The problems of this exercise strengthen the main concepts of the chapter. The Selina Solutions for Class 10 Maths is a one-stop solution to all the doubts and clarifications pertaining to solving problems of the book. The solutions PDF of the Concise Selina Solutions for Class 10 Maths Chapter 17 Circles Exercise 17(C) is available in the link given below.

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

**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}