ML Aggarwal ICSE Solutions Class 10 Math Fifteenth Chapter Circle

ML Aggarwal ICSE Solutions Class 10 Math 15th Chapter Circle

ML Aggarwal ICSE Solutions Class 10 Math Chapter 15 Circle

 

Exercise – 15.1

(1) Using the information, find the value of x in each of the following fig.

Solution:

(i) From fig, ∠ADB & ∠ACB are on the same segment

=> ∠ADB = ∠ACB = 50°

Now, In △ADB,

∠DAB + x + ∠ADB = 180° (Since sum of all three angles of a triangle is 180°)

42° + x + 50° = 180°

90° + x = 180

x = 180° – 92°

x = 88° is the required value of x.

(ii) From fig, we know that the sum of all three angles of a triangle is found to be 180°.

32° + 45° + x = 180°

77 + x = 180°

x = 180° is the required value of x.

(iii) From fig, we can write,

∠BAD = ∠BCD

But, ∠BAD = 20° => ∠BCD = 20°

And ∠CEA = ∠CED = 90°

In △CED, ∠CED + ∠BCD + ∠CDE = 180°

90° + 20° + x = 180°

110° + x = 180°

x = 70° is the required value of x.

 

(iv) From fig, in △ABC,

∠ABC + ∠ACB + ∠BAC = 180°

Since, sum of all three angles of a triangle is 180°

=> 180° = 69° + 31° + ∠BAC

= 180° = 100° + ∠BAC

∠BAC = 80°

Again, ∠BAC & ∠BAD are on the same segment.

=> ∠BAD = x° = 80° is the required value of x.

 

(2) If 0 is the centre of the circle, find the value of x in each of the following fig.

Solution:

(i) From fig, we can write, ∠ACB = ∠ADB

(∵ angles on the same segment of a circle)

But, ∠ADB = x° => ∠ABC = x°

In △ABC, ∠CAB + ∠ABC + ∠ACB = 180°

40° + 90° + x° = 180°

130° + x° = 180°

x = 50° is the required value of x.

(ii) From fig, ∠ACD = ∠ABD

∠ACD = x°

In, △OAC, OA = OC (Radii of the same circle)

=> ∠ACO = ∠AOC

= x° = 62° is the required value of x.

(iii) From fig, ∠AOB + ∠AOC + ∠BOC = 360°

∠AOB + 80° + 130° = 360°

∠AOB = 360° – 210° = 150°

∠AOB = 150°

But, ∠AOB = 2 ∠ACB

∠ACB = 1/2 ∠AOB = 1/2 × 150° = 75°

∠ACB = 75°

(iv) From fig, ∠ACB + ∠CBD = 180°

∠ABC + 75° = 180°

∠ABC = 180° – 75°

∠ABC = 105°

Reflex ∠AOC = 2∠ABC

= 2 x 105°

Reflex ∠AOC = 210°

 

(3) (a) In the fig (i), AD||BC, if ∠ACB = 35°. Find the m∠DBC.

(b) In the fig (ii), given below it is given that ‘o’ is the centre of the circle & ∠AOC = 130°. Find ∠ABC

Solution:

(a) In fig, we will join AB.

∠A = ∠C = 35° (∵∠A & ∠C are alternate angles)

=> ∠ABC = 36°

(b) ∠AOC + reflex ∠AOC = 360°

130° + reflex ∠AOC = 360°

Reflex ∠AOC = 360° – 130° = 230°

Now, arc BC extended so that reflex ∠AOC at centre & ∠ABC at the remaining part of circle.

=> Reflex ∠AOC = 2∠ABC

∠ABC = 1/2 reflex ∠AOC => ∠ABC = 115°

 

(4) (a) In the fig (i) given below, calculate the values of x & y.

(b) In the fig (ii) given below, O is the centre of the circle.

Calculate the values of x and y.

Solution:

(a) In fig (i)

ABCD is a cyclic quadrilateral

=> ∠B + ∠D = 180°

= y + 40° + 45° = 180°

y + 85° = 180°

y = 95°

∠ACB = ∠ADB

= x = 40°

These are the required values of x & y

(b) In fig, arc ADC intersects angle ∠AOC at the centre & ∠ABC at remaining portion of the circle.

= ∠AOC = 2 ∠ABC

= x = 60°

Again, ABCD is a cyclic quadrilateral

∠B + ∠O = 180°

60° + y° = 180°

y° = 120°

These are the values of x & y respectively.

 

(6) (a) In the fig (i) given below, O is the centre of the circle and ∠PBA = 42°. Calculate the value of ∠PQB.

(b) In the, (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:

(a) In fig (i)

In △APB,

∠APB = 90°

(Since angle subtended in a semi-circle)

But, ∠A + ∠APB + ∠ABP = 180°

∠A + 90° + 42° = 180°

∠A = 180° – 132° = 48° => ∠A = ∠PQB

Since, angles in the same segment of a circle.

=> ∠PQB = 48° is the required angle.

(b) In fig (2)

In, △EDC, arc CF subtends ∠COF at the centre and ∠CDF at the remaining part of the circle.

Now, ∠COF = 2 ∠CDF = 2 ∠CDF

= 2×32°

= 2 ∠CDE

= 2×32°

∠COF = 64° is the required angle.

 

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

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

Solution:

(a) In fig (i), (i) ∠PRB = ∠BAP

Since, angles in the same segment of the circle.

=> ∠PRB = 35°

(ii) In △PRQ, (Exterior angle) = ∠PRQ + ∠PQR

= ∠PRB + ∠Q

= 35° + 25°

∠APR = 60° is the required angle

But, ∠APB = 90° which is the angle in a semi-circle

=> ∠BPR = ∠APB – ∠APR

= 90° – 60°

∠BPR = 30° is the required angle.

(iii) ∠APR = ∠ABR

= ∠ABR = 60°

In △PBQ,

Exterior ∠PBR = ∠Q + ∠BPQ

= 25° + 90°

∠PBR = 115° is the required angle.

