ML Aggarwal Solutions for Class 10 Maths Chapter 15 Circles

ML Aggarwal Solutions for Class 10 Maths Chapter 15 Circles are provided here in PDF format, which can be downloaded for free. The ML Aggarwal Solutions for the chapter Circles have been designed accurately by mathematics experts at BYJU’S. These serve as reference tools for the students to do their homework and assignments, as well. These solutions for Class 10, containing the exercise-wise answers for all the chapters, are very useful study material for the students studying in Class 10. Chapter 15 of ML Aggarwal Solutions for Class 10 Maths Circles explains circles, constructions of circles and their applications. A circle is a special kind of ellipse in which the eccentricity is zero, and the two foci are coincident. A circle is also termed as the locus of the points drawn at an equidistant from the centre. The distance from the centre of the circle to the outer line is its radius. The diameter is the line which divides the circle into two equal parts and is also equal to twice the radius. We, in our aim to help students, have devised detailed chapter-wise solutions for them to understand the concepts easily.

ML Aggarwal Solutions for Class 10 Maths Chapter 15 :

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Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 15 – Circles

Exercise 15.1

1. Using the given information, find the value of x in each of the following figures:

Solution:

(i)
∠ADB and ∠ACB are in the same segment.

∠ADB = ∠ACB = 50°

Now in
∆ADB,

∠DAB + X + ∠ADB = 180°

= 42^o + x + 50^o = 180^o

= 92^o + x = 180^o

x = 180^o – 92^o

x = 88^o

(ii) In the given figure, we have

= 32^o + 45^o + x = 180^o

= 77^o + x = 180^o

x = 103^o

(iii) From the given number, we have

∠BAD = ∠BCD

Because angles in the same segment

But ∠BAD = 20^o

∠BAD = 20^o

∠BCD = 20^o

∠CEA = 90^o

∠CED = 90^o

Now in triangle CED,

∠CED +
∠BCD + ∠CDE = 180^o

90^o + 20^o + x = 180^o

= 110^o + x = 180^o

x = 180^o – 110^o

x = 70^o

(iv) In
∆ABC

∠ABC + ∠ABC + ∠BAC = 180^o

(Because sum of a triangle)

69^o + 31^o + ∠BAC = 180^o

∠BAC = 180^o – 100^o

∠BAC = 80^o

Since ∠BAC and ∠BAD are in the same

Segment.

∠BAD = x^o = 80^o

(v) Given ∠CPB = 120^o , ∠ACP = 70^o

To find, x^o i,e., ∠PBD

Reflex ∠CPB = ∠BPO + ∠CPA

120⁰ = ∠BPD + ∠BPD

(BPD = CPA are vertically opposite ∠s)

2∠BPD = 120^o ∠PBD = 120⁰/2 = 60^o

Also ∠ACP and PBD are in the same segment

∠PBD + ∠ACP = 70⁰

Now, In ∆PBD

∠PBD + ∠PDB + ∠BPD = 180^o

(sum of all ∠s in a triangle)

70⁰ + x^o + 60⁰ = 180^o

x = 180^o – 130^o

x = 50^o

(vi) ∠DAB = ∠BCD

(Angles in the same segment of the circle)

∠DAB = 25⁰ (∠BCD = 25⁰ given)

In ∆DAP,

Ex, ∠CDA = ∠DAP + ∠DPA

x^o = ∠DAB + ∠DPA

x^o = 25^o + 35 ^o

x^o = 60^o

2. If O is the center of the circle, find the value of x in each of the following figures (using the given information):

Solution:

(i) ∠ACB = ∠ADB

(Angles in the same segment of a circle)

But ∠ADB = x°

∠ABC = x^o

Now in ∆ABC

∠CAB + ∠ABC + ∠ACB = 180^o

40^o + 90⁰ + x^o = 180^o

(AC is the diameter)

130^{o +} x^o = 180^o

x^o = 180⁰ – 130^o = 50^o

(ii) ∠ACD = ∠ABD

(angles in the same segment)

∠ACD = x^o

Now in triangle OAC,

OA = OC

(radii of the same circle)

∠ACO = ∠AOC

(opposite angles of equal sides)

Therefore, x^o = 62^o

(iii) ∠AOB + ∠AOC + ∠BOC = 360^o

(sum of angles at a point)

∠AOB + 80^o + 130^o = 360^o

∠AOB + 210^o = 360^o

∠AOB = 360^o – 210^o = 150⁰

Now arc AB subtends ∠AOB at the centre ∠ACB at the remaining part of the circle

∠AOB = 2 ∠ACB

∠ACB = ½ ∠AOB = ½ × 150^o = 75^o

(iv)
∠ACB + ∠CBD = 180^o

∠ABC + 75^o = 180^o

∠ABC = 180^o – 75^o‑ = 105^o

Now arc AC Subtends reflex ∠AOC at the centre and ∠ ABC at the remaining part of the circle.

