆ ACB = 70^{o}/ 2 = 35^{o}

Now, OC = OA [Radii of same circle]

Thus,

âˆ OCA = âˆ 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 = Â½ x 130^{o}

Thus, b = 65^{o}

Now,

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

a = 180^{o} â€“ 65^{o} = 115^{o}

(ii) Here, c = Â½ x Reflex (112^{o})

Thus, c = Â½ x (360^{o} â€“ 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, âˆ BAD = 90^{o} [Angle in a semi-circle]

So, âˆ BDA = 90^{o} â€“ 35^{o} = 55^{o}

And,

a = âˆ ACB = âˆ BDA = 55^{o}

(ii) Here, âˆ DAC = âˆ CBD = 25^{o}

And, we have

120^{o} = b + 25^{o}

b = 95^{o}

(iii) âˆ AOB = 2âˆ AOB = 2 x 50^{o} = 100^{o}

Also, OA = OB

âˆ OBA = âˆ OAB = c

c = (180^{o}– 100^{o})/ 2 = 40^{o}

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

âˆ BAP = 90^{o} â€“ 45^{o} = 45^{o}

Now, d = âˆ BCP = âˆ 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. O _{1}Â and O_{2}Â are the centers of two circles.**

**Solution: **

Itâ€™s seen that,

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

So, adding both

âˆ DBA + âˆ 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) âˆ BCD,**

**(ii) âˆ ADC,**

**(iii) âˆ ABC.**

**Show steps of your working.**

**Solution: **

From the given fig, itâ€™s seen that

In cyclic quadrilateral ABCD, DC || AB

And given, âˆ DAB = 105^{o}

(i) So,

âˆ BCD = 180^{o} â€“ 105^{o} = 75^{o}

^{ }[Sum of opposite angles in a cyclic quadrilateral is 180^{o}]

(ii) Now,

âˆ ADC and âˆ DAB are corresponding angles.

So,

âˆ ADC + âˆ DAB = 180^{o}

âˆ ADC = 180^{o} â€“ 105^{o}

Thus,

âˆ ADC = 75^{o}

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

So,

âˆ ADC + âˆ DAB +âˆ BCD + âˆ ABC = 360^{o}

75^{o }+ 105^{o} + 75^{o} + âˆ ABC = 360^{o}

âˆ ABC = 360^{o }â€“ 255^{o}

Thus,

âˆ ABC = 105^{o}

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

**find:**

**(i) âˆ ACB, **

**(ii) âˆ OBC, **

**(iii) âˆ OAB, **

**(iv) âˆ CBA.**

**Solution: **

Given, âˆ AOB = 140^{o} and âˆ OAC = 50^{o}

(i) Now,

âˆ ACB = Â½ Reflex (âˆ AOB) = Â½ (360^{o} â€“ 140^{o}) = 110^{o}

(ii) In quadrilateral OBCA,

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

âˆ OBC + 110^{o} + 50^{o} + 140^{o} = 360^{o}

Thus, âˆ OBC = 360^{o} â€“ 300^{o} = 60^{o}

(iii) In âˆ†AOB, we have

OA = OB (radii)

So, âˆ OBA = âˆ OAB

Hence, by angle sum property of a triangle

âˆ OBA + âˆ OAB + âˆ AOB = 180^{o}

2âˆ OBA + 140^{o} = 180^{o}

2âˆ OBA = 40^{o}

âˆ OBA = 20^{o}

(iv) We already found, âˆ OBC = 60^{o}

And, âˆ OBC = âˆ CBA + âˆ OBA

60^{o} = âˆ CBA + 20^{o}

Therefore,

âˆ CBA = 40^{o}

**9. Calculate: **

**(i) âˆ CDB,**

**(ii) âˆ ABC, **

**(iii) âˆ ACB.**

**Solution: **

Here, we have

âˆ CDB = âˆ BAC = 49^{o}

âˆ ABC = âˆ ADC = 43^{o}

Now, by angle sum property of a triangle we have

âˆ ACB = 180^{o} â€“ 49^{o} â€“ 43^{o} = 88^{o}

**10. In the figure given below, ABCD is a cyclic quadrilateral in whichÂ âˆ BAD = 75 ^{o};Â âˆ ABD = 58^{o}Â andÂ âˆ ADC = 77^{o}. **

