These AP 9th Class Maths Important Questions 9th Lesson Circles will help students prepare well for the exams.

## AP Board Class 9 Maths 9th Lesson Circles Important Questions

### 9th Class Maths Circles 2 Marks Important Questions

Question 1.

Define the terms.

i) arc

ii) chord

Solution:

i) The part between any two points on j the circle is called an arc.

PXQ is minor arc whereas PYQ is major arc.

ii) A line segment joining any two points on the circle is called a chord.

MN is the chord.

Question 2.

Define a ‘Sector’.

Solution:

The area enclosed by an arc and the two radii joining the centre to the end points of an arc is called a sector.

The shaded portion is minor sector and the unshaded portion is major sector.

Question 3.

In the figure, if O is the centre of the circle and ∠CBA = 35°, find the value of x.

Solution:

∠CBA = 35° (Given)

∠CAB = x° (Given)

∠BCA = 90° | Angle in a semi-circle is 90°

In Δ ABC,

∠CBA + ∠CAB + ∠BCA = 180° [Sum of the angles in a triangle is 180°]

⇒ 35°+ x° + 90° = 180° ⇒ x° + 125° = 180°

⇒ x° = 180° – 125° ⇒ x° = 55° ⇒ x = 55

Question 4.

In figure, ΔABC is an equilateral triangle. Find the value of x.

Solution:

∵ ΔABC is an equilateral triangle.

∴ ∠ABC = ∠B AC = ∠ACB = 60° …….. (1)

| Each angle of an equilateral triangle is 60°

∠BDC = ∠BAC [Angles in the same segment of a circle are equal]

⇒ x° = 60° [From (1)]

⇒ x = 60

Question 5.

Define a ‘Cyclic Quadrilateral’.

Solution:

If all the vertices of a quadrilateral lie on the same circle, then the quadrilateral is called cyclic quadrilateral. ABCD is a cyclic quadrilateral.

Question 6.

In the given figure,

1) What is the shaded portion called ?

2) Write arc DCB symbolically ?

3) What is ∠ACB called ? What is its value ?

Solution:

1) The shaded portion is called minor segment.

2) \(\overparen{\mathrm{DCB}}\)

3) ∠ACB is called the angle in the semicircle. ∠ACB = 90°.

Question 7.

Find the value of ∠AOB.

Solution:

∠APB = ∠QPR [Vertically Opposite Angles]

⇒ ∠APB = 35° ………….. (1)

∠AOB = 2 ∠APB [The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle.]

⇒ ∠AOB = 2(35°) [From(1)]

⇒ ∠AOB = 70°

Question 8.

O is the centre of a circle that passes through P, Q, R and S, as shown in the figure. SR is produced to X. If ∠QRX = 133°, find x.

Solution:

∠QRX = ∠SPQ

An exterior angle of a cyclic quadrilateral is equal to its interior opposite angles

⇒ 133° = (4x + 13)° ⇒ (4x + 13)° = 133°

⇒ 4x° = 133° – 13° ⇒ 4x° = 120°

⇒ x° = \(\frac{120^{\circ}}{4}\) ⇒ x° = 30° ⇒ x = 30

Question 9.

In the given figure, find the value of x.

Solution:

∵ ABCD is a cyclic quadrilateral.

∴ ∠BAD + ∠BCD = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

⇒ (2x + 4)° + (4x – 64)° = 180°

⇒ (2x + 4x – 60)° = 180°

⇒ (6x – 60)° = 180°

⇒ 6x – 60° = 180

⇒ 6x = 240

⇒ x = 40

Question 10.

In the given figure, O is the centre of the Circle. Find the value of x.

Solution:

∠DBC = ∠DAC (Angles in- the same segment of a circle are equal)

⇒ ∠DBC = x° ………… (1)

⇒ ∠DCB = 90° ………… (2)

[Angle in a semi-circle is 90°]

In Δ BCD,

∠DCB + ∠BDC + ∠DBC = 180° [The sum of the angles of a triangle is 180°]

⇒ 90° + 2x° + x° = 180° ,

⇒ 3x° = 90°

⇒ x° = 30° ⇒ x = 30

### 9th Class Maths Circles 4 Marks Important Questions

Question 1.

In the given figure, ‘O’ is the centre of the circle and AB, CD are equal chords. If ∠AOB = 70°. Find the angles of ΔOCD.

Solution:

‘O’ is the centre of the circle.

AB, CD are equal chords

⇒ They subtend equal angles at the centre.

