ML Aggarwal Circles Exe-15.1 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-15.1 Questions for Circles as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Circles Exe-15.1 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-15 | Circles |

Writer / Book | Understanding |

Topics | Solutions of Exe-15.1 |

Academic Session | 2024-2025 |

### Circles Exe-15.1

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 320)

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

**Answer :**

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

∠ADB = ∠ACB = 50°

Now in ∆ADB,

∠DAB + X + ∠ADB = 180°

= 42° + x + 50° = 180°

= 92° + x = 180°

x = 180° – 92°

x = 88°

**(ii)** In the given figure we have

= 32° + 45° + x = 180°

= 77° + x = 180°

x = 103°

**(iii)** From the given number we have

∠BAD = ∠BCD

∠BAD = 20°

∠BCD = 20°

∠CEA = 90°

∠CED = 90°

Now in triangle CED,

∠CED + ∠BCD + ∠CDE = 180°

⇒ 90° + 20° + x = 180°

⇒ 110° + x = 180°

⇒ x = 180° – 110°

⇒ x = 70°

**(iv)** In ∆ABC

∠ABC + ∠ABC + ∠BAC = 180°

⇒ 69° + 31°^{ }+ ∠BAC = 180°

⇒ ∠BAC = 180°^{ }– 100°

⇒ ∠BAC = 80°

Since, ∠BAC and ∠BAD are in the same

Segment.

∠BAD = x°^{ }= 80°

**(v)** Given ∠CPB = 120°, ∠ACP = 70°

To find, x°^{ }i,e., ∠PBD

Reflex ∠CPB = ∠BPO + ∠CPA

⇒ 120°^{ }= ∠BPD + ∠BPD

**(BPD = CPA are vertically opposite ∠s)**

⇒ 2∠BPD = 120°

⇒ ∠PBD = 120°/2 = 60°

Also ∠ACP and PBD are in the same segment

∠PBD + ∠ACP = 70°

Now, In ∆PBD

∠PBD + ∠PDB + ∠BPD = 180°

**(sum of all ∠s in a triangle)**

70°^{ }+ x° + 60° = 180°

⇒ x = 180°^{ }– 130°

⇒ x = 50°

**(vi)** ∠DAB = ∠BCD

**(Angles in the same segment of the circle)**

∠DAB = 25° **(∠BCD = 25° ^{ }given)**

In ∆DAP,

Ex. ∠CDA = ∠DAP + ∠DPA

⇒ x°^{ }= ∠DAB + ∠DPA

⇒ x°^{ }= 25°^{ }+ 35°

⇒ x° = 60°

**Circles Exe-15.1**

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 321)

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

**Answer :**

**(i) ∠ACB = ∠ADB**

(Angles in the same segment of a circle)

But ∠ADB = x°

∠ABC = x°

Now in ∆ABC

∠CAB + ∠ABC + ∠ACB = 180°

⇒ 40°^{ }+ 90°^{ }+ x°^{ }= 180°

**(AC is the diameter)**

⇒ 130°^{ + }x°^{ }= 180°

⇒ x°^{ }= 180°^{ }– 130° = 50°

**(ii)** ∠ACD = ∠ABD

**(angles in the same segment)**

∠ACD = x°

Now in triangle OAC,

OA = OC

**(radii of the same circle)**

∠ACO = ∠AOC

**(opposite angles of equal sides)**

Therefore, x° = 62°

**(iii)** ∠AOB + ∠AOC + ∠BOC = 360°

**(sum of angles at a point)**

∠AOB + 80° + 130°^{ }= 360°

⇒ ∠AOB + 210°^{ }= 360°

⇒ ∠AOB = 360° – 210° = 150°

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

∠AOB = 2 ∠ACB

⇒ ∠ACB = ½ ∠AOB = ½ × 150° = 75°

**(iv)** ∠ACB + ∠CBD = 180°

⇒ ∠ABC + 75° = 180°

⇒ ∠ABC = 180°^{ }– 75° = 105°

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

Reflex ∠AOC = 2 ∠ABC

= 2 × 105°^{ }

= 210°

**(v)** ∠AOC + ∠COB = 180°

⇒ 135° + ∠COB = 180°

⇒ ∠COB = 180° – 135° = 45°

Now arc BC Subtends reflex ∠COB at the center 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 center 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})**

∠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}

**Question 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**

**Answer :**

**(a) Construction: Join AB**

∠A = ∠C = 35° [∵ Alt angles]

