ML Aggarwal Circles Exe-15.2 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-15.2 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-15.2 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-15 | Circles |

Writer / Book | Understanding |

Topics | Solutions of Exe-15.2 |

Academic Session | 2024-2025 |

ML Aggarwal Class 10 ICSE Maths Solutions

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

**Answer :**

##### in figure

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

Ext. ∠DCE = ∠BAD

∠BAD = x^{o}

Now arc BD subtends ∠BOD at the center

And ∠BAD at the remaining part of the circle.

∠BOD = 2∠BAD = 2 x

⇒ 2x = 150^{o }**(x = 75°)**

**(ii)** ∠BCD + ∠DCE = 180^{o }**(Linear pair)**

⇒ ∠BCD + 80° = 180^{o}

⇒ ∠BCD = 180°^{ }– 80°^{ }= 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 }– 115°^{ }=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}

**Question 2.**

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

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

**Answer :**

##### (a) Given, ∠AOC = 150° and AD = CD

We know that an angle subtends by an arc of a circle at the centre 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 angels in a cyclic quadrilateral is 180 ^{o})**

⇒ 75^{o} + ∠ADC = 180^{o}

⇒ ∠ADC + 180^{o }– 75^{o}

⇒ ∠ADC = 105^{o}

**(b) (i)** as 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}

**Circles Exe-15.2**

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 338)

**Question 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)**

**Answer :**

**(a)** ∠DBC = 58°

BD is diameter

∠DCB = 90° **(Angle in 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}

**Question 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 centre of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC.**

**Answer :**

##### (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}

**(b)** In the given figure, O is the centre of the circle.

∠AOE = 150°, ∠DAO = 51°

To find ∠BEC and ∠EBC

ABED is a cyclic quadrilateral

∴ Ext. ∠BEC = ∠DAB = 51°

∵ ∠AOE = 150°

∴ Ref. ∠AOE = 360° – 150° = 210°

Now arc ABE subtends ∠AOE at the centre and ∠AE at the remaining part of the circle.

∴ ∠ADE = ½ Ref. ∠AOE = 1/2 ×210° = 105°

But, Ext. ∠EBC = ∠ADE = 105°

Hence, ∠BEC = 51° and ∠EBC = 105°

**Question 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.**

**Answer :**

(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°

** ****Question 6.**

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

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

(i) ∠QBC (ii) ∠BCP

Figure see in your text book

**Answer :**

(a)

##### (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}

AQPD is a cyclic quadrilateral

∠A + ∠QPD = 180^{o}

**(b)** Join PQ

AQPD is a cyclic quadrilateral

∴ ∠A + ∠QPD = 180^{o}

⇒ 80° + ∠QPD = 180°

⇒ ∠QPD = 180° – 80°

= 100°

And ∠D + ∠AQP = 180°

⇒ 84° + ∠AQP = 180°

⇒ ∠AQP = 180° – 84° = 96°

Now, PQBC is a cyclic quadrilateral,

∴ Ext. ∠QPD = ∠QBC

⇒ ∠QBC = 100°

And ext. ∠AQP = ∠BCP

⇒ ∠BCP = 96°

**Circles Exe-15.2**

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 339)

**Question 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)**

**Answer :**

**(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 = 62^{0}

Now, in ∆AEC

∠ACE + ∠CAE + ∠AEC = 180^{o}

⇒ 43^{o }+ 62^{o }+ ∠AEC = 180^{o}

⇒ 105^{o }+ ∠AEC = 180^{o}

⇒ ∠AEC = 180^{o }– 105°^{ }= 75^{o}

But ∠ABD + ∠AED = 180°^{ }(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)**

**Question 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.**

**Answer :**

(a) Construction: Join BC, and AC then

ABCD is a cyclic quadrilateral.

