NCERT Class 9 Mathematics Chapter 9 Circles Solutions, NCERT Solutions For Class 9 Maths, NCERT Class 9 Mathematics Chapter 9 Circles Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 9 Mathematics Chapter 9 Circles Notes and select needs one.
NCERT Class 9 Mathematics Chapter 9 Circles
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 9 Mathematics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 9 Mathematics Chapter 9 Circles Solutions for All Subject, You can practice these here.
Circles
Chapter – 9
| Exercise 9.1 |
Q.1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of a congruent circle subtend equal angles at their centres.
Ans: Given: AB and CD are two equal chords of congruent circles with centres O and O’ respectively.
To Prove: ∠AOB= ∠CO’D
Proof: In ∆AOB and ∆ OCD,
OA = OC | Radii of congruent circles
OB = OD | Radii of congruent circles
AB = CD | Given
∴ ∆ OAB ≅ OCD | SSS Rule
∴ ∠AOB ≅ ∠ COD | c.p.c.t.
∴ ∠AOB= ∠ COD
Q.2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Ans: ∠ AOB and ∠ COD are the two equal angles subtended by the chords AB and CD of two congruent circles with centres O and O respectively.
To Prove: AB = CD
Prove: In ΔΟΑΒ and ∆OCD,
OA = O’C | Radii of congruent circles
OB = O’D | Radii of congruent circles
∠AOB = ∠CO’D | Given
∴ ΔΟΑΒ ≅ O’CD | SAS Rule
∴ AB = CD | c.p.c.t.
| Exercise 9.2 |
Q.1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Ans: We know that if two circles intersect each other at two points, then line joining their centres is the perpendicular bisector of their common chord.
∴ Length of the common chord
= PQ = 2O’P
= 2 x 3cm = 6 cm
Q.2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Ans: Given: A circle with centre O.
Its two equal chords AB and CD intersect at E.
To Prove: AE = DE and CE = BE
Construction: Draw ON⊥CD. Join OE
Proof: In ∆OME and ∆ONE,
OM = ON
| ∵ Equal chords of a circle are equidistant from the centre
OE = OE | Common
∴ ∆ OME ≅ ∆ONE | R.H.S.
∴ ME = NE | c.p.c.t.
⇒ AM + ME = DN + NE | AM = DN = 1/2 AB = 1/2 CD
⇒ AB – AE = CD – DE | ∵ AB = CD
⇒ BE = CE | Given
Q.3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Ans: Given: Two equal chords AB and CD of a circle with centre O intersect within the circle. Their point of intersection is E.
To Prove: ∠OEA = ∠ OED
Construction: Join OA and OD.
Proof: In ∆OEA and ∆OED,
OE = OE | Common
OA = OD | Radii of a circle
AE = DE | Proved in Example 2 above
∴ ∆ OEA ≅ ∆ OED | SSS Rule
∴ ∠OEA = ∠OED | c.p.c
Q.4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D. Prove that
AB = CD [see figure]
Ans: Given: A line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.
To Prove: AB = CD
Construction: Draw OM⊥BC
Proof: The perpendicular drawn from the centre of a circle to a chord bisects the chord.
∴ AM=DM …(1)
and BM = CM …(2)
Subtracting (2) from (1), we get
AM-BM-DM-CM
⇒ AB = CD
Q.5. There girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma and Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Ans: In ∆NOR and ∆NOM,
ON=ON | Common
∠NOR = ∠ΝΟΜ
|∵ Equal chords of a circle subtend equal angles at the centre
OR=OM | Radii of a circle
∴ ∆NOR≅∆NOM | SAS Rule
∴ ∠ONR = ∠ONM | c.p.c.t.
and NR = NM | c.p.c.t.
But ∠ONR +∠ONM=180⁰ | Linear Pair Axiom
∴ ∠ONR = ∠ONM = 90⁰
∴ ON is the perpendicular bisector of RM
Draw bisector SN of ∠RSM to intersect the chord RM in N.
In ∆RSM and ∆MSN,
RS = MS (= 6 cm each)
SN=SN | Common
∠RSN = ∠MSN | By construction
∴ ∆RSN≅ ∆MSN | SAS Rule
∴ ∠RNS=∠MNS | c.p.c.t.
and RN = MN | c.p.c.t.
But ∠RNS +∠MNS = 180⁰ | Linear Pair Axiom
∴ ∠RNS = ∠MNS = 90⁰
∴ SN is the perpendicular bisector of RM and therefore passes through O when produced.
Let ON = xm
Then SN = (5 – x)m
In right triangle ONR,
x²+ RN² = 5² …(1) | By Pythagoras Theorem In right triangle SNR,
(5 – x)² + RN² = 6² …(2) | By Pythagoras Theorem
From (1),
RN² = 5²- x²
From (2),
RN² = 6²- (5 – x)²
Equating the two values of RN² ,we get
5² – x² = 6² – (5 – x)²
⇒ 25-x² = 36-(25-10x+x)²⇒ 25 – x² = 36 – 25 + 10x – x²
⇒ 25 – x² = 11+10x-x²⇒ 25-11=10x
⇒14 = 10x ⇒10 x = 14
Put x = 1.4 in (1), we get
(1.4)² + RN² = 5²
⇒ RN² = 5² – (1.4)²
⇒ RN² = 25 – 1.96
⇒ RN² = 23.04
⇒ RN = √(23.04)
⇒ RN = 4.8
∴ RM = 2RN = 2×4.8m = 9.6m
Hence the distance between Reshma and Mandip is 9.6 m.
