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 BYJUâ€™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

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Exercise 17(A) Page No: 257

**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}

^{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}

**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}

(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.

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:

Itâ€™s seen that chord AD subtends âˆ ABD and chord BC subtends âˆ BDC at the circumference of the circle.

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

So, Chord AD must be equal to chord BC

AD = BC

Now, in âˆ†ADC and âˆ†BCD

DC = DC [Common]

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

AD = BC [Proved above]

Hence, by SAS criterion of congruence

âˆ†ADC â‰… âˆ†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 CDÂ areÂ equal. IfÂ âˆ DEF = 110 ^{o}, calculate:**

**(i)Â âˆ AFE, (ii)Â âˆ FAB.**

**Solution: **

Join AE, OB and OC.

(i) As AOD is the diameter

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

But, given âˆ DEF = 110^{o}

So,

âˆ AEF = âˆ DEF – âˆ AED = 110^{o} â€“ 90^{o} = 20^{o}

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

So,

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

But,

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

Thus,

âˆ AOB = âˆ BOC = âˆ COD = 60^{o}

Now, in âˆ†OAB we have

OA = OB [Radii of same circle]

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

But, by angle sum property of âˆ†OAB

âˆ OAB + âˆ OBA = 180^{o} – âˆ AOB

= 180^{o} â€“ 60^{o}

= 120^{o}

Therefore, âˆ OAB = âˆ OBA = 60^{o}

Now, in cyclic quadrilateral ADEF

âˆ DEF + âˆ DAF = 180^{o}

âˆ DAF = 180^{o} – âˆ DEF

= 180^{o} â€“ 110^{o}

= 70^{o}

Thus,

âˆ FAB = âˆ DAF + âˆ OAB

= 70^{o} + 60^{o} = 130^{o}

**3. If two sides of aÂ cycli-quadrilateral are parallel; proveÂ that:**

**(i)Â itsÂ other two sides are equal.**

**(ii)Â itsÂ 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)

âˆ DCA = âˆ CAB [Alternate angles]

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

So,

âˆ DCA = âˆ CAB

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

(ii) Now, in âˆ†ABC and âˆ†ADB

AB = AB [Common]

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

BC = AD [Proved above]

Hence, by SAS criterion of congruence

âˆ†ACB â‰… âˆ†ADB

Therefore, by CPCT

AC = BD

**4. The given figure show a circle with centre O. Also, PQ = QR = RS andÂ âˆ PTS = 75Â°. **

**Calculate:**

**(i)Â âˆ POS,**

**(ii)Â âˆ QOR,**

**(iii)Â âˆ PQR.**

**Solution: **

Join OP, OQ, OR and OS.

Given, PQ = QR = RS

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

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

Thus,

âˆ POS = 2 x âˆ PTS = 2 x 75^{o} = 150^{o}

âˆ POQ + âˆ QOR + âˆ ROS = 150^{o}

âˆ POQ = âˆ QOR = âˆ ROS = 150^{o}/ 3 = 50^{o }

In âˆ†OPQ we have,

OP = OQ [Radii of the same circle]

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

But, by angle sum property of âˆ†OPQ

âˆ OPQ + âˆ OQP + âˆ POQ = 180^{o}

âˆ OPQ + âˆ OQP + 50^{o} = 180^{o}

âˆ OPQ + âˆ OQP = 130^{o}

2 âˆ OPQ = 130^{o}

âˆ OPQ = âˆ OPQ = 130^{o}/ 2 = 65^{o}

Similarly, we can prove that

In âˆ†OQR,

âˆ OQR = âˆ ORQ = 65^{o}

And in âˆ†ORS,

âˆ ORS = âˆ OSR = 65^{o}

Hence,

(i) âˆ POS = 150^{o}

(ii) âˆ QOR = 50^{o} and

(iii) âˆ PQR = âˆ PQO + âˆ 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)Â **âˆ **AOB,**

**(ii)Â **âˆ **ACB,**

**(iii)Â **âˆ **ABC.**

**Solution:**

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

âˆ ACB = Â½ âˆ AOB

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

âˆ AOB = 60^{o}

(ii) Now,

âˆ ACB = Â½ (60^{o}) = 30^{o}

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

âˆ AOC = 360^{o}/ 8 = 45^{o}

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

âˆ ABC = Â½ âˆ AOC

âˆ 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,

âˆ ABD = âˆ ACD = 30^{o} [Angles in the same segment]

In âˆ†ADB, by angle sum property we have

âˆ BAD + âˆ ADB + âˆ ABD = 180^{o}

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

âˆ 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 whichÂ âˆ 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:

âˆ BOC = 2âˆ BAC = 2 x 30^{o} = 60^{o}

Now, in âˆ†OBC

OB = OC [Radii of same circle]

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

^{}

And in âˆ†OBC, by angle sum property we have

âˆ OBC + âˆ OCB + âˆ BOC = 180^{o}

âˆ OBC + âˆ OBC + 60^{o} = 180^{o}

2 âˆ OBC = 180^{o} â€“ 60^{o}Â = 120^{o}

âˆ OBC = 120^{o}/ 2 = 60^{o}

So, âˆ OBC = âˆ OCB = âˆ BOC = 60^{o}

Thus, âˆ†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: **

Letâ€™s consider âˆ†ABC, AB = AC and circle with AB as diameter is drawn which intersects the side BC and D.

