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

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

Solution:

Now,

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

In âˆ†ADB, by angle sum property we have

âˆ BAD + âˆ ADB + âˆ ABD = 180o

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

âˆ ADB = 90o

So,

x + 90o + 30o = 180o

x = 180o – 120o

Hence, x = 60o

2. In the given figure, ABC is a triangle in whichÂ âˆ BAC = 30o. 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 30o = 60o

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 = 180o

âˆ OBC + âˆ OBC + 60o = 180o

2 âˆ OBC = 180o â€“ 60oÂ = 120o

âˆ OBC = 120o/ 2 = 60o

So, âˆ OBC = âˆ OCB = âˆ BOC = 60o

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.

Proof:

Itâ€™s seen that,

âˆ ADB = 90o [Angle in a semi-circle]

And,

âˆ ADC + âˆ ADB = 180o [Linear pair]

Thus, âˆ ADC = 90o

Now, in right âˆ†ABD and âˆ†ACD

AB = AC [Given]

âˆ ADB = âˆ ADC = 90o

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 = 65o, 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 65o = 130o

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 = 180o [Angles of a triangle]

âˆ OCE + âˆ OCE + âˆ EOC = 180o

2 âˆ OCE + 130o = 180o

2 âˆ OCE = 180o – 130o

âˆ OCE = 50o/ 2 = 25o

And, AC || ED [Given]

âˆ DEC = âˆ OCE [Alternate angles]

Thus,

âˆ DEC = 25o

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 = 180o â€¦. (i)

And, in âˆ†BQC

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

Adding (i) and (ii), we get

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

But,

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

= Â½ x 360o = 180o

Therefore,

âˆ APD + âˆ BQC = 360o – 180o = 180o [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 = 180o

58Â° + 90Â° + âˆ BDC = 180o

148o + âˆ BDC = 180o

âˆ BDC = 180o – 148o

Thus, âˆ BDC = 32o

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

So, in cyclic quadrilateral BECD

âˆ BEC + âˆ BDC = 180o

âˆ BEC + 32o = 180o

âˆ BEC = 148o

(iii) In cyclic quadrilateral ABEC,

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

âˆ BAC + 148o = 180o

âˆ BAC = 180o – 148o

Thus, âˆ BAC = 32o

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,

AD = AE [Given]

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

Now, in âˆ†ABC we have

Hence, DE || BC [Converse of BPT]

So,

âˆ ADE = âˆ B [Corresponding angles]

(180o – âˆ EDB) = âˆ B

âˆ B + âˆ EDB = 180o

But, itâ€™s proved above that

âˆ B = âˆ C

So,

âˆ C + âˆ EDB = 180o

Thus, opposite angles are supplementary.

Similarly,

âˆ B + âˆ CED = 180o

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 = 92o, âˆ FAE = 20o; determine âˆ 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 = 92o and âˆ FAE = 20o

So,

âˆ B + âˆ D = 180o

âˆ B + 92o = 180o

âˆ B = 88o

As AF || CB, âˆ FAB = âˆ B = 88o

But, âˆ FAD = 20o [Given]

Ext. âˆ BAE = âˆ BAF + âˆ FAE

= 88o + 22o = 108o

But, Ext. âˆ BAE = âˆ BCD

Therefore,

âˆ BCD = 108o

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 = 66o andÂ âˆ ABC =Â 80o. Calculate:

(i) âˆ DBC,

(ii) âˆ IBC,

(iii) âˆ BIC

Solution:

Join DB and DC, IB and IC.

Given, if âˆ BAC = 66o andÂ âˆ ABC =Â 80o, 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 66o = 33o

Thus, âˆ DBC = 33o

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

Therefore,

âˆ IBC = Â½ âˆ ABC = Â½ x 80o = 40o

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

âˆ ACB = 180o â€“ (âˆ ABC + âˆ BAC)

âˆ ACB = 180o â€“ (80o + 66o)

âˆ ACB = 180o â€“ 156o

âˆ ACB = 34o

And since, IC bisects âˆ C

Thus, âˆ ICB = Â½ âˆ C = Â½ x 34o = 17o

Now, in âˆ†IBC

âˆ IBC + âˆ ICB + âˆ BIC = 180o

40o + 17o + âˆ BIC = 180o

57o + âˆ BIC = 180o

âˆ BIC = 180o â€“ 57o

Therefore, âˆ BIC = 123o

10. In the given figure, AB = AD = DC = PB andÂ âˆ DBC = xo. 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 = xo

Join AC and BD.

Proof:

âˆ DAC = âˆ DBC = xo [Angles in the same segment]

And, âˆ DCA = âˆ DAC = xo [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 xo = âˆ APB + âˆ APB = 2âˆ APB

2âˆ APB = 2xo

So, âˆ APB = xo

Thus, âˆ APB = âˆ DBC = xo

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 = 180o â€¦.. (i) [Opposite angles of a cyclic quad. are supplementary]

Similarly, in cyclic quad.BCQE

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

Adding (i) and (ii), we have

âˆ APE + âˆ ABE + âˆ CQE + âˆ CBE = 180o + 180o = 360o

âˆ APE + âˆ ABE + âˆ CQE + âˆ CBE = 360o

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

âˆ APE + âˆ CQE + 180o = 360o

âˆ APE + âˆ CQE = 180o

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 32o, 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 32o = 64o

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

âˆ AOC + âˆ BOC = 180o

64o + âˆ BOC = 180o

âˆ BOC = 180o â€“ 64o

Therefore, âˆ BOC = 116o