Q.6. A circular part of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
Ans: Length of the string of each phone = Given equal distance.
| Exercise 9.3 |
Q.1. In figure, A, B and C are three points on a circle with centre O such that angle BOC = 30⁰ and ∠ AOB = 60⁰ If D is a point on the circle other than the arc ABC, find ∠ADC.
Ans:
| The angle subtended by an the angle subtended by it at any point on the remaining part of the circle
Q.2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Ans: ∵ OA = OB = AB | Given
∴ ∆AOB is equilateral
∴ ∠ AOB = 60⁰
| The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
Now, ∵ ADBC is a cyclic quadrilateral.
∴ ∠ADB + ∠ACB = 180⁰
| The sum of either pair of opposite angles of a cyclic quadrilateral is 180⁰
⇒∠ADB + 30⁰= 180⁰
⇒ ∠ ADB = 180⁰- 30⁰
⇒ ∠ADB = 150⁰
Q.3. In figure, ∠ PQR = 100⁰, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Ans: Take a point S in the major arc. Join PS and RS.
∵ PQRS is a cyclic quadrilateral.
∵ ∠ PQR + ∠PSQ = 180⁰
| The sum of either pair of opposite angles of a cyclic quadrilateral is 180⁰
⇒ 100⁰ + ∠PSR = 180⁰
⇒ ∠ PSR = 180⁰ – 100⁰
⇒∠PSR = 80⁰ …(1)
Now, ∠POR = 2∠PSR = 2 × 80⁰ = 160⁰ … (2) | Using (1)
| The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle In, DOPR
∵ OP=OR | Radii of a circle
∴ ∠OPR = ∠ORP …(3)
| Angles opposite to equal sides of a triangle are equal In ∆OPR,
∠OPR + ∠ORP + ∠POR = 180⁰
Sum of all the angles of a triangle is 180⁰
⇒ ∠OPR + ∠OPR + 160 = 180⁰ | Using (2) and (1)
⇒ 2 ∠OPR + 160⁰ = 180⁰
⇒ 2 ∠OPR = 180⁰ – 160⁰ = 20⁰
⇒ ∠ OPR = 10⁰
Q.4. In figure,
∠ ABC = 69⁰,∠ACB = 31⁰ find ∠BDC
Ans: In ∆ ABC,
| Sum of all the angles of a triangle is 180⁰
∠ BAC +∠ABC + ∠ ACB = 180⁰
⇒ ∠BAC +69° + 31°= 180°
⇒ ∠ BAC + 100⁰ = 180⁰
⇒ ∠BAC = 180⁰- 100⁰= 80⁰
Now, ∠ BDC = ∠ BAC = 80⁰
| Angles in the same segment of a circle are equal
Q.5. In figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130⁰ and ∠ BCD = 20⁰ Find ∠ BAC.
Ans: ∠CED+ ∠ BEC = 180⁰ | Linear Pair Axiom
⇒∠CED + 130⁰ = 180⁰
⇒∠CED = 180⁰- 130⁰ = 50⁰ …(1)
∠ ECD = 20⁰ …(2)
In ∆CED,
∠CED + ∠ECD + ∠CDE = 180⁰
| Sum of all the angles of a triangle is 180⁰
⇒ 50⁰+20⁰ +∠CDE = 180⁰ | Using (1) and (2)
⇒ 70⁰+∠CDE = 180⁰
⇒ ∠CDE = 180⁰ – 130⁰ = 50⁰ …(1)
∠ ECD = 20⁰ …(2)
In ∆CED,
∠CED + ∠ECD + ∠CDE = 180⁰
| Sum of all the angles of a triangle is 180⁰
⇒ 50⁰ + 20⁰ +∠CDE = 180⁰ | Using (1) and (2)
⇒ 70⁰ +∠CDE = 180⁰
⇒ ∠CDE = 180⁰ – 70⁰
⇒∠CDE = 110⁰
Now, ∠BAC= ∠CDE | Angles in the same segment of a circle are equal
= 110⁰ | Using (3)
Q.6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If angle ∠DBC = 70⁰, ∠BAC is 30⁰, find ∠BCD. Further, if AB = BC, find ∠ECD.