And, join AD

Proof:

Itâ€™s seen that,

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

And,

âˆ ADC + âˆ ADB = 180^{o} [Linear pair]

Thus, âˆ ADC = 90^{o}

Now, in right âˆ†ABD and âˆ†ACD

AB = AC [Given]

AD = AD [Common]

âˆ ADB = âˆ ADC = 90^{o}

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

âˆ†ABD â‰… âˆ†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. GivenÂ âˆ CBE = 65 ^{o}, calculate âˆ DEC.**

**Solution: **

Join OE.

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

âˆ EOC = 2âˆ EBC = 2 x 65^{o} = 130^{o}

Now, in âˆ†OEC

OE = OC [Radii of the same circle]

So, âˆ OEC = âˆ OCE

But, in âˆ†EOC by angle sum property

âˆ OEC + âˆ OCE + âˆ EOC = 180^{o} [Angles of a triangle]

âˆ OCE + âˆ OCE + âˆ EOC = 180^{o}

2 âˆ OCE + 130^{o} = 180^{o}

2 âˆ OCE = 180^{o} – 130^{o}

âˆ OCE = 50^{o}/ 2 = 25^{o}

And, AC || ED [Given]

âˆ DEC = âˆ OCE [Alternate angles]

Thus,

âˆ 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 âˆ A, âˆ B, âˆ C and âˆ D.

Required to prove: PQRS is a cyclic quadrilateral.

Proof:

By angle sum property of a triangle

In âˆ†APD,

âˆ PAD + âˆ ADP + âˆ APD = 180^{o} â€¦. (i)

And, in âˆ†BQC

âˆ QBC + âˆ BCQ + âˆ BQC = 180^{o} â€¦. (ii)

Adding (i) and (ii), we get

âˆ PAD + âˆ ADP + âˆ APD + âˆ QBC + âˆ BCQ + âˆ BQC = 180^{o} + 180^{o} = 360^{o} â€¦â€¦ (iii)

But,

âˆ PAD + âˆ ADP + âˆ QBC + âˆ BCQ = Â½ [âˆ A + âˆ B + âˆ C + âˆ D]

= Â½ x 360^{o} = 180^{o}

Therefore,

âˆ APD + âˆ 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,Â âˆ DBC = 58Â°. BD is a diameter of the circle. Calculate:**

**(i)Â âˆ BDC**

**(ii)Â âˆ BEC**

**(iii)Â âˆ BAC**

**Solution: **

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

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

So, âˆ BCD = 90Â°

Also given that,

âˆ DBC = 58Â°

In âˆ†BDC,

âˆ DBC + âˆ BCD + âˆ BDC = 180^{o}

58Â° + 90Â° + âˆ BDC = 180^{o}

148^{o} + âˆ BDC = 180^{o}

âˆ BDC = 180^{o} – 148^{o}

Thus, âˆ BDC = 32^{o}

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

So, in cyclic quadrilateral BECD

âˆ BEC + âˆ BDC = 180^{o}

âˆ BEC + 32^{o} = 180^{o}

âˆ BEC = 148^{o}

(iii) In cyclic quadrilateral ABEC,

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

âˆ BAC + 148^{o} = 180^{o}

âˆ BAC = 180^{o} – 148^{o}

Thus, âˆ 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,

âˆ†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 âˆ†ABC,

AB = AC [Given]

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

Similarly,

In âˆ†ADE,

AD = AE [Given]

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

Now, in âˆ†ABC we have

AD/AB = AE/AC

Hence, DE || BC [Converse of BPT]

So,

âˆ ADE = âˆ B [Corresponding angles]

(180^{o} – âˆ EDB) = âˆ B

âˆ B + âˆ EDB = 180^{o}

But, itâ€™s proved above that

âˆ B = âˆ C

So,

âˆ C + âˆ EDB = 180^{o}

Thus, opposite angles are supplementary.