**Find:**

**(i) âˆ BDC,**

**(ii) âˆ BCD, **

**(iii) âˆ BCA. **

**Solution: **

(i) By angle sum property of triangle ABD,

âˆ ADB = 180^{o} â€“ 75^{o} â€“ 58^{o} = 47^{o}

Thus, âˆ BDC = âˆ ADC – âˆ ADB = 77^{o} â€“ 47^{o} = 30^{o}

(ii) âˆ BAD + âˆ BCD = 180^{o}

^{o}]

Thus, âˆ BCD = 180^{o} – 75^{o} = 105^{o}

(iii) âˆ BCA = âˆ ADB = 47^{o}

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

**Find:**

**(i) âˆ ADB, (ii) âˆ AEB**

**Solution: **

**(i) **As, itâ€™s seen that** **âˆ ACB and âˆ ADB are in the same segment,

So,

âˆ ADB = 2 âˆ ACB = 60^{o}

(ii) Now, join OA and OB.

And, we have

âˆ AEB = Â½ Reflex (âˆ AOB) = Â½ (360^{o }â€“ 120^{o}) = 120^{o}

** **[Angle at the center is double the angle at the circumference subtend by the same chord]

**12. Given: âˆ CAB = 75 ^{o} and âˆ CBA = 50^{o}. Find the value of âˆ DAB + âˆ ABD. **

**Solution: **

Given, âˆ CAB = 75^{o} and âˆ CBA = 50^{o}

In âˆ†ABC, by angle sum property we have

âˆ ACB = 180^{o} â€“ (âˆ CBA + âˆ CAB)

= 180^{o} â€“ (50^{o} + 75^{o}) = 180^{o} â€“ 125^{o}

= 55^{o}

And,

âˆ ADB = âˆ ACB = 55^{o}

Now, taking âˆ†ABD

âˆ DAB + âˆ ABD + âˆ ADB = 180^{o}

âˆ DAB + âˆ ABD + 55^{o} = 180^{o}

âˆ DAB + âˆ ABD = 180^{o} – 55^{o}

âˆ DAB + âˆ ABD = 125^{o}

**13. ABCD is a cyclic quadrilateral in a circle with centre O. If âˆ ADC = 130 ^{o}, findÂ âˆ BAC.**

**Solution: **

From the fig. its seem that,

âˆ ACB = 90^{o} [Angle in a semi-circle is 90^{o}]

Also,

âˆ ABC = 180^{o} – âˆ ADC = 180^{o} – 130^{o} = 50^{o}

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

âˆ BAC = 90^{o} – âˆ ABC

= 90^{o} â€“ 50^{o}

Thus, âˆ BAC = 40^{o}

**14. In the figure given alongside, AOB is a diameter of the circle andÂ âˆ AOC = 110 ^{o}, findÂ âˆ BDC.**

**Solution: **

Letâ€™s join AD first.

So, we have

âˆ ADC = Â½ âˆ AOC = Â½ x 110^{o} = 55^{o}

Also, we know that

âˆ ADB = 90^{o}

Therefore,

âˆ BDC = 90^{o} – âˆ ADC = 90^{o} â€“ 55^{o}

âˆ BDC = 35^{o}

**15. In the following figure, O is the centre of the circle;** âˆ **AOB = 60 ^{o}Â andÂ **âˆ

**BDC = 100**

^{o}, findÂ âˆ OBC.**Solution: **

Form the figure, we have

âˆ ACB = Â½ âˆ AOB = Â½ x 60^{o} = 30^{o}

Now, by applying angle sum property in âˆ†BDC,

âˆ DBC = 180^{o }– 100^{o} â€“ 30^{o} = 50^{o}

Therefore,

âˆ OBC = 50^{o}

**16. In ABCD is a cyclic quadrilateral in which âˆ DAC = 27 ^{o}, âˆ DBA = 50^{o}Â andÂ âˆ ADB = 33^{o}. Calculate (i)Â âˆ DBC, (ii)Â âˆ DCB, (iii) âˆ CAB.**