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)**

**Circles Exe-15.2**

### ML Aggarwal Class 10 ICSE Maths Solutions

(Page 340)

**Question 9.**

** ****(a) In the figure (i) given below, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E. If ∠ADE = 70° and ∠OBA = 45°, calculate**

**(i) ∠OCA (ii) ∠BAC**

**(b) In figure (ii) given below, ABF is a straight line and BE || DC. If ∠DAB = 92° and ∠EBF = 20°, find :**

**(i) ∠BCD (ii) ∠ADC.**

**Answer :**

(a) ABCD is a cyclic quadrilateral

**Question 10.**

**(a) In the figure (ii) given below, PQRS is a cyclic quadrilateral in which PQ = QR and RS is produced to T. If ∠QPR = 52°, calculate ∠PST.**

**(b) In the figure (ii) given below, O is the centre of the circle. If ∠OAD = 50°, find the values of x and y.**

**Answer :**

**(a) PQRS is a cyclic quadrilateral in which**

PQ = QR

** ****Question 11.**

**(a) In the figure (i) given below, O is the centre of the circle. If ∠COD = 40° and ∠CBE = 100°, then find :**

**(i) ∠ADC**

**(ii) ∠DAC**

**(iii) ∠ODA**

**(iv) ∠OCA.**

**(b) In the figure (ii) given below, O is the centre of the circle. If ∠BAD = 75° and BC = CD, find :**

**(i) ∠BOD**

**(ii) ∠BCD**

**(iii) ∠BOC**

**(iv) ∠OBD (2009)**

**Answer :**

**(a) (i) ∴ ABCD is a cyclic quadrilateral.**

∴ Ext. ∠CBE = ∠ADC

⇒ ∠ADC = 100°

(ii) Arc CD subtends ∠COD at the centre

and ∠CAD at the remaining part of the circle

∴ ∠COD = 2 ∠CAD

= 2**∠OBD = 30°**

**= ∠OBD = 15°**

**Circles Exe-15.2**

ML Aggarwal Class 10 ICSE Maths Solutions

(Page 341)

**Question 12.**** ****In the adjoining figure, O is the centre and AOE is the diameter of the semicircle ABCDE. If AB = BC and ∠AEC = 50°, find :**

**(i) ∠CBE**

**(ii) ∠CDE**

**(iii) ∠AOB.**

**Prove that OB is parallel to EC.**

**Answer :**

In the given figure,

O is the centre of the semi-circle ABCDE

and AOE is the diameter. AB = BC, ∠AEC = 50°

**Question 13.**

**(a) In the figure (i) given below, ED and BC are two parallel chords of the circle and ABE, ACD are two st. lines. Prove that AED is an isosceles triangle.**

**(b) In the figure (ii) given below, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = RS.**

**Answer :**

(a) Given: Chord BC || ED,

ABE and ACD are straight lines.

To Prove: ∆AED is an isosceles triangle.

Proof: BCDE is a cyclic quadrilateral.

Ext. ∠ABC = ∠D …(i)

But BC || ED (given)

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

**Answer :**

In the given figure,

∆ABC is an isosceles triangle in which AB = AC.

A circle passing through B and C intersects

sides AB and AC at D and E.

To prove: DE || BC

Construction : Join DE.

∵ AB = AC

∠B = ∠C (angles opposite to equal sides)

But BCED is a cyclic quadrilateral

Ext. ∠ADE = ∠C

= ∠B (∵ ∠C = ∠B)

But these are corresponding angles

DE || BC

Hence proved.

**Question 15**

**(a) Prove that a cyclic parallelogram is a rectangle.**

**(b) Prove that a cyclic rhombus is a square.**

**Answer :**

(a) ABCD is a cyclic parallelogram.

To prove: ABCD is a rectangle

Proof: ABCD is a parallelogram

∠A = ∠C and ∠B = ∠D

**Question 16. ****In the adjoining figure, chords AB and CD of the circle are produced to meet at O. Prove that triangles ODB and OAC are similar. Given that CD = 2 cm, DO = 6 cm and BO = 3 cm, area of quad. CABD**

**Answer :**

In the given figure, AB and CD are chords of a circle.

They are produced to meet at O.

To prove : (i) ∆ODB ~ ∆OAC

If CD = 2 cm, DO = 6 cm, and BO = 3 cm

To find : AB and also area of the

**quad ABCD/area of ΔABC**

Construction : Join AC and BD

— : End of ML Aggarwal Circles Exe-15.2 Class 10 ICSE Maths Solutions : –

Return to: ML Aggarwal Solutions for ICSE Class-10

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