∴ ∠AOB = ∠COD = 70°

Now in ΔOCD

∠OCD = ∠ODC [∵ OC = OD; radii angles opp. to equal sides]

∴ ∠OCD + ∠ODC + 70° = 180°

⇒ ∠OCD +∠ODC = 180° – 70° = 110°

∴ ∠OCD + ∠ODC = \(\frac{110^{\circ}}{2}\) = 55°

Question 2.

In the figure, ‘O’ is the centre of the circle. OM = 3 cm and AB = 8 cm. Find the radius of the circle.

Solution:

‘O’ is the centre of the circle.

OM bisects AB.

∴ AM = \(\frac{\mathrm{AB}}{2}\) = \(\frac{8}{2}\) = 4 cm

OA^{2} = OM^{2} + AM^{2} [∵ Pythagoras theorem]

OA = \(\sqrt{3^2+4^2}\) = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5 cm

Question 3.

In the figure, ‘O’ is the centre of the circle and ∠POR = 120°. Find ∠PQR and ∠PSR.

Solution:

‘O’ is the centre; ∠POR = 120°

∠PQR = \(\frac{1}{2}\)∠POR [∵ angle made by an arc at the centre is twice the angle made by it on the remaining part]

∠PSR = \(\frac{1}{2}\) [Angle made by \(\widehat{\mathrm{PQR}}\) at the centre]

∴ ∠PSR = \(\frac{1}{2}\)[360° – 120°] from the fig.

= \(\frac{1}{2}\) × 240 = 120°

Question 4.

For each of the following, draw a circle and inscribe the figure given. If a polygon of the given type can’t be inscribed, write not possible.

a) Obtuse triangle

b) Non-rectangular parallelogram

c) Acute isosceles triangle

d) A quadrilateral PQRS with \(\overline{\mathbf{P R}}\) as diameter.

Solution:

a) Obtuse triangle

b) Non-rectangular parallelogram (Not possible)

c) Acute isosceles triangle

d) A quadrilateral PQRS with \(\overline{\mathbf{P R}}\) as diameter

Question 5.

Two chords PQ and RS of a circle are parallel to each other and AB is the perpendicular bisector of PQ. Without using any construction, prove that AB bisects RS.

Solution:

Given : Two chords PQ and RS of a circles are parallel to each other and AB is the perpendicular bisector of PQ.

To Prove : AB bisects RS.

Proof : ∵ AB is the perpendicular bisector of PQ.

∴ AB passes through the centre ‘O’ of the circle

The perpendicular bisector of a chord of a circle passes through the centre of the circle

and AB ⊥ PQ

But PQ || RS

∴ AB ⊥ RS

⇒ OB ⊥ RS

∴ OB bisects RS

The perpendicular drawn from the centre of a circle to a chord bisects the chord

⇒ AB bisects RS.

Question 6.

In figure, AB and AC are two equal chords of a circle whose centre is O. If OD ⊥ AB and OE ⊥ AC, prove that ADE is an isosceles triangle.

Solution:

Given : AB and AC are two equal chords of a circle whose centre is O. OD ⊥ AB and OE ⊥ AC.

To Prove : ∠ADE is an isosceles triangle.

Proof : AB = AC | Given

∴ OD = OE ………… (1)

Equal chords of a circle are equidistant from the centre

In Δ ODE,

OD = OE | From(1)

∴ ∠ OED = ∠ ODE

Angles opposite to equal sides of a triangle are equal

⇒ 90° – ∠OED = 90° – ∠ODE

⇒ ∠OEA – ∠ OED = ∠ODA – ∠ODE

⇒ ∠AED = ∠ADE

∴ AD = AE

Sides opposite to equal angles of a triangle are equal

∴ ∠ADE is an isosceles triangle.

Question 7.

The circumcentre of Δ ABC is O. Prove that ∠OBC + ∠BAC = 90°.

Solution:

Given : The circumcentre of ΔABC is O

To Prove : ∠OBC + ∠BAC = 90°

Proof : OB = OC [Radii of the same circle]

∴ ∠OCB = ∠OBC [Angles opposite to equal sides of a triangle are equal)

= θ (say)

In Δ OBC,

∠ OBC + ∠ OCB + ∠ BOC = 180° [The sum of the angles of a triangle is 180°]

⇒ θ + θ + ∠BOC = 180° [From (1)].