Ans: ∠CDB=∠BAC | Angles in the same segment of a circle are equal
= 30° …(1)
∠ DBC = 70⁰ …(2)
In ∆BCD,
∠BCD +∠BDC + ∠CDB = 180⁰
| Sum of the angles of triangle is 180⁰
⇒ ∠BCD + 70⁰+ 30⁰ = 180⁰ | Using (1) and (2)
⇒ ∠BCD + 100⁰ = 180⁰
⇒ ∠BCD = 180⁰ – 100⁰
⇒ ∠BCD = 80⁰ …(3)
In ∆ABC, AB = BC
∴ ∠BCA = ∠BAC | Angles opposite to equal sides of A a triangle are equal
= 30⁰ …(4) | ∵ ∠ BAC=30⁰ (given)
Now, ∠ BCD = 80⁰ | From (3)
⇒ ∠BCA+ ∠ECD = 80⁰
⇒ 30⁰ + ∠ECD = 80⁰
⇒ ∠ECD = 80⁰ – 30⁰
⇒ ∠ECD = 50⁰
Q.7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Ans: In and ∆OCD,
OA = OC | Radii of a circle
OB = OD | Radii of a circle
∠AOB= ∠COD | Vertically Opposite Angles
∴ ΔΑΟΒ ≅ ∆OCD | SAS Rule
∴ AB = CD | c.p.c.t.
⇒ Arc AB = Arc CD …(1)
Similarly, we can show that
ArcAD = Arc CB …(2)
Adding (1) and (2), we get
Arc AB + ArcAD + Arc CB
⇒ Arc BAD Arc BCD
⇒ BD divides the circle into two equal parts (each semicircle)
∴ ∠ A = 90⁰ ∠C = 90⁰ | Angle of a semicircle is 90⁰
Similarly, we can show that
∠B = 90⁰, ∠D=90⁰
∴ ∠A= ∠B= ∠C= ∠D = 90⁰
∴ ABCD is a rectangle.
Q.8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Ans: Given: ABCD is cyclic.
Construction: Draw BE || AD
Proof: ∵ AB || DE | Given
AB || BE | By construction
∴ Quadrilateral ABCD is a parallelogram.
∴ ∠BAD = ∠BED …(1) | Opp. ∠s of all || gm
and AD = BE …(2) | Opp. sides of a || gm
But AD = BC …(3) | Given
From (2) and (3), BE = BC
∴ ∠BEC = ∠BCE …(4) | Angles opposite to equal sides
∠BEC+ ∠BED = 180⁰ | Linear Pair Axiom
⇒ ∠BCE +∠BAD = 180⁰ | From (4) and (1)
⇒ Trapezium ABCD is cyclic.
∵ If a pair of opposite angles of a quadrilateral is 180⁰, then the quadrilateral is cyclic.
Q.9 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P. Q respectively (see figure). Prove that ∠ACP = ∠QCD.
Ans: Given: Two circles intersect at two points B and C.
Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P,Q respectively.
To Prove: ∠ACP = ∠QCD
Proof: ∠ACP = ∠ABP…(1) | Angles in the same segment of a circle are equal
∠QCD = ∠QBD…(2) | Angles in the same segment of a circle are equal
∠ABD = ∠QBD…(3) | Vertically Opposite Angles
From (1), (2) and (3), ∠ACP = ∠QCD
Q.10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Ans: Given: Circles are described with sides AB and AC of a triangle ABC as diameters. They intersect in a point D.
To Prove: D lies on the third side BC of ∆АВС.
Construction: Join AD.
Proof: ∵ Circle described on AB as diameter intersects BC in D.
∴ ∠ADB=90⁰ | Angle in a semi-circle
But ∠ADB + ∠ADC = 180⁰ | Linear Pair Axiom
∴ ∠ADC=90⁰
Hence, the circle described on AC as diameter must pass through D.
Thus, the two circles intersect in D.
Now, ∠ADC=180⁰.
∴ Points B, D, C are collinear.
∴ D lies on BC.
Q.11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD
Ans: Given: ABC and ADC are two right triangles with common hypotenuse AC.
To Prove: ∠CAD = ∠CBD
Proof: ∵ AC is the common hypotenuse and ABC and ADC are two right triangles.
∴ ∠ABC = 90⁰ = ∠ADC
⇒ Both the triangles are in the same semi-circle.
∴ Points A, B, D and C are concyclic.
DC is a chord
∴ ∠CAD = ∠CBD | Angles in the same segment are equal
Q.12. Prove that a cyclic parallelogram is a rectangle.
Ans: Given: ABCD is a cyclic parallelogram.
To Prove: ABCD is a rectangle.
Proof: ∵ ABCD is a cyclic quadrilateral
∴ ∠1+ ∠2 = 180⁰ …(1)
| ∵ Opposite angles of a cyclic quadrilateral are supplementary
∴ ABCD is a parallelogram
∴ ∠1+ ∠2 = …(2) | Opp. angles of a || gm
From (1) and (2), ∠1 = ∠2 = 90⁰
∴ || gm ABCD is a rectangle.
Hi! my Name is Parimal Roy. I have completed my Bachelor’s degree in Philosophy (B.A.) from Silapathar General College. Currently, I am working as an HR Manager at Dev Library. It is a website that provides study materials for students from Class 3 to 12, including SCERT and NCERT notes. It also offers resources for BA, B.Com, B.Sc, and Computer Science, along with postgraduate notes. Besides study materials, the website has novels, eBooks, health and finance articles, biographies, quotes, and more.
