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 a reference tool for the students to do their homework and assignments, as well. These solutions for Class 10 contain the exercise-wise answers for all the chapters, thus being a 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 its 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 center of the circle to the outer line is its radius. 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 :-Click Here to Download PDF

 

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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:

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 1

Solution:

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

∠ADB = ∠ACB = 50°

Now in
∆ADB,

∠DAB + X + ∠ADB = 180°

= 42o + x + 50o = 180o

= 92o + x = 180o

x = 180o – 92o

x = 88o

(ii) In the given figure we have

= 32o + 45o + x = 180o

= 77o + x = 180o

x = 103o

(iii) From the given number we have

∠BAD = ∠BCD

Because angles in the same segment

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 2

But ∠BAD = 20o

∠BAD = 20o

∠BCD = 20o

∠CEA = 90o

∠CED = 90o

Now in triangle CED,

∠CED +
∠BCD + ∠CDE = 180o

90o + 20o + x = 180o

= 110o + x = 180o

x = 180o – 110o

x = 70o

(iv) In
∆ABC

∠ABC + ∠ABC + ∠BAC = 180o

(Because sum of a triangle)

69o + 31o + ∠BAC = 180o

∠BAC = 180o – 100o

∠BAC = 80o

Since ∠BAC and ∠BAD are in the same

Segment.

∠BAD = xo = 80o

(v) Given ∠CPB = 120o , ∠ACP = 70o

To find, xo i,e., ∠PBD

Reflex ∠CPB = ∠BPO + ∠CPA

1200 = ∠BPD + ∠BPD

(BPD = CPA are vertically opposite ∠s)

2∠BPD = 120o ∠PBD = 1200/2 = 60o

Also ∠ACP and PBD are in the same segment

∠PBD + ∠ACP = 700

Now, In ∆PBD

∠PBD + ∠PDB + ∠BPD = 180o

(sum of all ∠s in a triangle)

700 + xo + 600 = 180o

x = 180o – 130o

x = 50o

(vi) ∠DAB = ∠BCD

(Angles in the same segment of the circle)

∠DAB = 250 (∠BCD = 250 given)

In ∆DAP,

Ex, ∠CDA = ∠DAP + ∠DPA

xo = ∠DAB + ∠DPA

xo = 25o + 35 o

xo = 60o

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

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 3

Solution:

(i) ∠ACB = ∠ADB

(Angles in the same segment of a circle)

But ∠ADB = x°

∠ABC = xo

Now in ∆ABC

∠CAB + ∠ABC + ∠ACB = 180o

40o + 900 + xo = 180o

(AC is the diameter)

130o + xo = 180o

xo = 1800 – 130o = 50o

(ii) ∠ACD = ∠ABD

(angles in the same segment)

∠ACD = xo

Now in triangle OAC,

OA = OC

(radii of the same circle)

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 4

∠ACO = ∠AOC

(opposite angles of equal sides)

Therefore, xo = 62o

(iii) ∠AOB + ∠AOC + ∠BOC = 360o

(sum of angles at a point)

∠AOB + 80o + 130o = 360o

∠AOB + 210o = 360o

∠AOB = 360o – 210o = 1500

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

∠AOB = 2 ∠ACB

∠ACB = ½ ∠AOB = ½ × 150o = 75o

(iv)
∠ACB + ∠CBD = 180o

∠ABC + 75o = 180o

∠ABC = 180o – 75o‑ = 105o

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

Reflex
∠AOC = 2 ∠ABC

= 2× 105o =210o

(v) ∠AOC + ∠COB = 180o

135o + ∠COB = 180o

∠COB = 1800 – 135o = 45o

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 × 45o = 45o/2 = 22 1/2o

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

∠AOD = 2 ∠ACB

∠ACB = ½ ∠AOD = ½ × 70o = 35o

∠CMO = 90o

∠AMC = 90o

(∠AMC + ∠CMO = 180o)

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 5

Now in ∆ACM

∠ACM + ∠AMC + ∠CAM = 180o

35o + 90o + xo = 180o

125o + xo = 180o

Xo = 180 – 125o = 55o

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

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 6

Solution:

(a) Construction: Join AB

∠A = ∠C = 350 (Alt Angles)

∠ABC = 35o

(b) ∠AOC + reflex ∠AOC = 360o

130o + Reflex ∠AOC = 360o

Reflex
∠AOC = 360o – 130o = 230o

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 × 230o = 115o

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 7

Solution:

(a) ABCD is cyclic Quadrilateral

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 8

∠B + ∠D = 1800

Y + 400 + 45o = 180o

(y + 85o = 180o)

Y = 180o – 85o = 95o

∠ACB = ∠ADB

xo = 40

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

∠AOC = 2 ∠ABC

xo = 60o

Again ABCD is a Cyclic quadrilateral

∠B + ∠D = 180o

(60o + yo = 180o)

y = 180o – 60o = 120o

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°.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 9

In
∆YNB,

NYB + ∠YNB + YBN = 180o

50o + 20o + ∠YBN = 180o

YBN + 70o = 180o

YBN = 180o – 70o = 110o

But MAN = YBN

(Angles in the same segment)

MAN = 110o

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 × 110o =220o

(b) (i)
AOB + reflex AOB = 360o

(Angles at the point)

140o + reflex AOB = 360o

Reflex AOB = 360o – 140o = 220o

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 10

Now major arc AB subtends AOB + OBC = 360o

50o + 110o + 140o + OBC = 3600

300o + ∠OBC = 3600

∠300o + ∠OBC = 3600

∠OBC = 360o – 300o

∠OBC = 60o

(ii) In Quadrilateral .OACB

∠OAC + ∠ACB + ∠AOB + ∠OBC = 360o

50o + 110o + 140o + ∠OBC = 360o

300o + ∠OBC = 360o

∠OBC = 360o – 300o

∠OBC =60o

(iii) in ∆OAB,

OA = OB

(Radii of the same circle)

∠OAB + ∠OBA = 180o

2 ∠OAB = 180o – 140o = 40o

∠OAB = 40o/2 = 200

But ∠OBC = 60o

∠CBA = ∠OBC – ∠OBA

= 60o – 20o = 40o

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 11

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 = 48o

(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 × 32o = 2 ∠CDE

= 2 × 32o = 64o

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 12

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 13

(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 × 130o = 65o

But ∠ACB + ∠BCD = 180o

(Linear Pair)

65o + ∠BCD = 180o

∠BCD = 180o – 65o = 115o

Major arc BD subtends reflex ∠BQD at the

Centre and ∠BCD at the remaining part of

the circle

reflex ∠BQD = 2
∠BCD

=2 × 115o = 2300

But reflex ∠BQD + x = 360o

(Angles at a point)

230o + x = 360o

x = 360o – 230o = 130o

(b) Join OC

In ∆ABC,AC = BC

∠A = ∠B

But ∠A + ∠B + ∠C = 180o

∠A + ∠A + 560 = 1800

2 ∠A 180o – 560 = 124o

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 14

∠A = 124/2 = 62o or ∠CAB = 620

OC is the radius of the circle

OC bisects ∠ACB

∠OCA = ½ ∠ACB = ½ × 56o = 28o

Now in ∆OAC

OA = OC

(radii of the same Circle)

∠OAC = ∠OCA = 28o

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)

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 15

Solution:

From the figure

(i) ABCD is a cyclic quadrilateral

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 16

Ext. ∠DCE = ∠BAD

∠BAD = xo

Now arc BD subtends ∠BOD at the center

And ∠BAD at the remaining part of the circle.

∠BOD = 2 ∠BAD = 2 x

2 x = 150o (x = 750)

(ii) ∠BCD + ∠DCE = 180o

(Linear pair)

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 17

∠BCD + 800 = 180o

∠BCD = 1800 – 800 = 100o

Arc BAD subtends reflex ∠BOD at the

Centre and ∠BCD at the remaining part of the circle

Reflex ∠BOD = 2 ∠BCD

Xo = 2 × 100o = 200o

(iii) In ∆ACB,

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 18

∠CAB + ∠ABC + ∠ACB = 180o

(Angles of a triangle)

But
∠ACB = 90o

((Angles of a semicircle)