⇒ ∠BOC = 180° – 2θ ………….. (2)

Also, ∠BOC – 2∠BAC …………… (3) (The angle subtended by an arc of a circle at the contre is double the angle subtended by it at any point on the re-maining part of the circle)

From (2) and (3),

180° – 2θ = 2∠BAC

⇒ ∠BAC = 90° – θ [Dividinging by 2]

⇒ ∠BAC = 90° – ∠OBC (from 1)

⇒ ∠OBC + ∠BAC = 90°

Question 8.

D is a point on the circumference of circumeircle of ΔABC in which AB = AC uch that B and D are on opposite sides of AC. If CD is produced to a point E such that CE = BD, prove that AD = AE.

Solution:

Given : D is a point on the circumference of circumeircle of Δ ABC in which AB = AC such that B and D are on opposite sides of AC.

CD is produced to a point E such that CE – BD.

To Prove : AD = AE

Proof : In Δ ACE and Δ ABD,

∠ACE = ∠ ABD (Angles in the same segment of a circle are equal)

AC = AB (Given)

CE = BD (Given)

∴ Δ ACE ≅ Δ ABD (SAS Rule).

∴ AE = AD (CPCT)

⇒ AD = AE.

Question 9.

Based on the above information, answer the following questions :

i) In fig 1, ‘O’ is the centre of the circle. If x = 50, then y =

A) 75

B) 50

C) 25

D) 37

Answer:

C) 25

ii) In fig. 2, If a = 20, then b =

A) 20

B) 40

C) 60

D) cannot be determined

Answer:

A) 20

iii) In fig. 3, if p = 60, then q =

A) 60

B) 30

C) 90

D) 120

Answer:

D) 120

iv) In fig. 4, if a = 50, then b =

A) 100

B) 50

C) 130

D) cannot be determined

Answer:

B) 50

v) In fig. 5, AB is a diameter. If x = 60, then y =

A) 60

B) 30

C) 90

D) 45

Answer:

B) 30

### 9th Class Maths Circles 8 Marks Important Questions

Question 1.

In the figure, ‘O’ is the centre of the circle. ∠AOB = 100°, find ∠ADB.

Solution:

‘O’ is the centre.

∠AOB = 100°

Thus ∠ACB = \(\frac{1}{2}\)∠AOB [∵ angle made by an arc at the centre is twice the angle made by it on the remaining part]

= \(\frac{1}{2}\) × 100° = 50°

∠ACB and ∠ADB are supplementary [∵ Opp. angles of a cyclic quadrilateral]

∴ ∠ADB = 180° – 50° = 130°

[OR]

∠ADB is the angle made by the major arc \(\widehat{A C B}\) at D.

∴ ∠ADB = \(\frac{1}{2}\)∠AOB [where ∠AOB is the angle made by \(\widehat{A C B}\) at the centre]

= \(\frac{1}{2}\) [360° – 100°] [from the figure]

= \(\frac{1}{2}\) × 260° – 130°

Question 2.

In the figure, ∠BAD = 40°, then find ∠BCD.

Solution:

‘O’ is the centre of the circle.

∴ In ΔOAB; OA = OB (radii)

∴ ∠OAB = ∠OBA = 40° (∵ angles opp. to equal sides)

Now ∠AOB = 180° – (40° + 40°) (∵ angle sum property of ΔOAB)

= 180° – 80° = 100°

But ∠AOB = ∠COD = 100°

Also ∠OCD = ∠ODC [OC = OD]

= 40° as in ΔOAB

∴ ∠BCD = 40°

(OR)

In ΔOAB and ΔOCD

OA = OD (radii)

OB = OC (radii)

∠AOB = ∠COD (vertically opp. angles)

∴ ΔOAB ≅ ΔOCD

∴ ∠BCD – ∠OBA = 40° [∵ OB = OA ⇒ ∠DAB = ∠DBA]

Question 3.

If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

Let two circles with centre P and Q intersect at two distinct points say A and B.

Join A, B to form the common chord

\(\overline{\mathrm{AB}}\). Let ‘O’ be the midpoint of AB. Join ‘O’ with P and Q.

Now in ΔAPO and ΔBPO

AP = BP (radii)

PO – PO (common)

AO = BO (∵ O is the midpoint)

∴ ΔAPO ≅ ΔBPO (S.S.S. congruence)

Also ∠AOP = ∠BOP (CPCT)

But these are linear pair of angles.

