Selina Solutions Concise Maths Class 10 Chapter 18 Tangents and Intersecting Chords Exercise 18(C)

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Access Selina Solutions Concise Maths Class 10 Chapter 18 Tangents and Intersecting Chords Exercise 18(C)

1. Prove that, of any two chords of a circle, the greater chord is nearer to the center.

Solution:

Given: A circle with center O and radius r. AB and CD are two chords such that AB > CD. Also, OM ⊥ AB and ON ⊥ CD.

Required to prove: OM < ON

Proof:

Join OA and OC.

Then in right ∆AOM, we have

AO² = AM² + OM²

r² = (½AB)² + OM²

r² = ¼ AB² + OM² ….. (i)

Again, in right ∆ONC, we have

OC² = NC² + ON²

r² = (½CD)² + ON²

r² = ¼ CD² + ON² ….. (ii)

On equating (i) and (ii), we get

¼ AB² + OM² = ¼ CD² + ON²

But, AB > CD [Given]

So, ON will be greater than OM to be equal on both sides.

Thus,

OM < ON

Hence, AB is nearer to the centre than CD.

2. OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.

i) If the radius of the circle is 10 cm, find the area of the rhombus.

ii) If the area of the rhombus is 32√3 cm², find the radius of the circle.

Solution:

(i) Given, radius = 10 cm

In rhombus OABC,

OC = 10 cm

So,

OE = ½ x OB = ½ x 10 = 5 cm

Now, in right ∆OCE

OC² = OE² + EC²

10² = 5² + EC²

EC² = 100 – 25 = 75

EC = √75 = 5√3

Hence, AC = 2 x EC = 2 x 5√3 = 10√3

We know that,

Area of rhombus = ½ x OB x AC

= ½ x 10 x 10√3

= 50√3 cm² ≈ 86.6 cm²

(ii) We have the area of rhombus = 32√3 cm²

But area of rhombus OABC = 2 x area of ∆OAB

Area of rhombus OABC = 2 x (√3/4) r²

Where r is the side of the equilateral triangle OAB.

2 x (√3/4) r² = 32√3

√3/2 r² = 32√3

r² = 64

r = 8

Therefore, the radius of the circle is 8 cm.

3. Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.

Solution:

We know that,

If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, AB = (5 – 3) cm = 2 cm.

Also, the common chord PQ is the perpendicular bisector of AB.

Thus, AC = CB = ½ AB = 1 cm

In right ∆ACP, we have

AP² = AC² + CP² [Pythagoras Theorem]

5² = 1² + CP²

CP² = 25 – 1 = 24

CP = √24 cm = 2√6 cm

Now,

PQ = 2 CP

= 2 x 2√6 cm

= 4√6 cm

Therefore, the length of PQ is 4√6 cm.

4. Two chords AB and AC of a circle are equal. Prove that the center of the circle, lies on the bisector of the angle BAC.

Solution:

Given: AB and AC are two equal chords of C (O, r).

Required to prove: Centre, O lies on the bisector of ∠BAC.

Construction: Join BC. Let the bisector of ∠BAC intersects BC in P.

Proof:

In ∆APB and ∆APC,

AB = AC [Given]

∠BAP = ∠CAP [Given]

AP = AP [Common]

Hence, ∆APB ≅ ∆APC by SAA congruence criterion

So, by CPCT we have

BP = CP and ∠APB = ∠APC

And,

∠APB + ∠APC = 180
  [Linear pair]

2∠APB = 180
  [∠APB = ∠APC]

∠APB = 90

Now, BP = CP and ∠APB = 90

Therefore, AP is the perpendicular bisector of chord BC.

Hence, AP passes through the centre, O of the circle.

5. The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the center of the circle?

Solution:

We have, AB as the diameter and AC as the chord.

Now, draw OL ⊥ AC

Since OL ⊥ AC and hence it bisects AC, O is the centre of the circle.

Therefore, OA = 10 cm and AL = 6 cm

Now, in right ∆OLA

AO² = AL² + OL² [By Pythagoras Theorem]

10² = 6² + OL²

OL² = 100 – 36 = 64

OL = 8 cm

Therefore, the chord is at a distance of 8 cm from the centre of the circle.

6. ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110^o and angle BAC = 50^o. Find angle DAC and angle DCA.

Solution:

Given, ABCD is a cyclic quadrilateral in which AD || BC

And, ∠ADC = 110^o, ∠BAC = 50^o

We know that,

∠B + ∠D = 180^o [Sum of opposite angles of a quadrilateral]

∠B + 110^o = 180^o

So, ∠B = 70^o

Now in ∆ADC, we have

∠BAC + ∠ABC + ∠ACB = 180^o

50^o + 70^o + ∠ACB = 180^o

∠ACB = 180^o – 120^o = 60^o

And, as AD || BC we have

∠DAC = ∠ACB = 60^o [Alternate angles]

Now in ∆ADC,

∠DAC + ∠ADC + ∠DCA = 180^o

60^o + 110^o + ∠DCA = 180^o

Thus,

∠DCA = 180^o – 170^o = 10^o 

7. In the given figure, C and D are points on the semi-circle described on AB as diameter.

Given angle BAD = 70^o and angle DBC = 30^o, calculate angle BDC.

Solution:

As ABCD is a cyclic quadrilateral, we have

∠BCD + ∠BAD = 180
[Opposite angles of a cyclic quadrilateral are supplementary]

∠BCD + 70^o = 180

∠BCD = 180^o – 70^o = 110^o

And, by angle sum property of ∆BCD we have

∠CBD + ∠BCD + ∠BDC = 180^o

30^o + 110^o + ∠BDC = 180^o

∠BDC = 180^o – 140^o

Thus,

∠BDC = 40^o

8. In cyclic quadrilateral ABCD, ∠A = 3 ∠C and ∠D = 5 ∠B. Find the measure of each angle of the quadrilateral.

Solution:

Given, cyclic quadrilateral ABCD

So, ∠A + ∠C = 180^o [Opposite angles in a cyclic quadrilateral is supplementary]

3∠C + ∠C = 180^o [As ∠A = 3 ∠C]

∠C = 45^o

Now,

∠A = 3 ∠C = 3 x 45^o

∠A = 135^o 

Similarly,

∠B + ∠D = 180^o  [As ∠D = 5 ∠B]

∠B + 5∠B = 180^o

6∠B = 180^o

∠B = 30^o

Now,

∠D = 5∠B = 5 x 30^o

∠D = 150^o

Therefore,

∠A = 135^o, ∠B = 30^o, ∠C = 45^o, ∠D = 150^o 

9. Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Solution:

Let’s join AD.

And. AB is the diameter.

We have ∠ADB = 90º  [Angle in a semi-circle]

But, 

∠ADB + ∠ADC = 180º  [Linear pair]

So, ∠ADC = 90º

Now, in ∆ABD and ∆ACD we have

∠ADB = ∠ADC [each 90º]

AB = AC [Given]

AD = AD [Common]

Hence, ∆ABD ≅ ∆ACD by RHS congruence criterion

So, by C.P.C.T

BD = DC

Therefore, the circle bisects base BC at D.

10. Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = 90^o – ½ ∠A

Solution:

Join ED, EF and DF. Also join BF, FA, AE and EC.

∠EBF = ∠ECF = ∠EDF ….. (i) [Angle in the same segment]

In cyclic quadrilateral AFBE,

∠EBF + ∠EAF = 180^o …… (ii)

Similarly in cyclic quadrilateral CEAF,

∠EAF + ∠ECF = 180^o ……. (iii)

Adding (ii) and (iii) we get,

∠EBF + ∠ECF + 2∠EAF = 360^o

∠EDF + ∠EDF + 2∠EAF = 360^o [From (i)]

∠EDF + ∠EAF = 180^o

∠EDF + ∠1 + ∠BAC + ∠2 = 180^o

But, ∠1 = ∠3 and ∠2 and ∠4 [Angles in the same segment]

∠EDF + ∠3 + ∠BAC + ∠4 = 180^o

But, ∠4 = ½ ∠C, ∠3 = ½ ∠B

Thus, ∠EDF + ½ ∠B + ∠BAC + ½ ∠C = 180^o

∠EDF + ½ ∠B + 2 x ½ ∠A + ½ ∠C = 180^o

∠EDF + ½ (∠A + ∠B + ∠C) + ½ ∠A = 180^o

∠EDF + ½ (180^o) + ½ ∠A = 180^o

∠EDF + 90^o + ½ ∠A = 180^o

∠EDF = 180^o – (90^o + ½ ∠A)

∠EDF = 90^o – ½ ∠A

11. In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20^o, find angle AOD.

Solution:

Join OB.