Reflex
∠AOC = 2 ∠ABC

= 2× 105^o =210^o

(v) ∠AOC + ∠COB = 180^o

135^o + ∠COB = 180^o

∠COB = 180⁰ – 135^o = 45^o

Now arc BC Subtends reflex ∠COB at the centre and ∠ CDB at the remaining part of the circle.

∠COB = 2 ∠CDB

∠CDB = ½ ∠COB

= ½ × 45^o = 45^o/2 = 22 1/2^o

(vi) Arc AB subtends ∠AOD at the centre and ∠ACD at the remaining part of the Circle

∠AOD = 2 ∠ACB

∠ACB = ½ ∠AOD = ½ × 70^o = 35^o

∠CMO = 90^o

∠AMC = 90^o

(∠AMC + ∠CMO = 180^o)

Now in ∆ACM

∠ACM + ∠AMC + ∠CAM = 180^o

35^o + 90^o + x^o = 180^o

125^o + x^o = 180^o

X^o = 180 – 125^o = 55^o

3. (a) In the figure (i) given below, AD || BC. If ∠ACB = 35°. Find the measurement of ∠DBC.

(b) In the figure (ii) given below, it is given that O is the centre of the circle and ∠AOC = 130°. Find ∠ ABC

Solution:

(a) Construction: Join AB

∠A = ∠C = 35⁰ (Alt Angles)

∠ABC = 35^o

(b) ∠AOC + reflex ∠AOC = 360^o

130^o + Reflex ∠AOC = 360^o

Reflex
∠AOC = 360^o – 130^o = 230^o

Now arc BC Subtends reflex ∠AOC at the centre and ∠ ABC at the remaining part of the circle.

Reflex ∠AOC = 2 ∠ABC

∠ABC =1/2 reflex ∠AOC

= ½ × 230^o = 115^o

4. a) In the figure (i) given below, calculate the values of x and y.
(b) In the figure (ii) given below, O is the centre of the circle. Calculate the values of x and y.

Solution:

(a) ABCD is cyclic Quadrilateral

∠B + ∠D = 180⁰

Y + 40⁰ + 45^o = 180^o

(y + 85^o = 180^o)

Y = 180^o – 85^o = 95^o

∠ACB = ∠ADB

x^o = 40

(a) Arc ADC Subtends ∠AOC at the centre and
∠ ABC at the remaining part of the circle

∠AOC = 2 ∠ABC

x^o = 60^o

Again ABCD is a Cyclic quadrilateral

∠B + ∠D = 180^o

(60^o + y^o = 180^o)

y = 180^o – 60^o = 120^o

5. (a) In the figure (i) given below, M, A, B, N are points on a circle having centre O. AN and MB cut at Y. If ∠NYB = 50° and ∠YNB = 20°, find ∠MAN and the reflex angle MON.

(b) In the figure (ii) given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find
(i) ∠ACB
(ii) ∠OBC
(iii) ∠OAB
(iv) ∠CBA

Solution

(a) ∠NYB = 50°, ∠YNB = 20°.

In
∆YNB,

∠NYB + ∠YNB + ∠YBN = 180o

50^o + 20^o + ∠YBN = 180^o

∠YBN + 70^o = 180^o

∠YBN = 180^o – 70^o = 110^o

But ∠MAN = ∠YBN

(Angles in the same segment)

∠MAN = 110^o

Major arc MN subtend reflex ∠MON at the

Centre and ∠MAN at the remaining part of

the choice.