∴ ∠AOP = ∠BOP = 90°

Similarly in ΔAOQ and ΔBOQ

AQ = BQ (radii)

AO = BO (∵ O is the midpoint of AB)

OQ = OQ (common)

ΔAOQ ≅ ΔBOQ

Also ∠AOQ = ∠BOQ (CPCT)

Also ∠AOQ + ∠BOQ = 180° (linear pair of angles)

∴ ∠AOQ = ∠BOQ = \(\frac{180^{\circ}}{2}\) = 90°

Now ∠AOP + ∠AOQ = 180°

∴ PQ is a line.

Hence the proof.

Question 4.

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre, if the distance between AB and CD is 6 cm, find the radius of the circle.

Solution:

Let the radius of the circle be r cm. Let OM = x cm.

Then, ON = (6 – x) cm.

∵ OM ⊥ CD

∴ M is the mid-point of CD. (The perpendicular from the centre of a circle to a chord bisects the chord)

∴ MD = MC = \(\frac{1}{2}\)CD = \(\frac{1}{2}\)(11) cm = \(\frac{11}{2}\) cm

(The perpendicular from the centre of a circle to a chord bisects the chord)

∴ NB = AN = \(\frac{1}{2}\)AB = \(\frac{1}{2}\)(5) = \(\frac{5}{2}\) cm

In right triangle ONB,

OB^{2} = ON^{2} + NB^{2} (By Pythagoras Theorem)

⇒ r^{2} = (6 – x)^{2} + (\(\frac{5}{2}\))^{2} ……………. (1)

In right triangle OMD,

OD^{2} = OM^{2} + MD^{2} (By Pythagoras Theorem)

⇒ r^{2} = x^{2} + (\(\frac{11}{2}\))^{2} …………. (2)

From (1) and (2). we get

(6 – x)^{2} + (\(\frac{5}{2}\))^{2} = x^{2} + (\(\frac{11}{2}\))^{2}

⇒ 36 – 12x + x^{2} + \(\frac{25}{4}\) = x^{2} + \(\frac{121}{4}\)

⇒ 12x = 36 + \(\frac{25}{4}\) – \(\frac{121}{4}\)

⇒ 12x = 12

⇒ x = \(\frac{12}{12}\) = 1

Putting x = 1 in (2), we get

r^{2} + (1)^{2} + (\(\frac{11}{2}\))^{2} = 1 + \(\frac{121}{4}\) = \(\frac{125}{4}\)

r = \(\frac{5 \sqrt{5}}{2}\)

Hence, the radius of the circle is \(\frac{5 \sqrt{5}}{2}\) cm.

Question 5.

AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.

Solution:

Given : AC and BD are chords of a circle that bisect each other.

To Prove : i) AC and BD are diameters

ii) ABCD is a rectangle.

Construction : Join AB, BC, CD and DA.

Proof : Let the chords AC and BD intersect each other at O.

i) In Δ OAB and Δ OCD,

OA = OC (∵ O is the mid-point ol AC)

∠AOB = ∠COD [Vertically Opposite Angles]

OB = OD [∵ O is the midpoint of BD)

∴ Δ OAB ≅ Δ OCD [SAS congruence rule]

∴ AB = CD [CPCT]

\(\overparen{\mathrm{AB}}\) ≅ \(\overparen{\mathrm{CD}}\) …………….. (1)

If two chords of a circle are equal, then their corresponding arcs are congruent

In Δ OAD and Δ OCB,

OA = OC |∵ O is the mid-point of AC

∠AOD = ∠COB [Vertically Opposite Angles]

OD = OB [∵ O is the mid-point of BD]

∴ Δ OAD ≅ Δ OCB [SAS congruence rule]

∴ AD = CB [CPCT]

\(\overparen{\mathrm{AD}}\) ≅ \(\overparen{\mathrm{CB}}\) …………… (2)

(If two chords of a circle are equal, then their corresponding arcs are congruent)

From (1) and (2),

\(\overparen{\mathrm{AB}}\) + \(\overparen{\mathrm{AD}}\) = \(\overparen{\mathrm{CD}}\) = \(\overparen{\mathrm{CB}}\)

\(\overparen{\mathrm{DAB}}\) = \(\overparen{\mathrm{DCB}}\)

BD divides the circle into two equal parts.

BD is a diameter.

Similarly, we can show that AC is a diameter.

ii) BD is a diameter. [Proved in (i)]

∴ ∠A = ∠C = 90° (Angle in a semi-circle is 90°)

AC is a diameter [Proved in (i)]

∴ ∠B , ∠D = 90° (Angle in a semi-circle is 90°)

∴ ∠A = ∠B = ∠C = ∠D = 90°

⇒ ABCD is a rectangle.