(b) ∠B = ∠D are the angles in the same segment

= ∠C = 40°

∠ACD = 90° is the angle subtended in semi-circle

Now, In △ADC,

∠ACD + ∠D + ∠DAC = 180°

90° + 40° + ∠DAC = 180°

130° + ∠DAC = 180°

∠DAC = 180° – 130°

∠DAC = 50° is the required angle.

 

(8) (a) In the fig below, P & Q are centres of two circles intersecting at B and C. ACD is a straight line. Calculate value of x.

(b) In the fig, below, O is the circumcenter of a triangle ABC in which AC = BC. Given that, ∠ACB = 56°. Calculate (i) ∠CAB (ii) ∠OAC

Solution:

Given that,

(a) In fig (i), arc AB subtends ∠APB at the centre and ∠ACB at the remaining portion of circle.

∠ACB = 1/2 ∠APB = 1/2 × 130° = 65°

But, for linear pair angles => ∠ACB + ∠BCD = 180°

60° + ∠BCD = 180°

∠BCD = 180° – 65° = 115°

∠BCD = 115°

In fig, major arc BD subtends reflex ∠BQD at the centre & ∠BCD at the remaining portion of the circle.

Here, reflex ∠BQD = 2 ∠BCD

= 2×115°

∠BQD = 230°

But, reflex ∠BQD + x = 360°

230° + x = 360°

x = 360° – 230°

x = 130° is the required of x.

(b) In fig, we joined OC.

In, △ABC, AC = BC

∠A = ∠B

But, ∠A + ∠B + ∠C = 180°

∠A + ∠A + 56° = 180°

2 ∠A + 56° = 180°

∠A = 62°

Here, OC is the radius of the circle

And OC bisects ∠ACB

=> ∠OCA = 1/2 ∠ACB = 1/2×56° = 28°

In △OAC, OA = OC (Radii)

∠OAC = ∠OCA = 28° is the required angle

 

(9) (a) In fig (i), chord ED is parallel to  diameter AC of the circle Given ∠CBE = 65°, Calculate ∠DEC

(b) In fig (ii), C is a point on the mirror arc AB of the circle with centre O. Given ∠ACB = P°, ∠AOB = q°, express q in terms of p. calculate p if OACB is a parallelogram.

Solution:

(a) In fig (i)

∠CBE = ∠CAE

(Angles in the same segment of a circle)

=> ∠CAE = 65°

∠AEC = 90° is the angle in semi-circle.

In △AEC,

∠AEC + ∠CAE + ∠ACE = 180°

90° + 65° + ∠ACE = 180°

155° + ∠ACE = 180°

∠ACE = 180° – 155°

∠ACE = 25°

Given that, AC||ED

=> ∠ACE = ∠DEC (∵ alternate angles are always equal)

∠DEC = 25°

 

(b) In fig (ii)

Major arc AB subtends reflex ∠AOB at the centre and ∠ACB at the remaining portion of the circle.

=> Reflex ∠AOB = 2∠ACB = 2p —– (1)

But, Reflex ∠AOB = 360° – q —- (2)

From (1) & (2) => 2p = 360° – q

q = 360° – 2p = 2 (180° – p)

Given that, OACB is the parallelogram them

P = q

=> p = 360° – 2p

3p = 360°

p = 120 => p = q = 120° is the required values of P & q

 

(11) O is the circumcenter of the triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A

Solution:

In the fig, O is the centre of circumcentre of △ABC

Here; D is the mid-point of BC, BO, CO & OD are joined.

In fig, arc BC subtends ∠BOC on the centre and

∠A on the remaining portion of the circle.

=> ∠BOC = 2∠A

In △OBD and △BCO,

OD = OD ∵ Common side

BD = CD ∵ D is the mid-point of BC

OB = OC ∵ radii of same circle

△OBD ≅ △BCD

=> ∠BOD =∠COD

=> ∠BOD = 1/2 ∠BOC — (1)

But, 2∠A = ∠BDC

= ∠A = 1/2 ∠BOC —- (2)

From (1) & (2) => ∠BOD = ∠A (Hence Proved)

     

(12) In the following fig, AB and CD are equal chords. AD and BC intersects at K. prove that AE = CE and BE = DE

Solution:

In the given fig, AB & CD are two equal chords

Also, AD & BC intersects each other at point ‘c’

Now, in △AEB and △CED,

AB = CD (∵ Given)

∠A = ∠C (∵ angles in the same segment)

∠B = ∠D (∵ angles in the same segment)

= △AEB ≅ △CED

=> AE = CE and BE = DE (Hence Proved)

 

(14) In the following fig, O, is the centre of the given circle and OABC is a parallelogram. BC is produced to meet the circle at D.

Prove that: ∠ABC = 2 ∠OAD

Solution:

In fig, OABC is a parallelogram & point O is the centre of circle.

BC is extended to meet circle at point D.

To prove ∠ABC = 2 ∠OAD, we joined AD

Here, arc AC subtends ∠AOC at the centre and ∠ADC at the remaining portion of the circle.

=> ∠AOC = 2∠ADC

But, ∠OAD = ∠ADC (∵ alternate angles are always equal)

=> ∠AOC = 2 ∠OAD

But, ∠ABC = ∠AOC (∵ opposite angles of a parallelogram are always equal)

∠ABC = 2 ∠OAD (Hence Proved)

 

(Q15) (a) In fig (i), P is the point of intersection of the chords BC and AQ so that AB = AP. Prove that CP = CQ

(b) In fig, AB = AC = CD, ∠ADC = 38°

Calculate (i) ∠ABC

(ii) ∠BEC

Solution:

(a) Given that,

Two chords AQ and BC intersect each other point P inside the circle.