Similarly,

âˆ B + âˆ 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. IfÂ **âˆ **ADC = 92 ^{o}, **âˆ

**FAE = 20**âˆ

^{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 âˆ ADC = 92^{o} and âˆ FAE = 20^{o}

So,

âˆ B + âˆ D = 180^{o}

âˆ B + 92^{o} = 180^{o}

âˆ B = 88^{o}

As AF || CB, âˆ FAB = âˆ B = 88^{o}

But, âˆ FAD = 20^{o} [Given]

Ext. âˆ BAE = âˆ BAF + âˆ FAE

= 88^{o} + 22^{o} = 108^{o}

But, Ext. âˆ BAE = âˆ BCD

Therefore,

âˆ 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 âˆ BAC = 66 ^{o }andÂ âˆ ABC =Â 80^{o}. Calculate:**

**(i) âˆ DBC,**

**(ii) âˆ IBC, **

**(iii) âˆ BIC**

**Solution:**

Join DB and DC, IB and IC.

Given, if âˆ BAC = 66^{o }andÂ âˆ ABC =Â 80^{o}, I is the incentre of the âˆ†ABC.

(i) As itâ€™s seen that âˆ DBC and âˆ DAC are in the same segment,

So, âˆ DBC = âˆ DAC

But, âˆ DAC = Â½ âˆ BAC = Â½ x 66^{o} = 33^{o}

Thus, âˆ DBC = 33^{o}

(ii) And, as I is the incentre of âˆ†ABC, IB bisects âˆ ABC.

Therefore,

âˆ IBC = Â½ âˆ ABC = Â½ x 80^{o }= 40^{o}

(iii) In âˆ†ABC, by angle sum property

âˆ ACB = 180^{o} â€“ (âˆ ABC + âˆ BAC)

âˆ ACB = 180^{o} â€“ (80^{o} + 66^{o})

âˆ ACB = 180^{o} â€“ 156^{o}

âˆ ACB = 34^{o}

And since, IC bisects âˆ C

Thus, âˆ ICB = Â½ âˆ C = Â½ x 34^{o} = 17^{o}

Now, in âˆ†IBC

âˆ IBC + âˆ ICB + âˆ BIC = 180^{o}

40^{o} + 17^{o} + âˆ BIC = 180^{o}

57^{o} + âˆ BIC = 180^{o}

âˆ BIC = 180^{o} â€“ 57^{o}

Therefore, âˆ BIC = 123^{o}

**10. In the given figure, AB = AD = DC = PB andÂ âˆ DBC = x ^{o}. Determine, in terms of x:**

**(i) âˆ ABD, (ii) âˆ APB.**

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

**Solution: **

Given, AB = AD = DC = PB andÂ âˆ DBC = x^{o}

Join AC and BD.

Proof:

âˆ DAC = âˆ DBC = x^{o} [Angles in the same segment]

And, âˆ DCA = âˆ DAC = x^{o} [As AD = DC]

Also, we have

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

And, in âˆ†ABP

Ext. âˆ ABD = âˆ BAP + âˆ APB

But, âˆ BAP = âˆ APB [Since, AB = BP]

2 x^{o} = âˆ APB + âˆ APB = 2âˆ APB

2âˆ APB = 2x^{o}

So, âˆ APB = x^{o}

Thus, âˆ APB = âˆ 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 âˆ APE and âˆ CQE are supplementary.**

**Solution: **

Join EB.

Then, in cyclic quad.ABEP

âˆ APE + âˆ ABE = 180^{o} â€¦.. (i) [Opposite angles of a cyclic quad. are supplementary]

Similarly, in cyclic quad.BCQE

âˆ CQE + âˆ CBE = 180^{o} â€¦.. (ii) [Opposite angles of a cyclic quad. are supplementary]

Adding (i) and (ii), we have

âˆ APE + âˆ ABE + âˆ CQE + âˆ CBE = 180^{o }+ 180^{o} = 360^{o}

âˆ APE + âˆ ABE + âˆ CQE + âˆ CBE = 360^{o}

But, âˆ ABE + âˆ CBE = 180^{o} [Linear pair]

âˆ APE + âˆ CQE + 180^{o} = 360^{o}

âˆ APE + âˆ CQE = 180^{o}

Therefore, âˆ APE and âˆ CQE are supplementary.

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

**If âˆ ADC =Â 32 ^{o}, find angle BOC.**

**Solution: **

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

Thus, âˆ AOC = 2âˆ ADC

âˆ AOC = 2 x 32^{o} = 64^{o}

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

âˆ AOC + âˆ BOC = 180^{o}

64^{o} + âˆ BOC = 180^{o}

âˆ BOC = 180^{o} â€“ 64^{o}

Therefore, âˆ BOC = 116^{o}

*The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2020-21 will be updated soon.*