NCERT solution for class 9 maths circles ( Chapter 10)
Solution for Exercise 10.1
i) The centre of a circle lies in _______ of the circle. (Exterior\interior)
ii) A point, whose distance from the centre of a circle is greater than its radius lies in _______ of the circle. (Exterior\interior)
iii) The longest chord of a circle is a _______ of the circle.
iv) An arc is a ________ when its ends are the ends of a diameter.
v) Segment of a circle is the region between an arc and ________ of the circle.
vi) A circle divides the plane, on which it lies, in ______ parts.
Answer :
i) The centre of a circle lies in interior of the circle.
ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
iii) The longest chord of a circle is a diameter of the circle.
iv) An arc is a semi-circle when its ends are the ends of a diameter.
v) Segment of a circle is the region between an arc and chord of the circle.
vi) A circle divides the plane, on which it lies, in three parts.
i) Line segment joining the centre to any point on the circle is a radius of the circle.
ii) A circle has only finite number of equal chords.
iii) If a circle is divided into three equal arcs, each is a major arc.
iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
v) Sector is the region between the chord and its corresponding arc.
vi) A circle is a plane figure.
Answer :
i) True, Because all points are equidistant from the centre to the circle.
ii) False, Because circle has infinitely may equal chords can be drawn.
iii) False, Because all three arcs are equal, so there no difference between the major and
minor arcs.
iv) True, By the definition of diameter, that diameter is twice the radius.
v) False, Because the sector is the region between two radii and an arc.
vi) True, Because circle is a part of the plane figure.
Solution for Exercise 10.2
Answer :
Given: MN and PO are two equal chords of two congruent circles with centre at O and
O'.
To prove: \(\angle{MON}\) and \(\angle{PO'Q}\)
Proof: In \(\triangle{MON}\) and \(\triangle{PO'Q}\), we have,
MO = PO' ...(Radii of congruent circles)
NO = QO' ...(Radii of congruent circles)
and MN = PQ ...(Given)
By SSS criterion, we get,
\(\triangle{MON}\) \(\displaystyle \cong\) \(\triangle{PO'Q}\)
Hence, \(\angle{MON}\) = \(\angle{PO'Q}\) ...(By CPCT)
Answer :
Given: MW and PQ are two chords of congruent circles such that angles subtended by
these chords at the centres O and O' of the circles are equal i.e., \(\angle{MON}\) = \(\angle{PO'Q}\)
To prove: MN = PQ
Proof: In \(\triangle{MON}\) and \(\triangle{PO'Q}\), we have,
MO = PO' ...(Radii of congruent circles)
NO = QO' ...(Radii of congruent circles)
and \(\angle{MON}\) = \(\angle{PO'Q}\) ...(Given)
By SAS criterion, we get,
\(\triangle{MON}\) \(\displaystyle \cong\) \(\triangle{PO'Q}\)
Hence, MN = PQ ...(By CPCT)
Solution for Exercise 10.3
Answer :
Different pairs of circles are as follows:
| i) No points in common (externally) <\th> | (ii) One point in common (internally)<\th><\tr> | ||||||||
|---|---|---|---|---|---|---|---|---|---|
 <\th> |  <\th><\tr> | ||||||||
| iii) No points in common (internally)<\th> | iv) One point in common (externally)<\th><\tr> | ||||||||
 <\th> |  <\th><\tr> | ||||||||
| i) No points in common (externally) | (ii) One point in common (internally) |
|---|---|
| |
| iii) No points in common (internally) | iv) One point in common (externally) |
| |
| v) Two points in common | |
 |
From figures, it is obvious that these pairs many have 0 or 1 or 2 points in common.
Hence, a pair of circles cannot intersect each other at more than two points.
Answer :
Steps of construction:
1. Taking three points P, Q and R on the circle.
2. Join PQ and QR.
3. Draw MQ and NS, respectively the perpendicular bisectors of PQ and RQ, which intersect each other at O.
Hence, O is the centre of the circle.
Answer :
Given: Two circles with centres O and O' intersect at two points M and N so that MN is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO’ intersect MN at P.
To prove: OO' is the perpendicular bisector of MN.
Construction: Draw line segments OM, ON, O'M and O'N.