Here, AB and CQ are joined & AB = AP

Now, in △ABP and △CQP,

∠B = ∠Q (∵ angles in the same segment)

∠BAP = ∠PCQ (∵ angles in the same segment)

∠BPA = ∠CPQ (Vertically opposite angles are always equal)

∴ △ABP ~ △CPQ

= AB/CQ = AP/CP

But, AB = AP

= CQ = CP (Hence Proved)

(b) Given that, AB = AC = CD and ∠ADC = 38°

In △ACD, AC = CD

= ∠CAD = ∠ADC = 38°

Also, exterior ∠ACB = ∠CAD + ∠ADC

= 38° + 38°

Ext ∠ACB = 76°

In △ABC, AB = AC

∴ ∠ABC = ∠ACB = 76°

and ∠BAC = 180° – (∠ABC + ∠ACB)

= 180° – (76° + 76°)

= 180° – 152°

∠BAC = 28°

But, ∠BEC = ∠BAC

= ∠BEC = 28°

 

(16) In fig (i), CP bisects ∠ACB – Prove that DP bisects ∠ADB

(b) In the fig (ii), BD bisects ∠ABC. Prove that AB/BD = BE/BC

Solution:

(a) Given that,

∴ CP bisects ∠ACB

In fig, ∠ACB & ∠ADB are in the same segment of circle.

= ∠ACB = ∠ADB

Similarity, ∠ACP & ∠ADP are in the same segment

= ∠ACP = ∠ADP

But, ∠ACP = 1/2 ∠ACB and ∠ACB = ∠ADB

Thus, DP bisects ∠ADB (Hence Proved)

(b) Given that, BD bisects ∠ABC and △ABC inscribed in a circle.

Here, we joined CD in fig. In △ABE & △BCD

∠A = ∠D

∠ABE = ∠BDC

=> △ABE ~ △BAC

= AB/BD = BE/BC (Hence Proved)

 

(18) In the following fig, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5cm, BD = 4cm and CD = 9cm, find DE.

Solution:

In fig, AE and BC intersect each other at point D.

And we joined AB as shown.

∠CDE = 90°,  AB = 5cm, CD = 9cm

In right △ADB,

By Pythagoras theorem,

AB2 = AD2 + BD2

52 = AD2 + 42

25 = AD2 + 16

AD = 3cm

Chords AE and BC of the given circle intersect each other at point D inside the circle.

= AD × DE = BD × DC

DE = 4×9/3 = 36/3

DE = 12cm

 

(20) (a) In the fig (i), QPX is the bisector of ∠YXZ of the △XYZ. Prove that XY/XQ = XP/XZ

(b) In the fig (ii), chords BA and DC of a circle meet at P. Prove that (i) ∠PAD = ∠PCB

(ii) PA.PC = PC.PD

Solution:

(a) In fig (i)

△XYZ is inscribed in a circle as shown.

Bisector of ∠YXZ meets the circle at Q

Here, we joined QY

Now, in △XYQ and △XPZ,

∠Q = ∠Z (∵ angles in the same segment)

∠YXQ = ∠PXZ (∵ XQ is the bisector of ∠YXZ)

= △XYQ ~△XPZ

= XY/XP = XQ/XZ = XY/XQ = XP/XZ

= XY:XQ = XP:XZ (Hence Proved)

 

(b) Give that, two chords BA and DC meets each other at point P outside the circle. We joined AD and BC.

From fig, ∠PAD + ∠DAB = ∠PCB + ∠BCD

But, ∠DAB = ∠BCD

= ∠PAD = ∠PCB (Hence Proved)

Now, in △PBC & △PAD

∠PCB = ∠PAD

∠P = ∠P

∴ △PBC ~ △PAD

=> PC/PA = PB/PD

=> PA.PB = PC.PD (Hence Proved)

 

Exercise – 15.2

(1) If ‘o’ is the centre of the circle, find the value of x in each of the following fig.

Solution:

(i) In fig (i) ABCD is a cyclic quadrilateral.

Exterior ∠DCE = ∠BAD

= ∠BAD = x°

Also, arc BD subtends ∠BOD at the centre & ∠BAD at the remaining portion of the circle.

= ∠BOD = 2 ∠BAD = 2x

2x = 150°

x = 75° is the required value of x.

 

(ii) In fig (ii)

∠BCD & ∠DCE are linear pair angles.

=> ∠BCD + ∠DCE = 180°

∠BCD + 80° = 180°

∠BCD = 180° – 80°

∠BCD = 100°

Also, arc BAD subtends reflex ∠BOD at the centre and ∠BCD at the remaining portion of the circle.

=> Reflex ∠BOD = 2 ∠BCD

X° = 2×100 = 200°

x° = 200° is the required value of x.

(iii) In △ACB,

∠CAB + ∠ABC + ∠ACB = 180°

But, ∠ACB = 90° is the angle in a semi-circle

=> 25° + 90° + ∠ABC = 180°

115° + ∠ABC = 180°

∠ABC = 65°

But, ABCD is a cyclic quadrilateral.

Then, opposite angles of a cyclic quadrilateral are sums to 180°.

=> ∠ABC + ∠ADC = 180°

65° + x = 180°

x = 115° is the required value of x.

 

(2) (a) In fig (1), O is the centre of the circle. If ∠AOC = 150°, find (i) ∠ABC (ii) ∠ADC

(b) In the fig. (2), AC is a diameter of the given circle and ∠BCD = 75°. Calculate the size of (i) ∠ABC (ii) ∠EAF

Solution:

(a) In fig given that,

∠AOC = 150° and AD = CD

We already know that,

  • An angle substended by an arc of a circle at the centre is twice the angle substended by the same arc at any point on the remaining portion of the circle.

(i) ∠AOC = 2×∠ABC

∠ABC = ∠AOC/2 = 150/2 = 75° => ∠ABC = 75°

(ii) In fig. ABCD is a quadrilateral which is cyclic.

=> ∠ABC + ∠ADC = 180°

75° + ∠ADC = 180°

∠ADC = 105°

(b) (i) ∠ABC is the angle substended in a semi-circle.

