AP 9th Class Maths Chapter 9 Circles Important Questions

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
OA² = OM² + AM² [∵ 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,
OB² = ON² + NB² (By Pythagoras Theorem)
⇒ r² = (6 – x)² + (\(\frac{5}{2}\))² ……………. (1)
In right triangle OMD,
OD² = OM² + MD² (By Pythagoras Theorem)
⇒ r² = x² + (\(\frac{11}{2}\))² …………. (2)
From (1) and (2). we get
(6 – x)² + (\(\frac{5}{2}\))² = x² + (\(\frac{11}{2}\))²
⇒ 36 – 12x + x² + \(\frac{25}{4}\) = x² + \(\frac{121}{4}\)
⇒ 12x = 36 + \(\frac{25}{4}\) – \(\frac{121}{4}\)
⇒ 12x = 12
⇒ x = \(\frac{12}{12}\) = 1
Putting x = 1 in (2), we get
r² + (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.