Proof: In \(\triangle{OMO'}\) and \(\triangle{ONO'}\), we have,
OM = ON ...(Radii of congruent circles)
O'M = O'N ...(Radii of congruent circles)
and OO' = O'O ...(Common)
By SSS criterion, we get,
\(\triangle{OMO'}\) \(\displaystyle \cong\) \(\triangle{ONO'}\)
Hence, \(\angle{MOO'}\) = \(\angle{NOO'}\) ...(By CPCT)
Therefore, \(\angle{MOP}\) = \(\angle{NOP}\) ...(i)(Since, \(\angle{MOO'}\) = \(\angle{MOP}\), \(\angle{NOO'}\) = \(\angle{NOP}\))
In \(\triangle{MOP}\) and \(\triangle{NOP}\), we have,
OM = ON ...(Radii of congruent circles)
OM = OM ...(Common)
and \(\angle{MOP}\) = \(\angle{NOP}\) ...(from eq. (i))
By SAS criterion, we get,
\(\triangle{MOP}\) \(\displaystyle \cong\) \(\triangle{NOP}\)
Hence, MP = NP ...(By CPCT)
And \(\angle{MPO}\) = \(\angle{NPO}\) ...(ii)
But \(\angle{MPO}\) + \(\angle{NPO}\) = \(180^\circ\) ...(Since, MPN is a straight line)
Thus, 2 \(\angle{MPO}\) = \(90^\circ\) ...(from (ii))
Therefore, \(\angle{MPO}\) = \(90^\circ\)
We have, MP = NP and \(\angle{MPO}\) = \(\angle{NPO}\) = \(90^\circ\)
Hence, it is proved that OO' is the perpendicular bisector of MN.
Solution for Exercise 10.4
Answer :
Let O and O' be the centres of the circles of radii 5 cm and 3 cm, respectively.
Let AB be their common chord.
Given: OA = 5cm, O'A = 3cm and OO' = 4cm
In \(\triangle{OO'A}\), we have,
\({AO'}^2 + {OO'}^2 = {3}^2 + {4}^2 = 9 + 16 = 25\)
i.e., \({OA}^2 = {5}^2 = {AO'}^2 + {OO'}^2\)
Therefore, OO'A is a right angled traingle and right angled at O'
Now, Area of \(\triangle{OO'A}\) = 1\2 × O'A × OO'
= 1\2 × 3 × 4
Area of \(\triangle{OO'A}\) = 6 sq.cm ...(i)
Also, Area of \(\triangle{OO'A}\) = 1\2 × OO' × AM
= 1\2 × 4 × AM
Area of \(\triangle{OO'A}\) = 2 × AM ...(ii)
From eq. (i) and (ii), we get,
2 × AM = 6
Therefore, AM = 3 cm.
Since, when two circles intersect at two points, then their centres lie on the perpendicular bisector of the common chord.
Thus, AB= 2 × AM = 2 × 3 = 6cm.
Therefore, the length of the common chord is 6cm.
Answer :
Given: MN and AB are two chords of a circle with centre O, AB and MN intersect at P
and MN = AB.
To prove: MP = PB and PN = AP
Construction: Draw \(OD\perp{MN}\) and \(OC\perp{AB}\).Join OP
Proof: DM = DN = (1\2) MN ...(Since, Perpendicular from centre bisects the chord)
Similarly, AC = CB = (1\2) AB ...(Since, Perpendicular from centre bisects the chord)
Therefore, we get,
MD = BC and DN = AC ... (i)(Since, MN = AB)
In \(\triangle{ODP}\) and \(\triangle{OPC}\),
OD = OC ...(Equal chords of a circle are equidistant from the centre)
\(\angle{ODP}\) = \(\angle{OCP}\)
and OP = OP ...(Common side)
Therefore, by RHS criterion of congruence,
\(\triangle{ODP}\) \(\displaystyle \cong\) \(\triangle{OPC}\)
Also, DP = PC ...(By CPCT) ...(ii)
On adding Equations (i) and (ii), we get,
MD + DP = BC + PC
i.e., MP = PB
Also, On subtracting Equation (ii) and (i),we get,
DN + DP = AC + PC
i.e., PN = AP
Hence, MP = PB and PN = AP are proved.