=> ∠ABC = 90°

(ii) In fig. ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180°

∠BAD + 75° = 180°

∠BAD = 105°

But, ∠EAF & ∠BAD are vertically opposite angles

=> ∠EAF = 105°

 

(3) (a) In the fig (1), if ∠DBC = 58° and BD is a diameter of the circle, calculate (i) ∠BDC, (ii) ∠BEC, (iii) ∠BAC

(b) In the fig (2), AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find (i) ∠CAD, (ii) ∠CBD, (iii) ∠ADC

Solution:

(a) Given that, ∠DBC = 58° and BD is a diameter.

∠DCB is the angle subtended in a semi-circle.

= ∠DCB = 90°

(i) In △BDC

∠BDC + ∠DCB + ∠CBD = 180°

∠BDC = 180° – 90° – 58° = 32°

∠BDC = 32°

(ii) We have, Opposite angles of a cyclic quadrilateral sums to 180°

∠BEC = 180° – 320°

∠BEC = 148°

(iii) Also, ∠BAC & ∠BDC are angles in the same segment.

∠BAC = ∠BDC = 32°

(b)

In fig (ii), AB is parallel to DC

∠BCE = 80° and ∠BAC = 25°

Also, ABCD is a cyclic quadrilateral & DC is extended of E.

(i) Exterior ∠BCE = interior ∠A

80° = ∠BAC + ∠CAD

80° = 25° + ∠CAD

∠CAD = 55°

(ii) ∠CAD & ∠CBD are alternate angles

= ∠CAD = ∠CBD

= ∠CBD = 55°

(iii) ∠BAC & ∠BDC are the angles in the same segments

∠BAC = ∠BDC = ∠BDC

= ∠BDC = ∠BAC = 25°

Here, AB is parallel to DC & BD is transversal.

∠BDC = ∠ABD

∠ABD = 25°

∠ABC = ∠ABD + ∠CBD = 25° + 55° = 80°

But, opposite angles of a cyclic quadrilateral sums to 180°

∠ABC + ∠ADC = 180°

80° + ∠ADC = 180°

∠ADC = 100°

 

(5) In fig (i) Given, 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 fig (ii), ABCD is a cyclic trapezium in which AD is parallel to BC and ∠B = 70° find (i) ∠BAD (ii) DBCD

Solution:

(a) In fig, ADFE is a cyclic quadrilateral.

Also, exterior ∠FEB = ∠ADF

= ∠ADF = 80°

=> ABCD is a parallelogram.

= ∠B = ∠D = ∠ADF = 80° or ∠ABC = 80°

(b) ABCD is a trapezium, AD is parallel to BC

(i) ∠B + ∠A = 180°

70° + ∠A = 180°

∠A = 180° – 70°

∠A = 110°

= ∠BAD = 110°

(ii) ABCD is a cyclic quadrilateral

=> ∠A + ∠C = 180°

∠C = 180° – 110° = 70° => ∠BCD = 70°

 

(7) (a) In the fig, PQ is a diameter chord SR is parallel to PQ.

Given ∠PQR = 58°. Calculate

(i) ∠RPQ

(ii) ∠STP

(Where T is the point on the minor are SP)

(b) In the fig, if ∠ACE = 43° and ∠CAF = 62°, find the value of a, b and c.

Solution:

(a) In △PQR

∠PRQ = 90° is the angle in semi-circle and ∠PQR = 58°

=> ∠RPQ = 90° – ∠PQR

= 90° – 58°

∠RPQ = 32°

Given that, SR is parallel to PQ.

=> ∠RPQ = 90° – ∠PQR = 90° – 58° = 32°

∠SRP = ∠RPQ = 32° (Alternate angles)

PRST is a cyclic quadrilateral.

= ∠STP + ∠SRP = 180°

∠STP = 180° – 32° = 148° => ∠STP = 148°

(b) In fig, ∠ACE = 43° & ∠CAF = 62°

In △AEC, ∠ACE + ∠CAE + ∠AEC = 180°

43° + 62° + ∠AEC = 180°

105° + ∠AEC = 180°

∠AEC = 180° – 105°

∠AEC = 75°

But, ∠ABD & ∠AED are opposite angles of cyclic quadrilateral.

∠ABD + ∠AED = 180°

∠AED = ∠AEC

a + 75° = 180°

a = 180° – 75°

a = 105° is the required angle.

∠EDF = ∠BAE are the angles in the alternate segment.

In cyclic quadrilateral PQRS, exterior ∠PST = ∠PQR = 76°

Given that, ABCD is a cyclic quadrilateral

= ∠OAD + ∠BCD = 180°

50° + x = 180°

x = 180 – 50° = 130°

x = 130°

In △OAD, OA = OD (∵ radii of same circle)

=> ∠ODA = OAD = 50°

Also, exterior ∠BOD = ∠ODA + ∠OAD

y = 50° + 50°

y = 100° is the required angle

 

(12) In the following fig; O is the centre and AOE is the diameter of the semi-circle ABCDE. If AB = BC and ∠AEC = 50°.

Prove that: OB is parallel to EC

Solution:

In fig; given that,

O is the centre of the semi-circle ABCDE and AOE is the diameter of semi-circle.

=> AB = BC, ∠AEC = 50°

Here, AECB is a cyclic quadrilateral.

=> ∠AEC + ∠ABC = 180°

(Since, ∠AEC & ∠ABC are opposite angles of quadrilateral)

=> 50° + ∠ABE + ∠EBC = 180°

50° + 90° + ∠EBC = 180°

∠EBC = 40°

∵ ∠ABE is the angle in semi-circle.

=> ∠EBC = 40°

But, here BEDC is a cyclic quadrilateral.