⇒ ∠ABC = 35^{o}

**(b) ∠AOC + reflex ∠AOC = 360**^{o}

^{o}

⇒ 130^{o }+ Reflex ∠AOC = 360^{o}

⇒ Reflex ∠AOC = 360^{o }– 130^{o }= 230^{o}

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

Reflex ∠AOC = 2 ∠ABC

⇒ ∠ABC =1/2 reflex ∠AOC

= ½ × 230^{o }= 115^{o}

**Question 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.**

**Answer :**

**(a) ABCD is a cyclic quadrilateral**

∠B + ∠D = 180°

⇒ y + 40^{° }+ 45^{o }= 180^{o}

**(y + 85 ^{o }= 180^{o})**

⇒ y = 180^{o }– 85^{o} = 95^{o}

**(∠ACB = ∠ADB)**

⇒ x^{o }= 40

**(b) Arc ADC Subtends ∠AOC at the center 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}

### Circles Exe-15.1

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 322)

**Question 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**

**Answer :**

**(a) ∠NYB = 50°, ∠YNB = 20°.**

**∠**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 =360°

⇒ 300^{o }+ ∠OBC = 360°

⇒ ∠300^{o} + ∠OBC = 360°

⇒ ∠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 = 20°

But, ∠OBC = 60^{o}

∠CBA = ∠OBC – ∠OBA = 60^{o} – 20^{o }= 40^{o}

**Question 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.**

**Answer :**

**(a) 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,**

**(Exterior angle of a triangle is equal to the sum of its interior opposite angels)**

**(ii)** arc CF subtends ∠COF at the center and

∠CDF at the remaining part of the circle

∠COF = 2∠CDF = 2∠CDE

= 2×32^{o }= 2 ∠CDE

= 2×32^{o }

= 64

**Question 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.**

**Answer :**

**(a) (i) ∠PRB = ∠BAP**

(Angles in the same segment of the circle)

∴ ∠PRB = 35° (∵ ∠BAP = 35° given)

**(ii) In ∆PRQ,**

Ext. ∠APR = ∠PRQ + ∠PQR

= ∠PRB + ∠Q

= 35°+25° = 60°

But ∠APB = 90° **(Angle in a semi circle)**

∴ ∠BPR = ∠APB – ∠APR

= 90° -60° = 30°

**(iii)** ∠APR = ∠ABR

⇒ ∠ABR = 60°

In △PBQ,

Ext. ∠PBR = ∠Q + ∠BPQ

= 25° + 90° = 115°

**(b)** ∠B = ∠D **(Angles in the same segment)**

∴ ∠C = 40°

∠ACD = 90° **(Angle in the semi circle)**

Now in △ADC,

∠ACD + ∠D + ∠DAC = 180° **(Angles in a triangle)**

⇒ 90° + 40° + ∠DAC = 180°

⇒ 130° + ∠DAC = 180°

⇒ ∠DAC = 180° – 130° = 50°

### Circles Exe-15.1

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 323)

**Question 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**

**Answer :**

**Given that:**

(a) Arc AB subtends ∠APB at the centre 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} = 230°

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 + 56°^{ }= 180°

⇒ 2∠A = 180^{o }– 56°^{ }= 124^{o}

⇒ ∠A = 124/2 = 62^{o }or ∠CAB = 62°

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}

**Question 9.**

**(a) In the figure (i) given below, chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.**

**(b) In the figure (ii) given below, C is a point on the minor 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.**

**Answer :**

**(a) ∠CBE = ∠CAE**

(Angle in the same segment of a circle)

⇒ ∠CAE = 65°

∠AEC = 90° (Angle in a semi circle)

Now in ∆AEC

∠AEC + ∠CAE + ∠ACE = 180° (Angle of a triangle)

⇒ 90°+ 65° +∠ACE = 180°

so ⇒ 155° + ∠ACE = 180°

hence ⇒ ∠ACE = 180° – 155° – 25°

∵AC || ED (given)

∴∠ACE = ∠DEC (alternate angles)

∴∠DEC = 25°

**Question 10.**

**(a) In the figure (i) given below, straight lines AB and CD pass through the centre O of a circle. If ∠OCE = 40° and ∠AOD = 75°, find the number of degrees in :**

(i) ∠CDE

(ii) ∠OBE.

**(b) In the figure (ii) given below, I is the incentre of ∆ABC. AI produced meets the circumcircle of ∆ABC at D. Given that ∠ABC = 55° and ∠ACB = 65°, calculate**

(i) ∠BCD

(ii) ∠CBD

(iii) ∠DCI

(iv) ∠BIC.

**Answer :**

**(a) (i) ∠CED = 90° (Angle in semi-circle)**

In ∆CED

∠CED + ∠CDE + ∠DCE = 180°

so ⇒ 90° +∠CDE + 40° = 180°

therefore ⇒ 130° + ∠CDE = 180°

hence ⇒ ∠CDE = 180° – 130° = 50°

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

**Answer :**

In the given figure, O is the centre of circumcenter of ∆ABC.

