Gujarat Board Textbook Solutions Class 9 Maths Chapter 10 Circles
GSEB Solutions Class 9 Maths Chapter 10 Circles
Ex 10.1
Question 1.
Fill in the blanks:
- The centre of a circle lies in ______ of the circle. (exterior/interior).
- A point, whose distance from the centre of a circle is greater than its radius lies in ______ of the circle. (exterior/interior).
- The longest chord of a circle is a ______ of the circle.
- An arc is a ______ when its ends are the ends of a diameter.
- A segment of a circle is the region between an arc and,_____ of the circle.
- A circle divides the plane, on which it lies, in _____ parts.
Solution:
- The centre of a circle lies in the interior of the circle.
- A point, whose distance from the centre of a circle is greater than its radius lies in the exterior of the circle.
- The longest chord of a circle is a diameter of the circle.
- An arc is a semicircle when its ends are the ends of a diameter.
- A segment of a circle is the region between an arc and the chord of the circle.
- A circle divides the plane, on which it lies, in three parts.
Question 2.
Write True or Fiase: Give reasons for your answers.
- The line segment joining the centre to any point on the circle is a radius of the circle.
- A circle has an only a finite number of equal chords.
- If a circle is divided into three equal arcs, each is a major arc.
- A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
- The sector is the region between the chord and its corresponding arc.
- A circle is a plane figure.
Solution:
- True because all points on the circle are equidistant from the centre.
- False because there are infinitely many points on the circle. So for each point on the circle, a point can be determined on the circle at a given distance from that point resulting in greatly many equal chords.
- False because for each arc, the remaining arc will have a greater length.
- True because of the definition of diameter.
- False by virtue of its definition.
- True as it is a part of a plane.
Question 3.
- A round pizza is cut into 4 pieces. Each piece represents ______
- Circles with the same centre and different radii are called.
- A chord of a circle divides the circle into two parts known as _____
- A constant distance between the centre of a circle and any point on a circle is called
- Diameter = 2 x ______
- The boundary of a circle is called ______
Solution:
- sector
- concentric circles
- segment
- radius
- radius
- circumference
Ex 10.2
Question 1.
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Solution:
Given: AB and CD are two equal chords of congruent circles with centres O and O’ respectively.
To Prove: ∠AOB = ∠CO’D.
Proof: In ΔOAB and ΔO’CD,
OA = O’C [Radii of congruent circles]
OB = O’D [Radii of congruent circles]
AB = CD [Given]
ΔOAB = ΔO’CD [SSS Rule]
∠AOB = ∠CO’D. [CPCT]
Question 2.
Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:
Given: ∠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’ respectively.
To Prove: AB = CD
Proof: In ΔOAB and ΔO’CD
OA = O’C [Radii of congruent circles]
OB = O’D [Radii of congruent circles]
∠AOB = ∠CO’D [Given]
∴ ΔOAB = ΔO’CD [SAS Rule]
∴ AB = CD [CPCT]
Ex 10.3
Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
A pair of circles can be drawn in the following condition:
The maximum number of common points is two.
Question 2.
Suppose you are given a circle. Give the construction to find its centre.
Solution:
Steps of construction:
(i) Take any three points P, Q and R on the circle.
(ii) Join PQ and QR.
(iii) Draw the perpendicular bisectors of PQ and QR. Let these intersect at O. Then, O is the centre of the circle.
Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution:
Given: Two circles with centres O and P intersecting at A and B.
Prove: OP is the perpendicular bisector of AB.
Construction: Join OA, OB, PA and PB. Let OP intersect AB at M.