=> ∠EBC + ∠CDE = 180°

40° + ∠CDE = 180°

∠CDE = 140°

∵ AB = BC

= ∠AEB = ∠BEC = 1/2 ∠AEC

= 1/2 × 50°

∠AEB = 25°

Now, in △OBE,

OB = OE (∵ radii of same circle)

= ∠OBE = ∠OEB = 25°

Also, exterior ∠AOB = ∠OBE + ∠OEB

= 25° + 25°

∠AOB = 50°

=> ∠AOB = ∠OEC = 50°

But, ∠AOB and ∠OEC are corresponding angles

=> OB is parallel to EC (Hence Proved)

 

(13) In the fig, ED and BC are two parallel chords of the circle and ABE, ACD are two straight lines. Prove that AED is an isosceles triangle.

Solution:

Given that, chord BC||ED

ABE and ACD are straight lines.

Here, in fig BCDE is a cyclic quadrilateral.

Exterior ∠ABC = ∠D —– (1)

But, BC is parallel to ED

 

=> ∠ABC = ∠E (ABC & ∠E are corresponding angles) — (2)

From (1) & (2), => ∠D = ∠E

In △AED, ∠D = ∠E

= AE = AD

Sides opposite to equal angles

Thus, △AED is an isosceles triangle.

 

(14) In the following ABC is an isosceles triangle in which AB = AC and circle passing through B and C intersects sides AB and AC at points D and E. Prove that DE||BC.

Solution:

In given fig, △ABC is an isosceles triangle so that AB = AC

A circle passing through B and C intersects sides AB and AC at points D and E.

We joined DE => AB = A

But ∠B and ∠C are angles opposite to equal sides

= ∠B = ∠C

But, BCED is a cyclic quadrilateral.

Exterior ∠ADE = ∠C = ∠B

But these angles are corresponding angles

= DE is parallel to BC (Hence Proved)

 

(16) In the following chords AB & CD of the circle are produced to meet at O. Prove that triangle OBD and OAC are similar. Given that, CD = 2cm, DO = 6cm, and BO = 3cm, area of quadrilateral CABD. Calculate AB. Also find (area of quadrilateral of CABD)/area of △OAC

Solution:

In fig, given that AB and CD are chords of a circle.

Here, chords AB & CD are produced to meet at point ‘o’

Given, CD = 2cm, DO = 6cm, BO = 3cm

 

We joined AC and BO here.

(i) In △ODB and △OAC

Exterior ∠ODB = ∠A

(Exterior angle of a cyclic quadrilateral is equal to its interior opposite angle)

Similarly, ∠OBD = ∠C

∠O = ∠O

∴ △ODB = △OAC

(ii) OD/OA = OB/OC=> 6/OA = 3/6+2 => OA = 16cm

∴ AB = OA – OB = 16 – 3 = 13 cm

AB = 13 cm

(iii) As △OAC ~ △ODB

= area of △OAC/area of △ODB = OC2/OB2 = 82/32 = 64/9

Subtracting 1 from both sides of above equation,

=> A(△OAC)/A(△ODB) – 1 = 64/9 – 1

A (△OAC)-A(△ODB)/A(△ODB) = 64-9/9 = 55/9

A (Quadrilateral CABD)/A(△OAC) = 55/64 (Hence Proved)

 

Exercise 15.3

(1) Find the length of tangent draw to a circle of radius 3cm, from a point distant 5cm from the centre.

Solution:

In fig in a circle with centre ‘o’ and radius 3cm and P is at a distance of 5cm.

=> OT = 3cm, OP = 5cm

∵ OT is the radius of the circle O⊥ ⊥or PT

Now, in right angle △OTP,

 

By Pythagoras theorem,

OP2 = OT2 + PT2

52 = 32 + PT2

PT2 = 52 – 32 = 25 – 9 = 16 => PT = 4cm

 

(2) A point P is at a distance 13cm from the centre C of a circle and PT is a tangent to the given circle. If PT = 12cm. Find the radius of the circle.

Solution:

Given that, CT is the radius CP = 13 cm and tangent PT = 12cm

Here, in fig, CT is the radius and TP is the tangent.

=> CT ⊥or TP

 

 

Now, in △CPT, By Pythagoras theorem,

CP2 = CT2 + PT2

132 = CT2 + 122

CT2 = 169 – 144 = 25

CT = 5cm is the required radius of circle.

 

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

Solution:

Radius of the circle (r) = 6cm

Length of largest = 8cm

Let ‘OP’ be the distance I.e. OA = 6cm, AP = 8cm

Here, OA is the radius => OA ⊥or AP

Now, in right △OAP, By Pythagoras theorem,

OP2 = OA2 + AP2

OP2 = 62 + 82 = 36 + 64

OP2 = 100

OP = 10cm is the required distance.

 

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

Solution:

In fig. Given that,

Two concentric circles with centre ‘o’

OP and OB are the radii of the circles respectively.

Then, OP = 5cm, OB = 13cm

In fig. AB is the chord of outer circle which touches the inner circle at P

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

In right △OPB,

By Pythagoras theorem,

OB2 = OP2 + PB2

132 = 52 + PB2

PB2 = 169 – 25 = 144

PB = 12cm

But, here point P is the mid-point of AB

=> AB = 2PB = 24cm

 

(5) Two circles of radii 5cm and 2.8 cm touch each other. Find the distance between their centers if they touch:

(i) Externally

(ii) Internally

Solution:

Given that, radii of circles are 5cm & 2.8 cm respectively.

= OP = 5cm and CP = 2.8 cm

(i) In fig (1) circle touches externally & hence distance between their centers => OC = 5 + 2.8 = 7.8 cm

(ii) In fig (2) when the circle touches internally then the distance between their centers => OC = 5 – 2.8 = 2.2 cm

 

(8) In fig, O is the centre of the circle and AB is a tangent of B. If AB = 15cm and AC = 7.5 cm, find the radius of the circle.

Solution:

(i) Here in fig, we joined OB,

∠OBA = 90°

Then, By Pythagoras theorem

OB2 = OA2 – AB2

r2 = (r + 7.5)2 – 152

r2 = r2 + 56.25 + 15r – 225

15r = 168.75

r = 11.25 cm

This is the required radius of the circle.