Reflex ∠MAN at the remaining part of the circle

Reflex ∠MON = 2 ∠MAN = 2 × 110^o =220^o

(b) (i)
∠AOB + reflex ∠AOB = 360^o

(Angles at the point)

140^o + reflex ∠AOB = 360^o

Reflex ∠AOB = 360^o – 140^o = 220^o

Now major arc AB subtends ∠AOB + ∠OBC = 360^o

50^o + 110^o + 140^o + ∠OBC = 360⁰

300^o + ∠OBC = 360⁰

∠300^o + ∠OBC = 360⁰

∠OBC = 360^o – 300^o

∠OBC = 60^o

(ii) In Quadrilateral .OACB

∠OAC + ∠ACB + ∠AOB + ∠OBC = 360^o

50^o + 110^o + 140^o + ∠OBC = 360^o

300^o + ∠OBC = 360^o

∠OBC = 360^o – 300^o

∠OBC =60^o

(iii) in ∆OAB,

OA = OB

(Radii of the same circle)

∠OAB + ∠OBA = 180^o

2 ∠OAB = 180^o – 140^o = 40^o

∠OAB = 40^o/2 = 20⁰

But ∠OBC = 60^o

∠CBA = ∠OBC – ∠OBA

= 60^o – 20^o = 40^o

6. (a)In the figure (i) given below, O is the centre of the circle and ∠PBA = 42°. Calculate the value of ∠PQB
(b) In the figure (ii) given below, AB is a diameter of the circle whose centre is O. Given that ∠ECD = ∠EDC = 32°, calculate
(i) ∠CEF
(ii) ∠COF.

Solution:

In ∆APB,

∠APB = 90° (Angle in a semi-circle)

But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle)

∠A + 90° + 42°= 180°

∠A + 132° = 180°

⇒ ∠A = 180° – 132° = 48°

But
∠A = ∠PQB

(Angles in the same segment of a circle)

∠PQB = 48^o

(b) (i) in ∆EDC,

(Ext, angle of a triangle is equal to the sum

of its interior opposite angels)

(ii) arc CF subtends ∠COF at the centre and

∠CDF at the remaining part of the circle

∠COF = 2 ∠CDF = 2 ∠CDE

=2 × 32^o = 2 ∠CDE

= 2 × 32^o = 64^o

7. (a) In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, ∠A = 35°, ∠Q = 25°. Find (i) ∠PRB (ii) ∠PBR (iii) ∠BPR.

(b) In the figure (ii) given below, it is given that ∠ABC = 40° and AD is a diameter of the circle. Calculate ∠DAC.

Solution

(a) (i) ∠PRB = ∠BAP

(Angles in the same segment of the circle)

∴
∠PRB = 35° (∵ ∠BAP = 35° given)

8. (a) In the figure given below, P and Q are centers of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.

(b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate

(i)∠CAB

(ii)∠OAC

Solution:

Given that

(a) Arc AB subtends ∠APB at the center

and ∠ACB at the remaining part of the circle

∠ACB = ½ ∠APB = ½ × 130^o = 65^o

But ∠ACB + ∠BCD = 180^o

(Linear Pair)

65^o + ∠BCD = 180^o

∠BCD = 180^o – 65^o = 115^o

Major arc BD subtends reflex ∠BQD at the

Centre and ∠BCD at the remaining part of

the circle

reflex ∠BQD = 2
∠BCD

=2 × 115^o = 230⁰

But reflex ∠BQD + x = 360^o

(Angles at a point)

230^o + x = 360^o

x = 360^o – 230^o = 130^o

(b) Join OC

In ∆ABC, AC = BC

∠A = ∠B

But ∠A + ∠B + ∠C = 180^o

∠A + ∠A + 56⁰ = 180⁰

2 ∠A 180^o – 56⁰ = 124^o

∠A = 124/2 = 62^o or ∠CAB = 62⁰

OC is the radius of the circle

OC bisects ∠ACB

∠OCA = ½ ∠ACB = ½ × 56^o = 28^o

Now in ∆OAC

OA = OC

(radii of the same Circle)

∠OAC = ∠OCA = 28^o

Exercise 15.2

1. If O is the center of the circle, find the value of x in each of the following figures (using the given information)

Solution:

From the figure

(i) ABCD is a cyclic quadrilateral

Ext. ∠DCE = ∠BAD

∠BAD = x^o

Now arc BD subtends ∠BOD at the centre

And ∠BAD at the remaining part of the circle.