D is mid-point of BC. BO, CO and OD are joined.

**Circles Exe-15.1**

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 324)

**Question 12. ****In the adjoining figure, AB and CD are equal chords. AD and BC intersect at E. Prove that AE = CE and BE = DE.**

**Answer :**

In the given figure, AB and CD are two equal chords

AD and BC intersect each other at E.

To prove : AE = CE and BE = DE

Proof:

In ∆AEB and ∆CED

AB = CD (given)

∠A = ∠C (angles in the same segment)

∠B = ∠D (angles in the same segment)

∴ ∆AEB ≅ ∆CED (ASA axiom)

∴ AE = CE and BE = DE (c.p.c.t.)

**Question 13.**

**(a) In the figure (i) given below, AB is a diameter of a circle with centre O. AC and BD are perpendiculars on a line PQ. BD meets the circle at E. Prove that AC = ED.**

**(b) In the figure (ii) given below, O is the centre of a circle. Chord CD is parallel to the diameter AB. If ∠ABC = 25°, calculate ∠CED.**

**Answer :**

**(a) Given: AB is the diameter of a circle with centre O.**

AC and BD are perpendiculars on a line PQ,

such that BD meets the circle at E.

**Question 14. ****In the adjoining figure, 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.**

**Answer :**

**Given: In the figure,**

OABC is a || gm and O is the centre of the circle.

BC is produced to meet the circle at D.

To Prove : ∠ABC = 2∠OAD.

Construction: Join AD.

**Question 15.**

(a) In the figure (i) given below, P is the point of intersection of the chords BC and AQ such that AB = AP. Prove that CP = CQ.

(b) In the figure (i) given below, AB = AC = CD, ∠ADC = 38°. Calculate :

(i) ∠ABC (ii) ∠BEC.

**Answer :**

**(a) Given:**

Two chords AQ and BC intersect each other at P

inside the circle. AB and CQ are joined and AB = AP.

To Prove : CP = CQ

Construction : Join AC.

Proof: In ∆ABP and ∆CQP

∴ ∠B = ∠Q

**Circles Exe-15.1**

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 325)

**Question 16.**

**(a) In the figure (i) given below, CP bisects ∠ACB. Prove that DP bisects ∠ADB.**

**(b) In the figure (ii) given below, BD bisects ∠ABC. Prove that AB/BD=BE/BC**

**Answer :**

**(a)Given: In the figure, CP is the bisector of**

∠ACB meeting the circle at P.

PD is joined

**Question 17.**

**(a) In the figure (ii) given below, chords AB and CD of a circle intersect at E.**

**(i) Prove that triangles ADE and CBE are similar.**

**(ii) Given DC =12 cm, DE = 4 cm and AE = 16 cm, calculate the length of BE.**

**(b) In the figure (ii) given below, AB and CD are two intersecting chords of a circle. Name two triangles which are similar. Hence, calculate CP given that AP = 6cm, PB = 4 cm, and CD = 14 cm (PC > PD).**

**Answer :**

**(a) Given: Two chords AB and CD intersect each other**

at E inside the circle.

**Question 18. ****In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE. (2008)**

**Answer :**

In the figure, AE and BC intersect each other at D.

AB is joined.

**Question 19.**

** ****(a) In the figure (i) given below, PR is a diameter of the circle, PQ = 7 cm, QR = 6 cm and RS = 2 cm. Calculate the perimeter of the cyclic quadrilateral PQRS.**

**(b) In the figure (ii) given below, the diagonals of a cyclic quadrilateral ABCD intersect in P and the area of the triangle APB is 24 cm². If AB = 8 cm and CD = 5 cm, calculate the area of ∆DPC.**

**Answer :**

(a) PR is the diameter of the circle

PQ = 7 cm, QR = 6 cm, RS = 2 cm.

**Circles Exe-15.1**

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 326)

**Question 20.**

** ****(a) In the figure (i) given below, QPX is the bisector of ∠YXZ of the triangle XYZ. Prove that XY : XQ = XP : XZ,**

**(b) In the figure (ii) given below, chords BA and DC of a circle meet at P. Prove that:**

**(i) ∠PAD = ∠PCB**

**(ii) PA. PB = PC . PD.**

**Answer :**

**(a) Given: ∆XYZ is inscribed in a circle.**

Bisector of ∠YXZ meets the circle at Q.

QY is joined.

To Prove : XY : XQ = XP : XZ

— : End of ML Aggarwal Circles Exe-15.1 Class 10 ICSE Maths Solutions : –

Return to: ML Aggarwal Solutions for ICSE Class-10

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