 

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

Solution:

In fig. three circles with centres A, B and C touch each other externally at points P, Q & R respectively.

The radii of these circles are 2cm, 3cm and 4cm respectively.

After joining the centres of these circles triangle ABC is formed as shown.

 

AB = 2+3 = 5cm

BC = 3+4 = 7cm

CA = 4+2 = 6cm

Thus, Perimeter of △ABC = AB + BC + CA

= 5 + 7 + 6

P (△ABC) = 18cm is the required perimeter of the triangle.

 

(12) (a) In the fig (i) PQ = 24 cm, QR = 7cm and ∠PQR = 90°. Find the radius of the inscribed circle △PQR

(b) In the fig (ii) two concentric circles with centre O are of radii 5cm and 3cm from an exterior point P, tangents PA and PB are drawn to these circles. If AP = 12cm. Find —-.

Solution:

In fig, a circle is inscribed in △PQR which touches the sides.

‘O’ is the centre of the circle.

PQ = 24 cm, QR = 7cm, ∠PQR = 90°

Here, OM is joined

Given that

OL & OM are radii of circle.

=> OL ⊥or PQ and OM ⊥or BC and ∠PQR = 90°

Here, QLOM is a square formed.

=> OL = OM = QL = QM

In △PQR, ∠O = 90°

=> PR2 = PQ2 + QR2

= 242 + 72

= 576 + 49

PR2 = 625

=> PR = 25cm

Let us consider, OB = x => QM = QL = x

Given that, PL and PN are tangents to the circle

=> PL = PN

=> PQ – LQ = PR – RN

24 – x = 25 – PM (∵ RN = RM)

24 – x = 25 – (QR – QM)

24 – x = 25 – (7 – x)

24 – x = 25 – 7 + x

24 – 25 + 7 = 2x

2x = 6

x = 3cm is the required radius of in circle formed.

 

(b) Given that, radius of outer circle = 5cm

Radius of inner circle = 3cm

= OA = 5cm

OB = 3cm & AP = 12cm

Given that, OA is radius & A.P. is the tangent,

=> OA ⊥or AP

Now, in △OAP,

OP2 = OA2 + AP2

= 52 + 122

OP2 = 25 + 144 = 169

OP = 13 cm

Similarity, in △OBP,

OP2 = OB2 + BP2

132 = 32 + BP2

169 = 9 + BP2

BP2 = 169 – 9

BP = √169cm

BP = 4√10 cm

 

(13) In fig. AB = 8cm and M is the mid-point of AB. Semi-circle are drawn on AB, AM and MB as diameters. A circle with centre C touches all these semi-circles as shown, find its radius.

Solution:

Let us, consider ‘x’ be the radius of circle with center ‘c’.

Radius of semi-circle = 4/2 = 2cm

Here, CP = x + 2 and CM = 4 – x

Now, in right △PCM,

PC2 = PM2 + CM2

(x + 2)2 = 22 + (4 – x)2

x2 + 4x + 4 = 4 + 16 – 8x + x2

4x + 8x + 4 – 4 – 16 = 0

12x = 16

x = 4/3 cm is the required radius of circle.

 

(Q14) The length of the direct common tangent to two circles of radii 12cm and 4cm 15cm. Calculate the distance between their centers.

Solution:

Let us considers, R & r be the radii of the circle having centres A and B respectively.

Let TT’ be the common tangent to these circles.

= (TT’)2 = (AB)2 – (R-r)2

152 = (AB)2 – (12 – 4)2

= (15)2 = (AB)2 – (12-4)2

225 = AB2 – 64

(AB)2 = 225 + 64

= 289

(AB)2 = (17)2

=> AB = 17cm is the required distance between their centres.

 

(15) Calculate the length of a direct common tangent to two circles of radii 3cm and 8cm with their centers 13 cm a part.

Solution:

Let us consider, A and B are the centers of circle with radii 8cm and 3cm respectively.

Let TT’ be the length of the common tangent to the circles.

= AB = 13cm

In fig, (TT’)2 = (AB)2 – (R – r)2

= 132 – (8-3)2

= 169 – 52 = 169 – 25

= 144

TT’ = 12cm is the length of required common tangent.

 

(16) In fig., AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3cm respectively.

Given that, AB = 8cm, calculate PQ.

Solution:

Given that, AC is a transverse common tangent to two circles with centres P and Q and o radii 6cm & 3cm respectively.

AB = 8cm

In △PAB, By Pythagoras theorem,

PB2 = PA2 + AB2

= 62 + 82

PB2 = 36 + 64

PB2 = 102

PB = 10cm

In △PAB & △BCQ,

∴ ∠A = ∠C = 90°

∠ABP = ∠CBQ

(Vertically opposite angles are always equal)

=> △PAB ~ △BCQ

=> AP/CQ = PB/BQ => 6/3 = 10/BQ

BQ = 10×3/6 = 5cm

PQ = PB + BQ

= 10 + 5

PQ = 15 cm

 

(17) Two circles with centers A, B are of radii 6 cm & 3cm respectively. If AB = 15cm, Find the length of a transverse common tangent to these circles.

Solution:

Given that, Two circles with centers A, B are having radii 6cm & 3cm respectively.