∠BOD = 2 ∠BAD = 2 x

2 x = 150^o (x = 75⁰)

(ii) ∠BCD + ∠DCE = 180^o

(Linear pair)

∠BCD + 80⁰ = 180^o

∠BCD = 180⁰ – 80⁰ = 100^o

Arc BAD subtends reflex ∠BOD at the

Centre and ∠BCD at the remaining part of the circle

Reflex ∠BOD = 2 ∠BCD

X^o = 2 × 100^o = 200^o

(iii) In ∆ACB,

∠CAB + ∠ABC + ∠ACB = 180^o

(Angles of a triangle)

But
∠ACB = 90^o

((Angles of a semicircle)

25^o + 90^o + ∠ABC = 180^o

=115^o + ∠ABC = 180^o

∠ABC = 180^o – 115⁰=65^o

ABCD is a cyclic quadrilateral

∠ABC + ∠ADC = 180^o

(Opposite angles of a cyclic quadrilateral)

65^o + x^o =180^o

x^o = 180^o -65^o = 115^o

2. (a) In the figure (i) given below, O is the center of the circle. If ∠AOC = 150°, find (i) ∠ABC (ii) ∠ADC

(b) In the figure (i) given below, AC is a diameter of the given circle and ∠BCD = 75°. Calculate the size of (i) ∠ABC (ii) ∠EAF.

Solution:

(a) Given,
∠AOC = 150° and AD = CD

We know that an angle subtends by an arc of a circle

at the center is twice the angle subtended by the same arc

at any point on the remaining part of the circle.

(i) ∠AOC = 2 × ∠ABC

∠ABC = ∠AOC/2 = 150^o/2 = 75^o

(ii) From the figure, ABCD is a cyclic quadrilateral

∠ABC + ∠ADC = 180^o

(Sum of opposite angles in a cyclic quadrilateral

Is 180^o)

75^o + ∠ADC = 180^o

∠ADC + 180^o – 75^o

∠ADC = 105^o

(b) (i) AC is the diameter of the circle

∠ABC = 90^o (Angle in a semi-circle)

(ii) ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180^o

∠BAD + 75^o = 180^o

(∠BCD = 75^o)

∠BAD = 180^o -75^o = 105^o

But ∠EAF = ∠BAD

(Vertically opposite angles)

∠EAF = 105^o

3. (a) In the figure, (i) given below, if ∠DBC = 58° and BD is a diameter of the circle, calculate:

(i) ∠BDC (ii) ∠BEC (iii) ∠BAC

(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find:
(i) ∠CAD (ii) ∠CBD (iii) ∠ADC (2008)

Solution:

(a) ∠DBC = 58°

BD is diameter

∠DCB = 90° (Angle in a semi-circle)

(i) In ∆BDC

∠BDC + ∠DCB + ∠CBD = 180°

∠BDC = 180°- 90° – 58° = 32°

(ii) BEC = 180^o – 32^o = 148^o

(opposite angles of cyclic quadrilateral)

(iii) ∠BAC = ∠BDC = 32^o

(Angles in same segment)

(b) in the figure, AB ∥DC

∠BCE = 80^o and ∠BAC = 25^o

ABCD is a cyclic Quadrilateral and DC is

Production to E

(i) Ext, ∠BCE = interior ∠A

80^o = ∠BAC + ∠CAD

80^o = 25^o + ∠CAD

∠CAD = 80^o – 25^o = 55^o

(ii) But ∠CAD = ∠CBD

(Alternate angels)

∠CBD = 55^o

(iii) ∠BAC = ∠BDC

(Angles in the same segments)

∠BDC = 25^o

(∠BAC = 25^o)

Now AB ∥ DC and BD is the transversal

∠BDC = ∠ABD

∠ABD = 25^o

∠ABC = ∠ABD + ∠CBD = 25^o + 55^o = 80^o

But ∠ABC + ∠ADC = 180^o

(opposite angles of a cyclic quadrilateral)

80^o + ∠ADC = 180^o

∠ADC = 180^o – 80^o = 100^o

4. (a) In the figure given below, ABCD is a cyclic quadrilateral. If ∠ADC = 80° and ∠ACD = 52°, find the values of ∠ABC and ∠CBD.

(b) In the figure given below, O is the center of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC.

