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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 ∆ O’CD,
OA = O’C | Radii of congruent circles
OB = O’D | Radii of congruent circles
AB = CD | Given
∴ ∆ OAB ≅ O’CD | SSS Rule
∴ ∠AOB ≅ ∠ CO’D | c.p.c.t.
∴ ∠AOB= ∠ CO’D
Q.2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Ans: ∠ AOB and ∠ CO’D are the two equal angles subtended by the chords AB and CD of two congruent circles with centres O and O’ respec- tively.
To Prove: AB = CD
Prove: In ΔΟΑΒ and ∆O’CD,
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.
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 in-tersect each other at two points, then line joining their centres is the particular 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 inter- sect 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
⇒ AB – AE = CD – DE
⇒ 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 inter- sect 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 cong ∆ 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 cen- tre 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 = ΔΝΟΜ
|∵ 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≅AMSN | 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 9.2 |
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 + 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 in- tersect 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
Arc AD=Arc CB …(2)
Adding (1) and (2), we get
Arc AB+ Arc AD + 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 in- tersect at two points B and C.
Through B. two line segments ABD and PBQ are drawn to in- tersect 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 seg- ment 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.