AB = 15cm

AT = 6cm, BS = 3cm

Let us consider, AP = x => PB = 15 – x

△ATP & △SBP ∠T = ∠S = 90°

∠APT = ∠BPS (Vertically opposite angles are equal)

= △ATP ~ △SBP

=> AT/BS = AP/BP = 6/3 = x/(15-x)

3x = 90 – 6x

9x = 90 => x = 10

AP = 10cm => PB = 15 – 10 = 5cm

In △ATP, AP2 = AT2 + TP2 ∵ By Pythagoras theorem

(10)2 = 62 + TP2

100 = 36 + TP2

TP2 = 100 – 36 = 64

= TP = 8 cm

In △PSB, By Pythagoras theorem,

PB2 = BS2 + PS2

52 = 32 + PS2

25 – 9 = PS2

PS2 = 16 => PS = 4cm

Hence, TS = TP + PS = 8 + 4 = 12 cm

 

(19) In the given fig, AD is a diameter of a circle with centre ‘o’ & AB is tangent at A. C is a point on the circle such that DC produced intersects the tangent of B. If ∠ABC = 50°, find ∠AOC

Solution:

Given that, AB is the tangent to the circle of A and OA is the radius => OA ⊥or AB

In △ABD, ∠OAB + 90° + ∠ADC = 180

∠OAB + 90° + 50° = 180°

∠OAB + 140° = 180°

∠OAB = 40°

But, we have, (Angle at the centre of a circle) = (Double the angle at the remaining part of circle)

∠ABC = 2×40

∠ABC = 80°

 

(20) In fig. tangents PQ and PR are drawn from an external point P to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ, Find ∠RQS

Solution:

Given that, PQ & PR tangents to the circle with centres ‘o’ drawn from P.

∠RPQ = 30°

And, Chord RS Parallel to PQ drawn

=> PQ = PR

(Tangents to the circle)

∠PRQ = ∠PQR

But, ∠RPO = 30°

=> ∠PRO = ∠PQR = (180 – 30)/2 = 150/2 = 75°

Also, ∠RSQ = ∠QRS = 75° ∵ QR = QS

In △QRS, ∠RQS = 180° – (∠RSQ + ∠QRS)

= 180° – (75° + 75°)

= 180 – 150

∠RQS = 30°

 

(23) In the following, O is the centre of the circle. Tangents to the circle at A and B meet at C. If ∠ACO = 30° find

(i) ∠BCO

(ii) ∠AOB

(iii) ∠APB

Solution:

(i) ∠BCO & ∠ACO = 30°

Because, C is the intersecting, part of tangent AC and BC.

(ii) ∠OAC = ∠OBC = 90

Given that, ∠AOC = ∠BOC = 180° – (90° + 30°)

∠AOC = ∠BOC = 60°

Since, ∠AOB = ∠AOC + ∠BOC

∠AOB = 60° + 60°

∠AOB = 120°

(iii) ∠APB = 1/2 ∠AOB = 1/2 (120°) = 60°

∠APB = 60°

 

(26) In a triangle ABC, the in circle touches BC, CA and AB at point P, Q & R respectively.

Calculate: (i) ∠QOR, (ii) ∠QPR

Given that: ∠A = 60°

Solution:

In fig, OQ and OR are the radii and AC, AB are tangents to the circle.

=> OQ ⊥or AC & OR ⊥or AB

In quadrilateral AROQ,

∠A = 60°, ∠ORA = 90° & ∠OQA = 90°

∴ ∠ROQ = 360° – (∠A + ∠ORA + ∠OQA)

= 360° – (60° + 90° + 90°)

= 360° – 240°

∠ROQ = 120°

Here, arc RQ subtends ∠ROQ at the centre & ∠RPQ at the remaining part of the circle as shown.

∴ ∠ROQ = 2 RPQ

= ∠RPQ = 1/2 ∠ROQ = 1/2 (120)

∠RPQ = 60° is the required angle.

 

(28) (a) In fig. AB is a diameter. The tangent at C meets AB produced at Q, ∠CAB = 34°. Find (i) ∠CBA (ii) ∠CQA

(b) In the fig, AP and BP are tangents to the circle with centre ‘o’.

Given ∠APB = 60°. Calculate (i) ∠AOB (ii) ∠OAB (iii) ∠ACB

Solution:

(a) In fig (i), AB is the diameter

=> ∠ACB = 90° & ∠CAB = 34°

In △ABC,

∠ACB + ∠CAB + ∠CBA = 180°

90° + 34° + ∠CBA = 180°

∠CBA = 180° – 90° – 34°

∠CBA = 56°

 

(b) (i) In fig (2) AP and BP are the tangents to the circle and OA, OB are Radii on it.

=> OA ⊥or AP and OB ⊥or BP

=> ∠AOB = 180° – 60° = 120°

∠AOB = 120°

(ii) In △OAB,

OA = OB (∵ radii of the same circle)

And

∠OAB = ∠OBA = 1/2 (180 – 120)

= 1/2 × 60

∠OAB = ∠OBA = 30°

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining portion of the circle.

=> ∠AOB = 2 ∠ACB

∠ACB = 1/2 (∠AOB) = 1/2 × 120 = 60°

∠ACB = 60°

 

(29) Two chords AB, CD of a circle intersects internally at a point P, if (i) AP = 6cm, PB = 4cm and PD = 3cm find PC.

(ii) AB = 12cm, AP = 2cm, PC = 5cm, find PD

(iii) AP = 5cm, PB = 6cm and CD = 13 cm find CP

Solution:

(i) In given fig, two chords AB and C intersects each other at P internally.

=> A.P PB = CP.PD

=> CP = AP×PB/PD = 6×4/3 = 8cm

PC = 8cm

(ii) In fig. AB = 12cm, AP = 2cm, PC = 5cm

From fig. PB = AB – AP

= 12 – 2

PB = 10 cm

Here, AP × PB = CP × PD

2 × 10 = 5 × PD

PD = (2 × 10)/5 = 4cm

PD = 4cm

 

(iii) Given that, AP = 5cm, PB = 6cm, CD = 13cm

Let us take, CP = x cm and PD = CD – CP or PD = 13 – x

=> AP × PB = CP × P

5 × 6 = x (13 – x)

30 = 13x – x2

x2 – 13x + 30 = 0

x2 – 10x – 3x + 30 = 0

x (x – 10) – 3 (x – 10) = 0

(x – 10) (x – 3) = 0

=> x = 10 or x = 3

CP = 10cm or 3cm

 

(30) (a) In the fig, PT is a tangent to the circle. Find TP if AT = 16cm and AB = 12cm

(b) In the fig, diameter AB and chord CD of a circle meet at P PT is a tangent to the circle at T.