Proof: In ΔOAP and ΔOBP,
OA = OB [Radii of a circle]
PA = PB [Radii of a circle]
OP = OP [Common]
ΔOAP = ΔOBP [SSS Rule]
∴ ∠AOP = ∠BOP [CPCT]
= ∠AOM = ∠BOM ……….(1)
In ΔAOM and ΔBOM,
OA = OB [Radii of a circle]
∠AOM = ∠BOM (From (1))
OM = OM [Common]
∴ ΔAOM = ΔBOM [SAS Rule]
∴ AM = BM ……..(2) [CPCT]
and ∠AMO = ∠BMO ……..(3) ECPCTI
But ∠AMO + ∠BMO = 180° [Linear pair axiomi]
∠AMO = ∠BMO = 90° ……… (4)
∴ OM, i.e., OP is the perpendicular bisector of AB. [From (2) and (4)]
Ex 10.4
Question 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.
Solution:
Let two circles with centres O and O’ intersect each other at P and Q.
Therefore, OP = OQ = 5cm, O’P = O’Q = 3cm and OO’ = 4 cm
In ΔOOP, OO’2 + OP2 = 42 + 32
= 16 + 9 = 25
= 52 = OP2
∴ AOO’P = 90° (Using converse of Pythagoras Theorem)
But OO’ ⊥ PQ
As if two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of their common chord.
∴ OO’ coincides 0M i.e., PQ passes through the centre O’.
∴ Length of the common chord
PQ = 2O’P = 2 x 3 cm = 6 cm.
Question 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.
Solution:
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 OM ⊥ AB and ON ⊥ CD. Join OE.
Proof: In ΔOME and ΔONE,
OM = ON
[∴ Equal chords of a circle are equidistant from the centre.]
OE = OE [Common]
∴ ΔOME = ΔONE [RHSI
∴ ME = NE [CPCT]
⇒ AM + ME = DN + NE
|∴ AM = DN = 1/2 AB = 1/2 CD|
AE = DE
Now, AB – AE = CD – DE [Given AB = CD]
⇒ BE = CE
Question 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.
Solution:
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 ΔOEA and ΔOED,
OE = OE [Common]
OA = OD [Radii of a circle]
AE = DE [Proved in 2 above]
∴ ΔOEA. = ΔOED [SSS Rule]
∴ ∠OEA = ∠OED [CPCT]
Note: Also it can be drawn OP⊥ AB, OQ ⊥ CD and shown ΔOPE = ΔOQE to prove ∠OEP = ∠OEQ.
Question 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].
Solution:
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.
Question 5.
Three girls Reshma, Salina 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 to Mandip, Mandip to Reshma. If the distance between Reshma and Salina and between Salina and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Solution:
Let KR = xm
ar(ΔORS) = ar(ΔORK) + ar(ΔSRK)
Question 6.
A circular park of radius 20 m 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 each other. Find the length of the string of each phone.
Solution:
Let BD = xm
Then in right triangle ODB,
OB2 = OD2 + BD2 [By Pythagoras Theorem]
Take AB = BC = AC
Ex 10.5
Question 1.
Infigure,A,B and Carethreepointsona circle with centre O such that ∠BOC = 300 and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution:
∠ADC = 1/2∠AOC
[The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.]
= 1/2(∠AOB + ∠BOC)
= 1/2 (60° + 300)
= 1/2 (90°) = 45°
Question 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.
Solution:
∴ OA = OB = AB [Given]
∴ OAB is equilateral.
∴ ∠AOB = 60°
∠ACB = 1/2∠A0B
[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.]
∴ ∠ACB = 1/2 x 60° = 30°
Now, ADBC is a cyclic quadrilateral.
∴ ∠ADB + ∠ACB = 180°
[The sum of either pair of opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠ADB + 30°= 180°
⇒ ∠ADB = 180°- 30°
∴ ∠ADB = 150°
Question 3.
In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Solution:
Take a point S in the major arc. Join PS and RS.
∴ PQRS is a cyclic quadrilateral.