Answer :
Given: RQ and MN are chords of a circle with centre O. MN and RQ intersect at P and
MN = RQ.
To prove: \(\angle{OPC}\) = \(\angle{OPB}\)
Construction: Draw \(OC\perp{RQ}\) and \(OB\perp{MN}\). Join OP.
Proof: In \(\triangle{OCP}\) and \(\triangle{OBP}\),
OC = OB ...(Equal chords of a circle are equidistant from the centre)
\(\angle{OCP}\) = \(\angle{OBP}\) ...(Each angle is \(90^\circ\))
and OP = OP ...(Common side)
Therefore, by RHS criterion of congruence,
\(\triangle{OCP}\) \(\displaystyle \cong\) \(\triangle{OBP}\)
Also, \(\angle{OPC}\) = \(\angle{OPB}\) ...(By CPCT)
Answer :
Let OP be the perpendicular from O on linel . Since, the perpendicular from the centre
of a circle to a chord bisects the chords.
Now, BC is the chord of the smaller circle and \(OP\perp{BC}\).
Therefore, BC = PC ...(i)
Since, AD is a chord of the larger circle and \(OP\perp{AD}\).
Therefore, AP = PD ...(ii)
On subtracting Eq. (i) from Eq.(ii), we get,
AP - BP = PD - PC
Thus, AB = CD
Hence, it is proved.
If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Answer :
Let O be the centre of the circle and Reshma, Salma and Mandip are represented by
the points R, S and M, respectively. Let RP = x m.
Area of \(\triangle{ORS}\) = 1/2 × x × 5 = 5x/2 ...(i)(Since, in \(\triangle{ORM}\), RM is a chord therefore \(OP\perp{RM}\))
In right angled triangle \(\triangle{RON}\), \({OR}^2 = {RN}^2 + {NO}^2\)
i.e., \({5}^2 = {3}^2 + {NO}^2\)
i.e., \({NO}^2 = 25 - 9 = 16\)
Thus, NO = 4cm
So, Area of \(\triangle{ORS}\) = 1/2 × RS × ON = 1/2 × 6 × 4 = 12 ...(ii)
From Equations (i) and (ii), we get,
5x/2 = 12
Therefore, x = 24/5.
Since, P is the mid- Point of RM,
Thus, RM = 2RP = 2 × 24/5
i.e., RM = 48/5 = 9.6m
Hence, the distance between Reshma and Mandip is 9.6m.
Find the Length of the string of each phone.
Answer :
Let Ankur, Syed and David standing on the point P, Q and R.
Let PQ = QR = PR = X
Therefore,DPQR is an equilateral traingle.
Drawn altitudes PC, QD and RN from vertices to the sides of a traingle and intersect these altitudes at the centre of a circle M.
As PQR is an equilateral, therefore these altitudes bisects their sides.
In \(\triangle{PQC}\),
\({PQ}^2 = {PC}^2 + {QC}^2 \) ...(By Pythagoras theorem)
i.e., \({x}^2 = {PC}^2 + (x/2)^2 \)
i.e., \({PC}^2 = {x}^2 - (x/2)^2\)
i.e., \({PC}^2 = {x}^2 - {x}^2/4 = 3{x}^2/4\) ...(Since, QC = 1/2 QR = x/2)
Therefore, PC = \(\sqrt{3}\)x/2
Now, MC = PC - PM = \(\sqrt{3}\)x/2 - 20 ...(Since, PM = radius)
Now, in \(\triangle{QCM}\),
\({QM}^2 = {QC}^2 + {MC}^2 \) ...(By Pythagoras theorem)
i.e., \({20}^2 = {x/2}^2 + (\sqrt{3}x/2 - 20)^2 \) ...(Since, QM = radius)
i.e., \(400 = {x}^2/4 +( \sqrt{3}x/2)^2 - 20\sqrt{3}x + 400\)
i.e., \(0 = {x}^2 - 20\sqrt{3}x\)
i.e., \({x}^2 = 20\sqrt{3}x\)
Therefore, \(x = 20\sqrt{3}\)
Hence, PQ = QR = PR = \(20\sqrt{3}\)m.
Solution for Exercise 10.5
Let PQ = QR = PR = X
Answer :
Here, \(\angle{AOC}\) = \(\angle{AOB}\) + \(\angle{BOC}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\)
Therefore, Arc ABC makes \(90^\circ\) at the centre of the circle.