**Solution: **

(i) Itâ€™s seen that,

âˆ DBC = âˆ DAC = 27^{o}

(ii) Itâ€™s seen that,

âˆ ACB = âˆ ADB = 33^{o}

And,

âˆ ACD = âˆ ABD = 50^{o}

Thus,

âˆ DCB = âˆ ACD + âˆ ACB = 50^{o} + 33^{o} = 83^{o}

(iii) In quad. ABCD,

âˆ DAB + âˆ DCB = 180^{o}

27^{o} + âˆ CAB + 83^{o} = 180^{o}

Thus,

âˆ CAB = 180^{o} â€“ 110^{o} = 70^{o}

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

**Solution:**

(i) Form the fig. its seen that,

âˆ DCE = 90^{o} – âˆ CDE = 90^{o} â€“ 40^{o} = 50^{o}

Therefore,

âˆ DEC = âˆ OCB = 50^{o}Â

(ii) In âˆ†BOC, we have

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

âˆ OBC = 80^{o}Â – 50^{o}Â = 30^{o}Â [Given âˆ AOC = 80^{o}]

Therefore, âˆ 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, âˆ 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) ifÂ âˆ BAD = 96 ^{o}, findÂ âˆ BCD andÂ âˆ BFE.**

**(ii) Prove that AD is parallel to FE.**

**(b) ABCD is a parallelogram. A circle **

**Solution:**

(i) ABCD is a cyclic quadrilateral

So, âˆ BAD + âˆ BCD = 180^{o}

âˆ BCD = 180^{o} – 96^{o} = 84^{o}

And, âˆ BCE = 180^{o} – 84^{o} = 96^{o}_{ } [Linear pair of angles]

Similarly, BCEF is a cyclic quadrilateral

So, âˆ BCE + âˆ BFE = 180^{o}

âˆ BFE = 180^{o} – 96^{o }= 84^{o}

(ii) Now, âˆ BAD + âˆ 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) Letâ€™s assume that ABCD is a parallelogram which is inscribed in a circle.

So, we have

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

And âˆ BAD + âˆ BCD = 180^{o}

So, 2âˆ BAD = 180^{o}

Thus, âˆ BAD = âˆ 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) Letâ€™s assume that ABCD is a rhombus which is inscribed in a circle.

So, we have

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

And âˆ BAD + âˆ BCD = 180^{o}

So, 2âˆ BAD = 180^{o}

Thus, âˆ BAD = âˆ 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, âˆ B = âˆ C â€¦ (1)

[Angles opposite to equal sides are equal]And, DECB is a cyclic quadrilateral.

So, âˆ B + âˆ DEC = 180^{o}

âˆ C + âˆ DEC = 180^{o} â€¦. (Using 1)

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

DE || BC.

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

Thus, âˆ ADE = âˆ AED

AD = AE

AB â€“ 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 Oâ€™ 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, âˆ AQP = 90^{o} and âˆ BQP = 90^{o}

Now, adding both these angles we get

âˆ AQP + âˆ BQP = 180^{o}

âˆ AQB = 180^{o}

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

**23. The figure given below, shows a circle with centre O. Given:Â **âˆ **AOC = a andÂ **âˆ **ABC = b.**

**(i) Find the relationship between a and b**

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

**Solution:**

(i) Itâ€™s seen that,

âˆ ABC = Â½ Reflex (âˆ COA)

[Angle at the centre is double the angle at the circumference subtended by the same chord]So, b = Â½ (360^{o} – a)

a + 2b = 180^{o} â€¦.. (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

âˆ COA + âˆ OAB = 180^{o}

60^{o} + âˆ OAB = 180^{o}

Therefore,

âˆ 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: âˆ AOC + âˆ BOD = 2âˆ APC

OA, OB, OC and OD are joined.