∠PQR + ∠PSR = 180°
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°
100° + ∠PSR = 180°
∠PSR = 180° – 100°
⇒ ∠PSR = 80° …….(1)
Now, ∠POR = 2∠PSR
(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]
∴ ∠POR= 2 x 800 = 160° …….(2) Using (1)
In ΔOPR,
∴ OP = OR Radiiofacircle
∴ ∠OPR = ∠ORP …………(3)
Angles opposite to equal sides of a triangle are equal
Also,
∠OPR + ∠ORP + ∠POR = 180°
Sum of all the angles of a triangle is 1800
⇒ ∠OPR + ∠OPR + 160° = 180°
Using (2) and (1)
⇒ 2∠OPR + 160° = 180°
⇒ 2∠OPR = 180° – 160° = 20°
∠OPR = 10°
Question 4.
In figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
Solution:
In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180°
[Sum of all the angles of a triangle is 180°]
∠BAC + 69° + 31° = 180°
∠BAC + 100° = 1800
∠BAC = 180° = 100° = 80° ………..(1)
Now, ∠BDC = ∠BAC
[Angles in the same segment of a circle are equal]
∠BDC = 80° [Using (1)]
Question 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 = 130° and ∠ECD = 20°. Find ∠BAC.
Solution:
In ΔCDE ∠CDE + ∠DCE = ∠BEC
[Exterior angle property of a triangle]
⇒ ∠CDE + 20° = 130°
⇒ ∠CDE = 130° – 20° = 110° ………..(2)
Now, ∠BAC = ∠CDE
[Angles in the same segment of a circle are equal]
∴ ∠BAC = ∠110° [Usmg(1)]
Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, If AB = BC, find ∠ECD.
Solution:
∠CDB = ∠BAC
[Angles in the same segment of a circle are equal]
∠CDB = 30° ………..(1)
∠DBC = 70° ………..(2) (Given)
In ΔBCD,
∠BCD + ∠DBC + ∠CDB = 180°
(Sum of all the angles of a triangle is 180°]
⇒ ∠BCD + 70° + 300 = 180° [Using (1) and (2)1
⇒ ∠BCD + 100° = 180°
∠BCD = 180° – 100°
⇒ ∠BCD = 80° ………(3)
In ΔABC, AB = BC
∠BCA = ∠BAC
[Angles opposite to equal sides of a triangle are equal]
⇒ ∠BCA = 30° ………(4)
∠BAC = 30° (given]
Now, ∠BCD = 80° (From (3))
= ∠BCA + ∠ECD = 80°
⇒ 30°+∠ECD = 80°
⇒ ∠ECD = 80 – 30°
⇒ ∠ECD = 50°
Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
In ΔOAB and ΔOCD,
OA = OC [Radii of a circle]
OB = OD [Radii of a circle]
∠AOB = ∠COD [Vertically opposite angles]
∴ ΔOAB = ΔOCD [SAS rule]
∴ AB = CD …….(1) [CPCTI
Similarly, we can show that
AD = CB ……..(2)
Adding (1) and (2), we get
ABCD is a || gm.
(A quadrilateral having opposite sides equal is a parallelogram)
Now, diagonal BD is also a diameter
∴ ∠BAD = 1/2 BOD
= 1/2 x 180° = 90°
∴ ABCD is the rectangle. (A parallelogram with an angle 90° is a rectangel)
Question 8.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic. Prove that an isosceles trapezium is cyclic.
Solution:
Given: ABCD is a trapezium whose two non-parallel sides AD and BC are equal.
To Prove: Trapezium ABCD is cyclic.
Construction: Draw BE || AD
Proof: AB || DE [Given]
AD || BE [By construction]
∴ Quadrilateral ABED is a parallelogram.
∴ ∠BAD = ∠BED ……….(1) [Opp. ∠s of a || gm]
and AD = BE [Opp. sides of a gin]
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 180°, then the quadrilateral is cyclic.]
Question 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.
Solution:
∠ACP = ∠ABP …………(1)
[Angles in the same segment of a circle are equal]
And ., ∠QCD = ∠QBD …………(2)
[Angles in the same segment of a circle are equal]
But ∠ABP = ∠QBD [Vertically opposite angles]
∠ACP = ∠QCD. [From (1) and (2)]
Question 10.