Solution:

(a) In the given figure, ABCD is a cyclic quadrilateral

∠ADC = 80° and ∠ACD = 52°

To find the measure of ∠ABC and ∠CBD

ABCD is a Cyclic Quadrilateral

∠ABC + ∠ADC = 180^o

(Sum of opposite angles = 180^o)

∠ABC + 80^o = 180^o

∠AOE = 150^o, ∠DAO = 51^o

To find ∠BEC and ∠EBC

ABED is a cyclic quadrilateral

Ext. ∠BEC = ∠DAB = 51^o

∠ AOE = 150^o

Ref ∠AOE = 360^o – 150^o = 51^o

∠AOE = 150^o

Ref ∠AOE = 360^o – 150^o = 210^o

Now arc ABE subtends ∠AOE at the Centre

And ∠ADE at the remaining part of the circle.

∠ADE = ½ ref ∠AOE = ½ × 210^o = 105^o

But Ext ∠EBC = ∠ADE = 105^o

Hence ∠BEC = 51^o and ∠EBC = 105^o

5. (a) In the figure (i) given below, ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. Given that ∠BEF = 80°, find ∠ABC.

(b) In the figure (ii) given below, ABCD is a cyclic trapezium in which AD is parallel to BC and ∠B = 70°, find:

(i)∠BAD (ii) DBCD.

Solution:

(a) ADFE is a cyclic quadrilateral

Ext. ∠FEB = ∠ADF

⇒ ∠ADF = 80°

ABCD is a parallelogram

∠B = ∠D = ∠ADF = 80°

or ∠ABC = 80°

(b)In trapezium ABCD, AD || BC

(i) ∠B + ∠A = 180°

⇒ 70° + ∠A = 180°

⇒ ∠A = 180° – 70° = 110°

∠BAD = 110°

(ii) ABCD is a cyclic quadrilateral

∠A + ∠C = 180°

⇒ 110° + ∠C = 180°

⇒ ∠C = 180° – 110° = 70°

∠BCD = 70°

6. (a) In the figure given below, O is the center of the circle. If ∠BAD = 30°, find the values of p, q and r.

(a) In the figure given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate

(i) ∠QBC

(ii) ∠BCP

Solution:

1. (i) ABCD is a cyclic quadrilateral

∠A + ∠C = 180^o

30^o + p = 180^o

P=180^o – 30^o = 150^o

(ii) Arc BD subtends ∠BOD at the center

And ∠BAD at the remaining part of the circle

∠BOD = 2 ∠BAD

q = 2 × 30^o = 60^o

∠BAD = ∠BED are in the same segment of the circle

∠BAD = ∠BED

30^o = r

r = 30^o

1. Join PQ

AQPD is a cyclic quadrilateral

∠A + ∠QPD = 180^o

7. (a) In the figure given below, PQ is a diameter. Chord SR is parallel to PQ.Given ∠PQR = 58°, calculate (i) ∠RPQ (ii) ∠STP

(T is a point on the minor arc SP)

(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007)

Solution:

(a) In ∆PQR,

∠PRQ = 90° (Angle in a semi-circle) and ∠PQR = 58°

∠RPQ = 90° – ∠PQR = 90° – 58° = 32°

SR || PQ (given)

∠SRP = ∠RPQ = 32^o (Alternate angles)

Now PRST is a cyclic quadrilateral,

∠STP + ∠SRP = 180^o

∠STP = 180^o – 32^o = 148^o

(b) In the given figure,

∠ACE 43^o and ∠CAF = 62⁰

Now, in ∆AEC

∠ACE + ∠CAE + ∠AEC = 180^o

43^o + 62^o + ∠AEC = 180^o

105^o + ∠AEC = 180^o

∠AEC = 180^o – 105⁰ = 75^o

But ∠ABD + ∠AED = 180⁰

(sum of opposite angles of acyclic quadrilateral)

and ∠AED = ∠AEC

a + 75^o = 180^o

a = 180^o – 75^o – 105^o

but ∠EDF = ∠BAE

(Angles in the alternate segment)

8. (a) In the figure (i) given below, AB is a diameter of the circle. If ∠ADC = 120°, find ∠CAB.

(b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.

Solution:

(a) Construction: Join BC, and AC then

ABCD is a cyclic quadrilateral.