Also, AD is joined.

Now, itâ€™s seen that

âˆ AOC = 2âˆ ADC â€¦. (1)

[Angle at the centre is double the angle at the circumference subtended by the same chord]Similarly,

âˆ BOD = 2âˆ BAD â€¦. (2)

Adding (1) and (2), we have

âˆ AOC + âˆ BOD = 2âˆ ADC + 2âˆ BAD

= 2(âˆ ADC + âˆ BAD) â€¦.. (3)

And in âˆ†PAD,

Ext. âˆ APC = âˆ PAD + âˆ ADC

= âˆ BAD + âˆ ADC â€¦. (4)

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

âˆ AOC + âˆ BOD = 2âˆ APC

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

**Calculate: (i)Â âˆ RNM; (ii)Â âˆ NRM.**

**Solution:**

(i) Join RN and MS

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

So, by angle sum property of âˆ†RMS

âˆ RMS = 90^{o} â€“ 29^{o} = 61^{o}

And,

âˆ RNM = 180^{o} – âˆ RSM = 180^{o}Â â€“ 61^{o} = 119^{o}

(ii) Now as RS || NM,

âˆ NMR = âˆ MRS = 29^{o} [Alternate angles]

âˆ NMS = 90^{o} + 29^{o} = 119^{o}

Also, we know that

âˆ NRS + âˆ NMS = 180^{o}

âˆ NRM + 29^{o} + 119^{o} = 180^{o}

âˆ NRM = 180^{o} â€“ 148^{o}

Therefore,

âˆ NRM = 32^{o}

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

**Solution:**

Join AC and BD.

So, we have

âˆ CAD = 90^{o}Â and âˆ CBD = 90^{o}

And, AB || CD

So, âˆ BAD = âˆ ADC = 25^{o} [Alternate angles]

âˆ BAC = âˆ BAD + âˆ CAD = 25^{o} + 90^{o} = 115^{o}

Thus,

âˆ ADB = 180^{o} â€“ 25^{o} – âˆ BAC = 180^{o} â€“ 25^{o} â€“ 115^{o} = 40^{o}

Finally,

âˆ AEB = âˆ 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:**

Letâ€™s join AC, PQ and BD.

As ACQP is a cyclic quadrilateral

âˆ CAP + âˆ PQC = 180^{o} â€¦â€¦. (i)

Similarly, as PQDB is a cyclic quadrilateral

âˆ PQD + âˆ DBP = 180^{o} â€¦â€¦. (ii)

Again, âˆ PQC + âˆ PQD = 180^{o} â€¦â€¦ (iii) [Linear pair of angles]

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

âˆ CAP + âˆ DBP = 180^{o}

Or âˆ CAB + âˆ 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:**

Letâ€™s assume that ABCD be the given cyclic quadrilateral.

Also, PA = PD [Given]

So, âˆ PAD = âˆ PDA â€¦â€¦ (1)

[Angles opposite to equal sides are equal]And,

âˆ BAD = 180^{o} – âˆ PAD [Linear pair of angles]

Similarly,

âˆ CDA = 180^{o} – âˆ PDA = 180^{o} – âˆ PAD [From (1)]

As the opposite angles of a cyclic quadrilateral are supplementary,

âˆ ABC = 180^{o} – âˆ CDA = 180^{o} â€“ (180^{o} – âˆ PAD) = âˆ PAD

And, âˆ DCB = 180^{o} – âˆ BAD = 180^{o} â€“ (180^{o} – âˆ PAD) = âˆ PAD

Thus,

âˆ ABC = âˆ DCB = âˆ PAD = âˆ PDA

Which is only possible when AD || BC.