In ∆OBC, we have

BC = OD = OB [Radii of the same circle]

∠BOC = ∠BCO = 20^o

And ext. ∠ABO = ∠BCO + ∠BOC

Ext. ∠ABO = 20^o + 20^o = 40^o ….. (1)

Now in ∆OAB,

OA = OB [Radii of the same circle]

∠OAB = ∠OBA = 40^o [from (1)]

∠AOB = 180^o – 40^o – 40^o = 100^o

As DOC is a straight line,

∠AOD + ∠AOB + ∠BOC = 180^o

∠AOD + 100^o + 20^o = 180^o

∠AOD = 180^o – 120^o

Thus, ∠AOD = 60^o

12. Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.

Solution:

Let’s join OL, OM and ON.

And, let D and d be the diameter of the circumcircle and incircle.

Also, let R and r be the radius of the circumcircle and incircle.

Now, in circumcircle of ∆ABC,

∠B = 90^o

Thus, AC is the diameter of the circumcircle i.e. AC = D

Let the radius of the incircle be ‘r’

OL = OM = ON = r

Now, from B, BL and BM are the tangents to the incircle.

So, BL = BM = r

Similarly,

AM = AN and CL = CN = R

Now,

AB + BC + CA = AM + BM + BL + CL + CA

= AN + r + r + CN + CA

= AN + CN + 2r + CA

= AC + AC + 2r

= 2AC + 2r

= 2D + d

– Hence Proved

13. P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.

Solution:

First join AP and BP.

As TPS is a tangent and PA is the chord of the circle.

∠BPT = ∠PAB  [Angles in alternate segments]

But,

∠PBA = ∠PAB [Since PA = PB]

Thus, ∠BPT = ∠PBA

But these are alternate angles,

Hence, TPS || AB

14. In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent.

Prove that the line NM produced bisects AB at P.

Solution:

From P, AP is the tangent and PMN is the secant for first circle.

AP² = PM x PN …. (1)

Again from P, PB is the tangent and PMN is the secant for second circle.

PB² = PM x PN …. (2)

From (i) and (ii), we have

AP² = PB²

AP = PB

Thus, P is the midpoint of AB.

15. In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40^o and ∠ABD = 60^o, find:

i) ∠DBC

ii) ∠BCP

iii) ∠ADB

Solution:

PQ is a tangent and CD is a chord.

∠DCQ = ∠DBC [Angles in the alternate segment]

∠DBC = 40^o [As ∠DCQ = 40^o]

(ii) ∠DCQ + ∠DCB + ∠BCP = 180^o

40^o + 90^o + ∠BCP = 180^o [As ∠DCB = 90^o]

∠BCP = 180^o – 130^o = 50^o

(iii) In ∆ABD,

∠BAD = 90^o [Angle in a semi-circle], ∠ABD = 60^o [Given]

∠ADB = 180^o – (90^o + 60^o)

∠ADB = 180^o – 150^o = 30^o

16. The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90^o

Solution:

Let’s join OC.

BCD is the tangent and OC is the radius.

As, OC ⊥ BD

∠OCD = 90^o

∠OCD + ∠ACD = 90^o …. (i)

But, in ∆OCA

OA = OC [Radii of the same circle]

Thus, ∠OCA = ∠OAC

Substituting in (i), we get

∠OAC + ∠ACD = 90^o

Hence, ∠BAC + ∠ACD = 90^o

17. ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D. Prove that:

i) AC x AD = AB²

ii) BD² = AD x DC.

Solution:

i) In ∆ABC, we have

∠B = 90^o and BC is the diameter of the circle.

Hence, AB is the tangent to the circle at B.