If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lies on the third side.
Solution:
Given: Circles are described with sides AB and AC of a triangle ABC as diameters. They intersect at a point D.
To Prove: D lies on the third side BC of ∠ABC.
Construction: Join AD.
Proof: Circle described on AB as diameter intersects BC in D.
∴ ∠ADB = 90°
[Angle in a semicircle Similarly, the circle described on AC as diameter passes through D.]
∠ADC = 90° ………..(2)
Now, adding (1) and (2) we get
∠ADB + ∠ADC = 180°
∴ Points B, D, C are collinear.
∴ D lies on BC.
Question 11.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution:
∴ AC is the common hypotenuse of two right triangles ABC and ADC.
∴ ∠ABC = 90° = ∠ADC
Here, there are two cases arising i.e., B and D are either on the same side of AC or on opposite sides.
Case I – Both the triangles are in the same semicircle.
∴ Points A, B, D and C are concyclic.
Case II – Triangles are on opposite sides, then
∠ABC + ∠ADC = 1800
∴ A, B, C and D are concyclic.
Now, in both cases, DC is a chord
∴∠CAD = ∠CBD
[∴ Angles in the same segment are equal]
Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
Given: ABCD is a cyclic parallelogram.
To prove: ABCD is a rectangle.
Proof: ABCD is a cyclic quadrilateral.
∴ ∠1 + ∠2 = 180° ……….(1)
[∴ Opposite angles of a cyclic quadrilateral are supplementary]
∴ ABCD is a parailelograin.
∴ ∠1 = ∠2 ………(2) [Opp. angles of a || gm]
From (1) and (2),
∠1 = ∠2 = 90°
∴ || gm ABCD is a rectangle.
Ex 10.6
Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given: Two intersecting circles with centres A and B. Their points of intersection are P and Q.
To prove: ∠APB = ∠AQB
Proof: In ΔAPB and ΔAQB,
AP = AQ [Radii of a circle]
BP = BQ [Radiiofacircle]
AB = AB [Common]
∴ ΔAPB = ΔAQB [SSS Rule]
∴ ∠APB = ∠AQB [CPCT]
Question 2.
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 0M = x cm.
Then ON = (6 – x)cm
∴ OM ⊥ CD
Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Solution:
Case I: When the two chords are on the same side of the centre.
∴ OD = OB = 5cm [Radiiofacircle]
In right triangle OND,
OD2 = ON2 + ND2 [By Pythagoras theorem]
⇒ (5)2 = ON2 + (4)2
⇒ 25 = ON2 + 16
⇒ ON2 = 25 – 16
⇒ ON2 = 9
⇒ ON = √9 = 3 cm
Hence, the distance of the other chord from the centre is 3 cm.
Case II: When the two chords are on opposite sides of the centre.
As in case I:
ON = 3 cm.
Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
To prove: ∠ABC = 12(∠DOE – ∠AOC)
Proof: Let ∠ABC = x, ∠AOC = y and ∠DOE = z.
∠C’OD + ∠A’OE = z – y …………(1)
Let ∠C’OD = θ
then ∠A’OE = z – y – θ [From (1)]
∠AOD = n – (∠AOC + ∠C’OD)
= Ï€ – (y + θ)
∠COE = n – (∠C’OA’ + ∠A’OE)
= Ï€ – (y + z – y – θ)
= Ï€ – (z – θ)
AD = CE
∠AOD = ∠COE
[Equal chords subtend equal angles at the centre]
∴ Ï€ – (y + θ) = n – (z – θ)
y + θ = z – θ
2θ = z – y
ΔOAC, ΔOCE, ΔOED and ΔODA are isosceles triangles.