CD = 7.8 cm, PD = 5cm, PB = 4cm

Find (i) A, (ii) length of tangent PT

Solution:

(a) In fig, PT is the tangent to the circle & AT is secant as shown in fig.

=> PT2 = TA×TB

Here, TA = 16cm, AB = 12cm

TB = AT – AB = 16 – 12 = 4cm

PT2 = 16+4 = 68

PT = 8cm

 

(b) In fig, PT is the tangent & PDC is secant out to circle.

PT2 = PC×PD

PT2 = (5 + 7.8) × 5 = 12.8 × 5

PT2 = 64

=> PT = 8cm

In △OTP, By Pythagoras theorem,

PT2 + OT2 = OP2

82 + x2 = (x + 4)2

64 + x2 = x2 + 16 + 8x

64 – 16 = 8x

48 = 8x

x = 6cm is the required radius.

Now, AB = 2×6 = 12cm

 

(32) In following fig, CBA is a secant and CD is tangent to the circle. If AB = 7cm and BC = 9cm, then

(i) Prove △ACD ~ △DCB

(ii) Find length of CD

Solution:

In fig, △ACD & △DCB,

∠C = ∠C (∵ common angle)

∠CAD = ∠CDB

= △ACD ~ △DCB

= AC/DC = DC/BC => DC2 = AC × BC

= 16 × 9

DC2 = 144

DC = 12 cm

 

(33) In the fig, PAB is a secant and PT is tangent to a circle. If PA:PB = 1:3 and PT = 6cm, find the length of PB.

Solution:

In fig, PAB is secant and PT is the tangent to the circle.

=> PT2 = PA × PB

Given that, PT = 6cm and PA:AB = 1:3

Let us consider, PA = x => AB = 3x

=> PB = PA + AB

= x + 3x

PB = 4x

Also, PT2 = PA × PB

62 = x (4x)

36 = 4x2

x2 = 9

=> x = 3cm

=> PB = 4x

= 4 × 3

PB = 12 cm

 

(34) Two chords AB, CD of a circle intersects externally at a point P. If PA = PC, Prove that AB = CD

Solution:

In fig two chords AB and CD intersects each other at point P outside the circle.

=> PA = PC —- (1)

Now, PA × PB = PC × PD

PB = PD —– (2)

=> PA – PB = PC – PD

=> AB = CD (Hence Proved)

 

(35) In the fig, AT is tangent to a circle at A. If ∠BAT = 45° and ∠BAC = 65°, find ∠ABC

Solution:

In fig, AT is the tangent to the circle at point A and AB is the chord to the circle.

= ∠ACB = ∠BAT

(Angles in the same segment)

= ∠ACB = ∠BAT = 45°

Now, in △ABC,

∠ABC + ∠BAC + ∠ACB = 180°

∠ABC + 65° + 45° = 180°

∠ABC = 180 – 110

∠ABC = 70°

 

(36) In the following fig, △ABC is isosceles with AB = AC. Prove that, the tangent at A to the circumcircle of △ABC is parallel to BC.

Solution:

In fig, △ABC is an isosceles triangle with AB = AC

AT is the tangent to the circumcircle at A.

In △ABC, AB = AC

=> ∠C = ∠B

(Since, angles opposite to equal sides are also equal)

But, here AT is the tangent and AC is the chord.

=> ∠TAC = ∠B

=> ∠B = ∠C

∴ ∠TAC = ∠C are the alternate angles.

=> AT is parallel to BC (Hence Proved)

 

(37) If the sides of a rectangle touch a circle, prove that the rectangle is a square.

Solution:

In fig, A circle touches the sides, AB, BC, CD and DA of a rectangle ABCD at points P, Q, R and S respectively.

Tangents from a point to the circle are always equal.

=> AP = AS

Similarity, BP = BQ, CR = CQ, DR = DS

=> AP + BP + CR + DR = AS + BQ + CQ + DS

AP + BP + CR + DR = AS + DS + BQ + CQ

AB + CD = AD + BC

But, AB = CD and AD = BC

=> AB + AB = BC + BC

2AB = 2BC

Hence, ABCD is a square => AB = BC & CD = DA

 

(39) (a) In the fig (i) AB is a chord of the circle with centre O, BT is tangent to the circle, if ∠OAQ = 32°, find the values of x and y.

(b) In the fig (1), O and O’ centres of two circles touching each other externally at the point P. The common tangent at point P meets at a direct common tangent AB at M.

Prove (i) M bisects AB (ii) ∠APB = 90°

Solution:

In fig, AB is a chord of a circle with centre ‘o’.

And, BT is the tangent to the circle and ∠OAB = 32°

In △OAB,

OA = OB (∵ radii of same circle)

∠OAB = ∠OBA

x = 32°

 

∠AOB = 180° – (x + 32°)

= 180 – 64°

∠AOB = 116°

Now, we can sec arc AB subtends ∠AOB at the centre and ∠ACB at the remaining portion of the circle.

∠AOB = 2 ∠ACB

=> ∠ACB = 1/2 ∠AOB

= 1/2 × 116

∠ACB = 58°

Thus, x = 32° & y = 58° are the required values.

 

(b)

In fig, two circles with centres O and O’ touch each other at point P externally.

From P, a common tangent is drawn & meets the common direct tangent AB at M.

(i) From point M, MA and MP are the tangents drawn.

=> MP = MP

Similarity MB = MP

=> MA = MB à M is the mid-point of AB.

(ii) Also, from fig, we can write

∠MAP + ∠MPB = ∠MPA + ∠MPB

∠APB ≠ ∠MAP + ∠MBP

But, ∠APB + ∠MAP + ∠MBP = 180°

∠APB + ∠APB = 180°

2 ∠APB = 180°

∠APB = 90° is the required

Updated: February 20, 2023 — 4:45 pm

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