Thus, \(\angle{ADC}\) = (1/2) \(\angle{AOC}\) ...(Since, The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle)
i.e., = (1/2) \(90^\circ\) = \(45^\circ\)
Hence, \(\angle{ADC}\) = \(45^\circ\).
Answer :
Let BC be chord, which is equal to the radius. Join OB and OC.
Given: BC = OB = OC
Therefore, \(\triangle{OBC}\) is an equilateral traingle.
So, \(\angle{BOC}\) = \(60^\circ\)
But, we know that,
The angle subtended by an arc at the centre is double the angle subtended by it at any part of the circle.
So, \(\angle{BAC}\) = (1/2) \(\angle{BOC}\)
i.e., \(\angle{BAC}\) = (1/2) \(60^\circ\) = \(30^\circ\)
Also, here, ABMC is a cyclic quadrilateral.
Therefore, \(\angle{BAC}\) + \(\angle{BMC}\) = \(180^\circ\) ...(Since, in a cyclic quadrilateral the sum of opposite angles is \(180^\circ\)).
So, \(\angle{BMC}\) = \(180^\circ\) - \(30^\circ\) = \(150^\circ\)
Hence, the angle subtended by
the chord at a point on the minor arc and also at a point on the major arc is \(150^\circ\).
Answer :
Since, the angle subtended by the centre is double the angle subtended by
circumference, we get,
\(\angle{PQR}\) = 2 \(\angle{PQR}\) = \(200^\circ\) ...(given: \(\angle{PQR}\) = \(100^\circ\))
So, in \(\triangle{OPR}\),
\(\angle{POR}\) = \(360^\circ\) - \(200^\circ\) = \(160^\circ\) ...(i)
Again, in \(\triangle{OPR}\),
OP = OR ...(Radii of the circle)
\(\angle{OPR}\) = \(\angle{ORP}\) ...(By property of isosceles traingle)
In \(\triangle{POR}\),
\(\angle{OPR}\) + \(\angle{ORP}\) + \(\angle{POR}\) = \(180^\circ\) ...(ii)(Sum of internal angles of triangle)
So, from eq.(i) and (ii), we get,
\(\angle{OPR}\) + \(\angle{OPR}\) + \(160^\circ\) = \(180^\circ\)
i.e., 2 \(\angle{OPR}\) = \(180^\circ\) - \(160^\circ\) = \(20^\circ\)
Therefore, \(\angle{OPR}\) = \(10^\circ\).
Answer :
As we know that, the angles in the same segment are equal.
We get, \(\angle{BDC}\) = \(\angle{BAC}\) ...(i)
Now, in \(\triangle{ABC}\),
\(\angle{BAC}\) + \(\angle{ABC}\) + \(\angle{ACB}\) = \(180^\circ\)
i.e., \(\angle{BAC}\) + \(69^\circ\) + \(31^\circ\) = \(180^\circ\)
i.e., \(\angle{ABC}\) + \(100^\circ\) = \(180^\circ\)
i.e., \(\angle{ABC}\) = \(180^\circ\) - \(100^\circ\)
So, \(\angle{BAC}\) = \(80^\circ\)
Therefore, \(\angle{BDC}\) = \(80^\circ\) ...(from (i))
Answer :
In \(\triangle{ABC}\),
\(\angle{AED}\) = \(180^\circ\) -\(130^\circ\) = \(50^\circ\) ...(linear pair of angles)
So, \(\angle{CED}\) = \(\angle{AED}\) = \(50^\circ\) ...(Vertically opposite angles)
Also, \(\angle{ABD}\) = \(\angle{ACD}\) ...(Since, the angles in the same segment are equal)
i.e., \(\angle{ABE}\) = \(\angle{ECD}\)
Thus,
\(\angle{ABE}\) = \(180^\circ\) ...(ii)
Now, in \(\triangle{CDE}\),
\(\angle{BAC}\) + \(20^\circ\) + \(50^\circ\) = \(180^\circ\) ...(from eq.(i) and (ii))
i.e., \(\angle{BAC}\) + \(70^\circ\) = \(180^\circ\)
i.e., \(\angle{BAC}\) = \(180^\circ\) - \(70^\circ\)
Therefore, \(\angle{BAC}\) = \(110^\circ\)
Answer :
As, Angles in the same segment are equal, we get,
\(\angle{BDC}\) = \(\angle{BAC}\)
Therefore, \(\angle{BDC}\) = \(30^\circ\).