Now, as AB is tangent and ADC is the secant we have

AB² = AD x AC

ii) In ∆ADB,

∠D = 90^o

So, ∠A + ∠ABD = 90^o …… (i)

But in ∆ABC, ∠B = 90^o

∠A + ∠C = 90^o ……. (ii)

From (i) and (ii),

∠C = ∠ABD

Now in ∆ABD and ∆CBD, we have

∠BDA = ∠BDC = 90^o

∠ABD = ∠BCD

Hence, ∆ABD ~ ∆CBD by AA postulate

So, we have

BD/DC = AD/BD

Therefore,

BD² = AD x DC

18. In the given figure, AC = AE.

Show that:

i) CP = EP

ii) BP = DP

Solution:

In ∆ADC and ∆ABE,

∠ACD = ∠AEB  [Angles in the same segment]

AC = AE [Given]

∠A = ∠A [Common]

Hence, ∆ADC ≅ ∆ABE by ASA postulate

So, by C.P.C.T we have

AB = AD

But, AC = AE [Given]

So, AC – AB = AE – AD

BC = DE

In ∆BPC and ∆DPE,

∠C = ∠E  [Angles in the same segment]

BC = DE

∠CBP = ∠CDE [Angles in the same segment]

Hence, ∆BPC ≅ ∆DPE by ASA postulate

So, by C.P.C.T we have

BP = DP and CP = PE

19. ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120^o

Calculate:

i) ∠BEC

ii) ∠BED

Solution:

i) Join OC and OB.

AB = BC = CD and ∠ABC = 120^o [Given]

So, ∠BCD = ∠ABC = 120^o

OB and OC are the bisectors of ∠ABC and ∠BCD and respectively.

So, ∠OBC = ∠BCO = 60^o

In ∆BOC,

∠BOC = 180^o – (∠OBC + ∠BOC)

∠BOC = 180^o – (60^o + 60^o) = 180^o – 120^o

∠BOC = 60^o

Arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.

∠BEC = ½ ∠BOC = ½ x 60^o = 30^o

ii) In cyclic quadrilateral BCDE, we have

∠BED + ∠BCD = 180^o

∠BED + 120^o = 180^o 

Thus, ∠BED = 60^o

20. In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30^o, find:

(i) angle BCO

(ii) angle AOB

(iii) angle APB

Solution:

In the given fig, O is the centre of the circle and, CA and CB are the tangents to the circle from C. Also, ∠ACO = 30^o

P is any point on the circle. P and B are joined.

To find:

(i) ∠BCO 

(ii) ∠AOB

(iii) ∠APB

Proof:

(i) In ∆OAC and ∆OBC, we have

OC = OC [Common]

OA = OB [Radii of the same circle]

CA = CB [Tangents to the circle]

Hence, ∆OAC ≅ ∆OBC by SSS congruence criterion

Thus, ∠ACO = ∠BCO = 30^o

(ii) As ∠ACB = 30^o + 30^o = 60^o

And, ∠AOB + ∠ACB = 180^o

∠AOB + 60^o = 180^o

∠AOB = 180^o – 60^o

∠AOB = 120^o

(iii) Arc AB subtends ∠AOB at the center and ∠APB is the remaining part of the circle.

∠APB = ½ ∠AOB = ½ x 120^o = 60^o

21. ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centers. Find the radii of the three circles.

Solution:

Given: ABC is a triangle with AB = 10 cm, BC= 8 cm, AC = 6 cm. Three circles are drawn with centre A, B and C touch each other at P, Q and R respectively.

So, we need to find the radii of the three circles.

Let,

PA = AQ = x

QC = CR = y

RB = BP = z

So, we have

x + z = 10 ….. (i)

z + y = 8 …… (ii)

y + x = 6 ……. (iii)

Adding all the three equations, we have

2(x + y + z) = 24

x + y + z = 24/2 = 12 ….. (iv)

Subtracting (i), (ii) and (iii) from (iv) we get

y = 12 – 10 = 2

x = 12 – 8 = 4

z = 12 – 6 = 6

Thus, radii of the three circles are 2 cm, 4 cm and 6 cm.