∴∠1 = ∠2,∠3 = ∠4,∠5 = ∠6 and ∠7 = ∠8 ……….(1)
In MBC,
∠B + ∠CAB = ∠ACE and ∠B + ∠ACB = ∠CAD
(Exterior angle property)
Adding, 2∠B + ∠CAB + ∠ACB = ∠ACE + ∠CAP
⇒ 2∠B + ∠CED + ∠ADE = ∠ACE + ∠CAD
(∴ Exterior angle of a cyclic quadrilateral is interior opposite angle,
∴ ∠CAB = ∠CED and ∠ACB = ∠ADE)
⇒ 2∠B + (∠6 + ∠7) + (∠1 + ∠8) = (∠4 + ∠5) + (∠2 + ∠3) (Using eqn (1))
⇒ 2∠B + (∠7 + ∠8) = (∠3 + ∠4)
⇒ 2∠B + 1800 – ∠DOE = 1800 – ∠AOC (Angle sum property)
⇒ 2∠B = ∠DOE – ∠AOC
∴ ∠B = 1/2(∠DOE – ∠AOC)
Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Solution:
Given: ABDC is a rhombus. E is the point of intersection of its diagonals.
To Prove: The circle drawn with any side, say AB of rhombus ABCD, as a diameter passes through the point E.
Proof: ∠AEB = ∠AEC = ∠CED = ∠BED = 900
(As diagonals of a rhombus bisect each other at right angles)
Now, ∠AEB = 900 and AB is diameter so E must lie on the semicircle.
E is the point of intersection of diagonals.
Hence, the circle drawn with AB as diameter passes through point E.
Similarly, we can prove that the circle is drawn with AC as diameter passes through point E which is the point of intersection of its diagonals.
Question 6.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
Given ABCD is a parallelogram. The circle through A, B and C intersects CD (produced, if necessary) at E.
To Prove: AE = AD
Proof: In cyclic quadrilateral ABCE,
∠AED + ∠ABC = 180° ………….(1)
[∴ Opposite angles of a cyclic quadrilateral are supplementary]
Also, ∠ADE + ∠ADO = 180° [Linear Pair Axiom]
But ∠ADO = ∠ABC [Opposite angles of a || gm]
∴ ∠ADE + ∠ABC = 180° ………..(2)
From (1) and (2), we have
∠AED + ∠ABC = ∠ADE + ∠ABC
⇒ ∠AED = ∠ADE
∴ In triangle ADE,
AE = AD
[∴ Sides opposite to equal angles of a triangle are equal] Proved
Question 7.
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:
1. Let the chords AC and BD intersect each other at O.
Join AB, BC, CD and DA.
AC and BD bisect each other, so ABCD is a parallelogram
In ΔOAB and ΔOCD,
OA = OC [Given]
OB = OD [Given]
∠AOB = ∠COD [Vert. opp. ∠s]
∴ ΔOAB = ΔOCD [SAS]
∴ AB = CD [CPCT]
BD divides the circle into two equal parts (each a semicircle) so BD is a diameter
Similarly, we can show that AC is a diameter.
2. ∴ ∠A = 90° and ∠C = 90°the angle of a semicircle is 90°.
∠B = 90° and ∠D = 90°
∴ ∠A = ∠B = ∠C = ∠D = 90°
∴ ABCD is a rectangle.
Question 9.
Two congruent circles intersect each other at point A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Solution:
Given: Two congruent circles intersect each other at point A and B. A line through A meets the circles in P and Q.
To Prove: BP = BQ
Proof: AB is the common chord of the two congruent circles.
∴ ∠APB = ∠AQB
[∴ Angles subtended by equal chords are equal]
∴ BP = BQ
[Sides opposite to equal angles are equal]
Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Solution:
Given: Bisector AP of angle A of ΔABC and the perpendicular bisector PQ of its opposite side BC intersect at P.
To prove: P lies on the circumcircle of the triangle ABC, i.e., P, A, B and C are concyclic
Construction: Draw the circle through three non-collinear points A, B and C with centre O on PQ.
⇒ It is only possible when P lies on the circle passing through A, B, C.