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

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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,

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

In ∆ADB, by angle sum property we have

∠BAD + ∠ADB + ∠ABD = 180^o

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

∠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 which ∠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:

∠BOC = 2∠BAC = 2 x 30^o = 60^o

Now, in ∆OBC

OB = OC [Radii of same circle]

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

And in ∆OBC, by angle sum property we have

∠OBC + ∠OCB + ∠BOC = 180^o

∠OBC + ∠OBC + 60^o = 180^o

2 ∠OBC = 180^o – 60^o = 120^o

∠OBC = 120^o/ 2 = 60^o

So, ∠OBC = ∠OCB = ∠BOC = 60^o

Thus, ∆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:

Let’s consider ∆ABC, AB = AC and circle with AB as diameter is drawn which intersects the side BC and D.

And, join AD

Proof:

It’s seen that,

∠ADB = 90^o [Angle in a semi-circle]

And,

∠ADC + ∠ADB = 180^o [Linear pair]

Thus, ∠ADC = 90^o

Now, in right ∆ABD and ∆ACD

AB = AC [Given]

AD = AD [Common]

∠ADB = ∠ADC = 90^o

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

∆ABD ≅ ∆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. Given ∠CBE = 65^o, calculate ∠DEC.

Solution:

Join OE.

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

∠EOC = 2∠EBC = 2 x 65^o = 130^o

Now, in ∆OEC

OE = OC [Radii of the same circle]

So, ∠OEC = ∠OCE

But, in ∆EOC by angle sum property

∠OEC + ∠OCE + ∠EOC = 180^o [Angles of a triangle]

∠OCE + ∠OCE + ∠EOC = 180^o

2 ∠OCE + 130^o = 180^o

2 ∠OCE = 180^o – 130^o

∠OCE = 50^o/ 2 = 25^o

And, AC || ED [Given]

∠DEC = ∠OCE [Alternate angles]

Thus,

∠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 ∠A, ∠B, ∠C and ∠D.

Required to prove: PQRS is a cyclic quadrilateral.

Proof:

By angle sum property of a triangle

In ∆APD,

∠PAD + ∠ADP + ∠APD = 180^o …. (i)

And, in ∆BQC

∠QBC + ∠BCQ + ∠BQC = 180^o …. (ii)

Adding (i) and (ii), we get

∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 180^o + 180^o = 360^o …… (iii)

But,

∠PAD + ∠ADP + ∠QBC + ∠BCQ = ½ [∠A + ∠B + ∠C + ∠D]

= ½ x 360^o = 180^o

Therefore,

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

(i) ∠BDC

(ii) ∠BEC

(iii) ∠BAC

Solution:

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

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

So, ∠BCD = 90°

Also given that,

∠DBC = 58°

In ∆BDC,

∠DBC + ∠BCD + ∠BDC = 180^o

58° + 90° + ∠BDC = 180^o

148^o + ∠BDC = 180^o

∠BDC = 180^o – 148^o

Thus, ∠BDC = 32^o

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

So, in cyclic quadrilateral BECD

∠BEC + ∠BDC = 180^o

∠BEC + 32^o = 180^o

∠BEC = 148^o

(iii) In cyclic quadrilateral ABEC,

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

∠BAC + 148^o = 180^o

∠BAC = 180^o – 148^o

Thus, ∠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,

∆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 ∆ABC,

AB = AC [Given]

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

Similarly,

In ∆ADE,

AD = AE [Given]

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

Now, in ∆ABC we have

AD/AB = AE/AC

Hence, DE || BC [Converse of BPT]

So,

∠ADE = ∠B [Corresponding angles]

(180^o – ∠EDB) = ∠B

∠B + ∠EDB = 180^o

But, it’s proved above that

∠B = ∠C

So,

∠C + ∠EDB = 180^o

Thus, opposite angles are supplementary.

Similarly,

∠B + ∠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. If ∠ADC = 92^o, ∠FAE = 20^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 ∠ADC = 92^o and ∠FAE = 20^o

So,

∠B + ∠D = 180^o

∠B + 92^o = 180^o

∠B = 88^o

As AF || CB, ∠FAB = ∠B = 88^o

But, ∠FAD = 20^o [Given]

Ext. ∠BAE = ∠BAF + ∠FAE

= 88^o + 22^o = 108^o

But, Ext. ∠BAE = ∠BCD

Therefore,

∠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 ∠BAC = 66^o and ∠ABC = 80^o. Calculate:

(i) ∠DBC,

(ii) ∠IBC,

(iii) ∠BIC

Solution:

Join DB and DC, IB and IC.

Given, if ∠BAC = 66^o and ∠ABC = 80^o, I is the incentre of the ∆ABC.

(i) As it’s seen that ∠DBC and ∠DAC are in the same segment,

So, ∠DBC = ∠DAC

But, ∠DAC = ½ ∠BAC = ½ x 66^o = 33^o

Thus, ∠DBC = 33^o

(ii) And, as I is the incentre of ∆ABC, IB bisects ∠ABC.

Therefore,

∠IBC = ½ ∠ABC = ½ x 80^o = 40^o

(iii) In ∆ABC, by angle sum property

∠ACB = 180^o – (∠ABC + ∠BAC)

∠ACB = 180^o – (80^o + 66^o)

∠ACB = 180^o – 156^o

∠ACB = 34^o

And since, IC bisects ∠C

Thus, ∠ICB = ½ ∠C = ½ x 34^o = 17^o

Now, in ∆IBC

∠IBC + ∠ICB + ∠BIC = 180^o

40^o + 17^o + ∠BIC = 180^o

57^o + ∠BIC = 180^o

∠BIC = 180^o – 57^o

Therefore, ∠BIC = 123^o

10. In the given figure, AB = AD = DC = PB and ∠DBC = x^o. Determine, in terms of x:

(i) ∠ABD, (ii) ∠APB.

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

Solution:

Given, AB = AD = DC = PB and ∠DBC = x^o

Join AC and BD.

Proof:

∠DAC = ∠DBC = x^o [Angles in the same segment]

And, ∠DCA = ∠DAC = x^o [As AD = DC]

Also, we have

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

And, in ∆ABP

Ext. ∠ABD = ∠BAP + ∠APB

But, ∠BAP = ∠APB [Since, AB = BP]

2 x^o = ∠APB + ∠APB = 2∠APB

2∠APB = 2x^o

So, ∠APB = x^o

Thus, ∠APB = ∠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 ∠APE and ∠CQE are supplementary.

Solution:

Join EB.

Then, in cyclic quad.ABEP

∠APE + ∠ABE = 180^o ….. (i) [Opposite angles of a cyclic quad. are supplementary]

Similarly, in cyclic quad.BCQE

∠CQE + ∠CBE = 180^o ….. (ii) [Opposite angles of a cyclic quad. are supplementary]

Adding (i) and (ii), we have

∠APE + ∠ABE + ∠CQE + ∠CBE = 180^o + 180^o = 360^o

∠APE + ∠ABE + ∠CQE + ∠CBE = 360^o

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

∠APE + ∠CQE + 180^o = 360^o

∠APE + ∠CQE = 180^o

Therefore, ∠APE and ∠CQE are supplementary.

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

If ∠ADC = 32^o, find angle BOC.

Solution:

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

Thus, ∠AOC = 2∠ADC

∠AOC = 2 x 32^o = 64^o

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

∠AOC + ∠BOC = 180^o

64^o + ∠BOC = 180^o

∠BOC = 180^o – 64^o

Therefore, ∠BOC = 116^o