25o + 90o + ∠ABC = 180o

=115o + ∠ABC = 180o

∠ABC = 180o – 1150=65o

ABCD is a cyclic quadrilateral

∠ABC + ∠ADC = 180o

(Opposite angles of a cyclic quadrilateral)

65o + xo =180o

xo = 180o -65o = 115o

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 19

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 = 150o/2 = 75o

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

∠ABC + ∠ADC = 180o

(Sum of opposite angels in a cyclic quadrilateral

Is 180o)

75o + ∠ADC = 180o

∠ADC + 180o – 75o

∠ADC = 105o

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

∠ABC = 90o (Angle in a semi-circle)

(ii) ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180o

∠BAD + 75o = 180o

(∠BCD = 75o)

∠BAD = 180o -75o = 105o

But ∠EAF = ∠BAD

(Vertically opposite angles)

∠EAF = 105o

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

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 20

(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)

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 21

Solution:

(a) ∠DBC = 58°

BD is diameter

∠DCB = 90° (Angle in semi-circle)


(i) In ∆BDC

∠BDC + ∠DCB + ∠CBD = 180°

∠BDC = 180°- 90° – 58° = 32°

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 22

(ii) BEC = 180o – 32o = 148o

(opposite angles of cyclic quadrilateral)

(iii) ∠BAC = ∠BDC = 32o

(Angles in same segment)

(b) in the figure, AB ∥DC

∠BCE = 80o and ∠BAC = 25o

ABCD is a cyclic Quadrilateral and DC is

Production to E

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 23

(i) Ext, ∠BCE = interior ∠A

80o = ∠BAC + ∠CAD

80o = 25o + ∠CAD

∠CAD = 80o – 25o = 55o

(ii) But ∠CAD = ∠CBD

(Alternate angels)

∠CBD = 55o

(iii) ∠BAC = ∠BDC

(Angles in the same segments)

∠BDC = 25o

(∠BAC = 25o)

Now AB ∥ DC and BD is the transversal

∠BDC = ∠ABD

∠ABD = 25o

∠ABC = ∠ABD + ∠CBD = 25o + 55o = 80o

But ∠ABC + ∠ADC = 180o

(opposite angles of a cyclic quadrilateral)

80o + ∠ADC = 180o

∠ADC = 180o – 80o = 100o

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 24

(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

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 25

ABCD is a Cyclic Quadrilateral

∠ABC + ∠ADC = 180o

(Sum of opposite angles = 180o)

∠ABC + 80o = 180o

∠AOE = 150o, ∠DAO = 51o

To find ∠BEC and ∠EBC

ABED is a cyclic quadrilateral

Ext. ∠BEC = ∠DAB = 51o

∠ AOE = 150o

Ref ∠AOE = 360o – 150o = 51o

∠AOE = 150o

Ref ∠AOE = 360o – 150o = 210o

Now arc ABE subtends ∠AOE at the Centre

And ∠ADE at the remaining part of the circle.

∠ADE = ½ ref ∠AOE = ½ × 210o = 105o

But Ext ∠EBC = ∠ADE = 105o

Hence ∠BEC = 51o and ∠EBC = 105o

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 26

(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

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 27

∠A + ∠C = 180o

30o + p = 180o

P=180o – 30o = 150o

(ii) Arc BD subtends ∠BOD at the center

And ∠BAD at the remaining part of the circle

∠BOD = 2 ∠BAD

q = 2 × 30o = 60o

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

∠BAD = ∠BED

30o = r

r = 30o

  1. Join PQ

AQPD is a cyclic quadrilateral

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 28

∠A + ∠QPD = 180o

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)

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 29

(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 = 32o (Alternate angles)

Now PRST is a cyclic quadrilateral,

∠STP + ∠SRP = 180o

∠STP = 180o – 32o = 148o

(b) In the given figure,

∠ACE 43o and ∠CAF = 620

Now, in ∆AEC

∠ACE + ∠CAE + ∠AEC = 180o

43o + 62o + ∠AEC = 180o

105o + ∠AEC = 180o

∠AEC = 180o – 1050 = 75o

But ∠ABD + ∠AED = 1800

(sum of opposite angles of acyclic quadrilateral)

and ∠AED = ∠AEC

a + 75o = 180o

a = 180o – 75o – 105o

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 30

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 31

Solution:

(a) Construction: Join BC, and AC then

ABCD is a cyclic quadrilateral.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 32

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 = 180o

(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:

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 33

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

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 34

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 35

Solution:

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 center O

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

OP = 5 cm, OB = 13 cm.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 36

Ab is the chord of 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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 37

(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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 38

Solution:

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

∴ AQ = AP = 4cm

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 39

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 40

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 41

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, 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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 42

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 43

Solution:

(i) Join OB

∠OBA = 90°

(Radius through the point of contact is

perpendicular to the tangent)

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 44

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 45

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 46

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).

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 47

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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 48

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 

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 49

Solution:

(a) triangle ABC is an equilateral triangle

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 50

Each angle = 60o

∠A = 60o

But ∠A = ∠D

(Angles in the same segment)

∠D = 600

Now ABEC is a cyclic quadrilateral,

∠A = ∠E = 180o

60o + ∠E = 180o

600 + ∠E = 180o (∠E = 180o – 60o

∠E = 120o

Hence ∠BDC = 60o and ∠BEC = 120o

  1. AB is diameter of circle with centre O.

OD ⊥ AB and C is a point on arc DB.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 51

In ∆AOD, ∠AOD = 900

OA = OD (raddi of the semi – circle)

∠OAD = ∠ODA

But ∠OAD + ∠ODA = 90o

∠OAD + ∠ODA = 90o

2∠OAD = 90o

∠OAD = 90o/2 = 450

Or ∠BAD = 45o

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

∠ACD at the remaining part of the circle

∠AOD = 2 ∠ACD

90o = 2 ∠ACD (OD ⊥ AB)

∠ACD = 90o/2 = 45o

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

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 52

(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. 

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 53

Solution:

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

(Angles in a semi-circle)

∠BDC = ∠ADC – ∠ADB

= 1180 – 90o = 28o

∠ABCD is a cyclic quadrilateral)

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 54

∠ADC + ∠ABC = 180o

118o + ∠ABC = 180o

∠ABC = 180o – 118o = 62o

But in ∆AEB

∠AEB = 90o

(Angles in a semi-circle)

∠EAB = ∠ABE (AE = BE)

∠EAB + ∠ABE = 90o

∠EAB = 90o × ½ = 45o

∠CBE = ∠ABC + ∠ABE

= 62o + 45o = 107o

But AEBD is a cyclic quadrilateral

∠CAE + ∠CBE = 180o

∠CAE + 107o = 180o

∠CAE = 180o – 107o = 73o

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

With center O.AE = ED and ∠BCD = 140o

In cyclic quadrilateral EBCD.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 55

(i) ∠BCD + ∠BED = 180o

140o + ∠BED = 180o

∠BED = 180o – 140o = 400

But ∠AED = 90o

(Angles in a semi circle)

∠AED = ∠AEB + ∠BED

= 90o + 40o = 130o

(ii) Now in cyclic quadrilateral AEDB

∠AED + ∠DBA = 180o

130o + ∠DBA =180o

∠BDA = 180o – 130o = 50o

Chord AE = ED (given)

∠DBE = ∠EBA

But ∠DBE + ∠EBA = 50o

DBE + ∠DBE = 50o

2∠DBE = 50o

∠DBE = 25o or ∠EBD = 25o

In ∆OEB,OE = OB

(raddi 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

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 56

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 = 180o – (∠ACB + ∠BAC)….(i)

In the circle, arc AC subtends ∠AOC at

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

Reflex ∠AOC = 2 ∠ABC …(ii)

Reflex AOC = 2 { (180o – (ACB + BAC)}

But ∠AOC = 360o – 2(∠ACB + ∠BAC)

But ∠AOC = 360o – reflex ∠AOC

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 57

=360 – (360o – 2(∠ACB + ∠BAC)

=360o – 360o + 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.

ML Aggarwal Solutions for Class 10 Chapter 15 - Image 58

Proof : Arc BC subtends ∠AOB at the center 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

 

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