Now in ∆DCF

Ext. ∠2 = x + z

and in ∆CBE

Ext. ∠1 = x + y

Adding (i) and (ii)

x + y + x + z = ∠1 + ∠2

2 x + y + z = 180^o

(ABCD is a cyclic quadrilateral)

But x : y : z = 3 : 4 : 5

x/y = ¾ (y = 4/3 x)

x/z = 3/5 (z = 5/3 x.

Exercise 15.3

1. Find the length of the tangent drawn to a circle of radius 3cm, from a point distnt 5cm from the center.

Solution:

In a circle with center O and radius 3cm and p is at a distance of 5cm.

That is OT = 3 cm, OP = 5 cm

OT is the radius of the circle

OT ⊥ PT

Now in right ∆ OTP, by Pythagoras axiom,

OP² = OT² + PT²

(5)² = (3)² + PT²

PT² = (5)² – (3)² = 25 – 9 = 16 = (4)²

PT = 4 cm.

2. A point P is at a distance 13 cm from the center C of a circle and PT is a tangent to the given circle. If PT = 12 cm, find the radius of the circle.

Solution:

CT is the radius

CP = 13 cm and tangent PT = 12 cm

CT is the radius and TP is the tangent

CT is perpendicular TP

Now in right-angled triangle CPT,

CP² = CT² + PT² [using Pythagoras axiom]

(13)² = (CT)² + (12)²

169 = (CT)² + 144

(CT)² = 169 -144 =25 = (5)²

CT = 5 cm.

Hence the radius of the circle is 5cm

3. The tangent to a circle of radius 6 cm from an external point P, is of length 8 cm. Calculate the distance of P from the nearest point of the circle.

Solution:

The radius of the circle = 6 cm

and length of tangent = 8 cm

Let OP be the distance

i.e. OA = 6 cm, AP = 8 cm .

OA is the radius

OA ⊥ AP

Now In right ∆OAP,

OP² = OA² + AP²

(By Pythagoras axiom)

= (6)² + (8)²

=36 + 64

= 100

= (10)²

OP = 10 cm.

4. Two concentric circles are of the radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

Solution:

Two concentric circles with centre O

OP and OB are the radii of the circles, respectively, then

OP = 5 cm, OB = 13 cm.

Ab is the chord of the outer circle, which touches the inner circle at P.

OP is the radius and APB is the tangent to the inner circle.

In the right-angled triangle OPB, by Pythagoras axiom,

OB² = OP² + PB²

13² = 5² + PB²

169 = 25 + PB²

PB² = 169 – 25

= 144

PB = 12 cm

But P is the mid-point of AB.

AB = 2PB

= 24 cm

5. Two circles of radii 5 cm and 2-8 cm touch each other. Find the distance between their centers if they touch :

(i) externally

(ii) internally.

Solution:

Radii of the circles are 5 cm and 2.8 cm.

i.e. OP = 5 cm and CP = 2.8 cm.

(i) When the circles touch externally,

then the distance between their centers = OC = 5 + 2.8

= 7.8 cm.

(ii) When the circles touch internally,

then the distance between their centers = OC = 5.0 – 2.8

= 2.2 cm

6. (a) In figure (i) given below, triangle ABC is circumscribed, find x.
(b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x.

Solution:

(a) From A, AP and AQ are the tangents to the circle

∴ AQ = AP = 4cm

But AC =12 cm

CQ = 12 – 4 = 8 cm.

From B, BP and BR are the tangents to the circle

BR = BP = 6 cm.

Similarly, from C,

CQ and CR the tangents

CR = CQ = 8 cm

x = BC = BR + CR = 6 cm + 8 cm = 14 cm

(b) From C, CR and CS are the tangents to the circle.

CS = CR = 3 cm.

But BC = 7 cm.

BS = BC – CS = 7 – 3 =4 cm.

Now from B, BP and BS are the tangents to the circle.

BP = BS = 4 cm

From A, AP and AQ are the tangents to the circle.

AP = AQ = 5cm

x = AB = AP + BP = 5 + 4

= 9 cm

7. (a) In figure (i) given below, quadrilateral ABCD is circumscribed; find the perimeter of quadrilateral ABCD.

(b) In figure (ii) given below, quadrilateral ABCD is circumscribed and AD ⊥ DC ; find x if radius of incircle is 10 cm.

Solution:

(a) From A, AP and AS are the tangents to the circle

∴AS = AP = 6

From B, BP and BQ are the tangents

∴BQ = BP = 5

From C, CQ and CR are the tangents

CR = CQ

From D, DS and DR are the tangents

DS = DR = 4

Therefore, the perimeter of the quadrilateral ABCD

= 6 + 5 + 5 + 3 + 3 + 4 + 4 + 6

= 36 cm

(b) in the circle with center O, radius OS = 10 cm

PB = 27 cm, BC = 38 cm

OS id the radius and AD is the tangent.

Therefore, OS perpendicular to AD.

SD = OS = 10 cm.

Now from D,DR and DS are the tangents

To the circle

DR = DS = 10 cm

From B, BP and BQ are tangents to the circle.

BQ = BP = 27 cm.

CQ = CB – BQ = 38 – 27 = 11 cm.

Now from C, CQ and CR are the tangents to the circle

CR = CQ = 11 cm.

DC = x =DR + CR

= 10 + 11 = 21 cm

8. (a) In the figure (i) given below, O is the center of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, find the radius of the circle.

(b) In the figure (ii) given below, from an external point P, tangents PA and PB are drawn to a circle. CE is a tangent to the circle at D. If AP = 15 cm, find the perimeter of the triangle PEC.

Solution:

(i) Join OB

∠OBA = 90°

(Radius through the point of contact is

perpendicular to the tangent)

OB² = OA² – AB²

r² = (r + 7.5)² – 15²

r² = r² + 56.25 + 15r – 225

15r = 168.75

r = 11.25

Hence, radius of the circles = 11.25 cm

(ii) In the figure, PA and PB are the tangents

Drawn from P to the circle.

CE is tangent at D

AP = 15 cm

PA and PB are tangents to the circle

AP = BP = 15 cm

Similarly EA and ED are tangents

EA = ED

Similarly BC = CD

Now perimeter of triangle PEC,

= PE + EC + PC

= PE + ED + CD + PC

PE + EA + CB + PC

(ED = EA and CB = CD)

=AP + PB = 15 + 15

= 30 cm.

9. (a) If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by

r = /frac (a + b – c) – (2)

(b) In the given figure, PB is a tangent to a circle with center O at B. AB is a chord of length 24 cm at a distance of 5 cm from the center. If the length of the tangent is 20 cm, find the length of OP.

Solution:

(a) Let the circle touch the sides BC, CA and AB

of the right triangle ABC at points D, E and F respectively,

where BC = a, CA = b

and AB = c (as showing in the given figure).

As the lengths of tangents drawn from an

External point to a circle are equal

AE = AF, BD = BF and CD = DE

OD ⊥ BC and OE ⊥ CA

(tangents is ⊥ to radius)

OD ⊥ BC and OE ⊥ CA

(tangents is ⊥ to radius)

ODCE is a square of side r

DC = CE = r

AF = AE = AC – EC = b – r and

BF = BD = BC – DC = a – r

Now, AB = AF + BF

C = (b – r) + (a – r)

2r = a + b – c

r = a + b – c/2

OP² = 400 + 169

OP = a−√569cm

10. Three circles of radii 2 cm, 3 cm and 4 cm touch each other externally. Find the perimeter of the triangle obtained on joining the centers of these circles.

Solution:

Three circles with centers A, B and C touch each other externally

at P, Q and R respectively and the radii of these circles are

2 cm, 3 cm and 4 cm.

By joining the centers of triangle ABC formed in which,

AB = 2 + 3 = 5 cm

BC = 3 + 4 = 7 cm

CA = 4 + 2 = 6 cm

Therefore, perimeter of the triangle ABC = AB + BC + CA

= 5 + 7 + 6

= 18 cm

Chapter test

1. (a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC. (b) In the figure (ii) given below, AB is a diameter of a circle with center O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD

Solution:

(a) triangle ABC is an equilateral triangle

Each angle = 60^o

∠A = 60^o

But ∠A = ∠D

(Angles in the same segment)

∠D = 60⁰

Now ABEC is a cyclic quadrilateral,

∠A = ∠E = 180^o

60^o + ∠E = 180^o

60⁰ + ∠E = 180^o (∠E = 180^o – 60^o

∠E = 120^o

Hence ∠BDC = 60^o and ∠BEC = 120^o

1. AB is diameter of circle with centre O.

OD ⊥ AB and C is a point on arc DB.

In ∆AOD, ∠AOD = 90⁰

OA = OD (radii of the semi–circle)

∠OAD = ∠ODA

But ∠OAD + ∠ODA = 90^o

∠OAD + ∠ODA = 90^o

2∠OAD = 90^o

∠OAD = 90^o/2 = 45⁰

Or ∠BAD = 45^o

(ii) Arc AD subtends ∠AOD at the centre and

∠ACD at the remaining part of the circle

∠AOD = 2 ∠ACD

90^o = 2 ∠ACD (OD ⊥ AB)

∠ACD = 90^o/2 = 45^o

2. (a) In the figure given below, AB is a diameter of the circle. If AE = BE and ∠ADC = 118°, find (i) ∠BDC (ii) ∠CAE

(B) inthe figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.

Solution:

(a) Join DB, CA and CB. ∠ADC = 118° (given) and ∠ADB = 90°

(Angles in a semi-circle)

∠BDC = ∠ADC – ∠ADB

= 118⁰ – 90^o = 28^o

∠ABCD is a cyclic quadrilateral)

∠ADC + ∠ABC = 180^o

118^o + ∠ABC = 180^o

∠ABC = 180^o – 118^o = 62^o

But in ∆AEB

∠AEB = 90^o

(Angles in a semi-circle)

∠EAB = ∠ABE (AE = BE)

∠EAB + ∠ABE = 90^o

∠EAB = 90^o × ½ = 45^o

∠CBE = ∠ABC + ∠ABE

= 62^o + 45^o = 107^o

But AEBD is a cyclic quadrilateral

∠CAE + ∠CBE = 180^o

∠CAE + 107^o = 180^o

∠CAE = 180^o – 107^o = 73^o

(b) AB is the diameter of semi-circle ABCDE

With center O.AE = ED and ∠BCD = 140^o

In cyclic quadrilateral EBCD.

(i) ∠BCD + ∠BED = 180o

140o + ∠BED = 180o

∠BED = 180o – 140^o = 40⁰

But ∠AED = 90^o

(Angles in a semi circle)

∠AED = ∠AEB + ∠BED

= 90^o + 40^o = 130^o

(ii) Now in cyclic quadrilateral AEDB

∠AED + ∠DBA = 180^o

130^o + ∠DBA =180^o

∠BDA = 180^o – 130^o = 50^o

Chord AE = ED (given)

∠DBE = ∠EBA

But ∠DBE + ∠EBA = 50^o

DBE + ∠DBE = 50^o

2∠DBE = 50^o

∠DBE = 25^o or ∠EBD = 25^o

In ∆OEB,OE = OB

(radii of the same circle)

∠OEB = ∠EBO = ∠DBE

But these are ultimate angles

OE ∥ BD

3. a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC). (b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z

Solution :

(a) Given: O is the center of the circle. To Prove : ∠AOC = 2 (∠ACB + ∠BAC). Proof: In ∆ABC, ∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle)

∠ABC = 180^o – (∠ACB + ∠BAC)….(i)

In the circle, arc AC subtends ∠AOC at

The centre and ∠ABC at the remaining part of the circle.

Reflex ∠AOC = 2 ∠ABC …(ii)

Reflex AOC = 2 { (180^o – (ACB + BAC)}

But ∠AOC = 360^o – 2(∠ACB + ∠BAC)

But ∠AOC = 360^o – reflex ∠AOC

=360 – (360^o – 2(∠ACB + ∠BAC)

=360^o – 360^o + 2 (∠ACB + ∠BAC)

=2 (∠ACB + ∠BAC)

Hence ∠AOC = 2 (∠ACB + ∠BAC)

(b) Given : in the figure, O is the center of the circle

To Prove : × + y = z.

Proof: Arc BC subtends ∠AOB at the centre and ∠BEC at the remaining part of the circle.

∠BOC = 2 ∠BEC

But ∠BEC = ∠BDC

(Angles in the same segment)

∠BOC = ∠BEC + ∠BDC ……. (ii)

Similarly in ∆ABD

Ext. ∠BDC = x + ∠ABD

= x + ∠EBD ………….(iii)

Substituting the value of (ii) and (iii) in (i)

∠BOC = y – ∠EBD + x + ∠EBD = x + y

Z = x + y