In \(\triangle{DBC}\), we have,
\(\angle{BDC}\) + \(\angle{DBC}\) + \(\angle{BCD}\) = \(180^\circ\)
\(30^\circ\) + \(70^\circ\) + \(\angle{BCD}\) = \(180^\circ\)
i.e., \(100^\circ\) + \(\angle{BCD}\) = \(180^\circ\)
i.e., \(\angle{BCD}\) = \(180^\circ\) - \(100^\circ\)
Therefore, \(\angle{BCD}\) = \(80^\circ\)
If AB= BC, \(\angle{BCA}\) = \(\angle{BAC}\) = \(30^\circ\) ...(Since, Angles opposite to equal sides in a traingle are equal)
Now, \(\angle{ECD}\) = \(\angle{BCD}\) - \(\angle{BCA}\) = \(80^\circ\) - \(30^\circ\) ...(Given \(\angle{BCD}\) = \(80^\circ\) and \(\angle{BCA}\) = \(30^\circ\))
i.e., \(\angle{ECD}\) = \(30^\circ\)
Therefore, \(\angle{EDC}\) = \(30^\circ\)
Answer :
Given: Diagonals NP and QM of a cyclic quadrilateral are diameters of the circle
through the vertices M, P, Q and N of the quadrilateral NQPM.
To prove: Quadrilateral NQPM is a rectangle.
Proof: ON = OP = OQ = OM ...(Radii of circle)
Now, ON = OP = (1/2) NP
and OM = OQ = (1/2) MQ
Therefore, NP = MQ
Hence, the diagonals MPQN are equal and bisect each other.
Hence, it is proved that quadrilateral NQPM is a rectangle.
Answer :
Given: Non-parallel sides PS and QR of a trapezium PQRS are equal.
Construction: Draw \(SM\perp{PQ}\) and \(RN\perp{PQ}\).
To prove: ABCD is cyclic trapezium.
Proof: In \(\triangle{SMP}\) and \(\triangle{RNQ}\), we have,
SP = RQ ...(Given)
SM = RN ...(Distance between two parallel twolines is always equal)
and \(\angle{SMP}\) = \(\angle{RNQ}\) ...(Each \(90^\circ\))
By RHS criterion, we get,
\(\triangle{SMP}\) \(\displaystyle \cong\) \(\triangle{RNQ}\)
Hence, \(\angle{P}\) = \(\angle{Q}\) ...(By CPCT)
Also, \(\angle{PSM}\) = \(\angle{QRN}\)
Now, adding \(90^\circ\) on both sides, we get,
Thus, \(\angle{PSM}\) + \(90^\circ\) = \(\angle{QRN}\) + \(90^\circ\)
So, \(\angle{MSR}\) + \(\angle{PSM}\) = \(\angle{NRS}\) + \(\angle{QRN}\) ...(Since, \(\angle{MSR}\) = \(\angle{NRS}\) = \(90^\circ\))
So, \(\angle{PSR}\) = \(\angle{QRS}\)
i.e., \(\angle{S}\) = \(\angle{R}\)
Thus, \(\angle{P}\) = \(\angle{Q}\) and \(\angle{S}\) = \(\angle{R}\)
Since, Sum of the angles of a quadrilateral is \(360^\circ\)
We get, \(\angle{P}\) + \(\angle{Q}\) + \(\angle{S}\) + \(\angle{R}\) = \(360^\circ\)
From eq. (i), we get,
2[\(\angle{Q}\) + \(\angle{S}\)] = \(360^\circ\)
Therefore, [\(\angle{Q}\) + \(\angle{S}\)] = \(180^\circ\)
Hence, it is proved that PQRS is a cyclic trapezium.
Answer :
Given: Two circles intersect at two points B and C. Through B two line segment ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively.
To prove: \(\angle{ACP}\) = \(\angle{QCD}\)
Proof:
In circle I,
\(\angle{ACP}\) = \(\angle{ABP}\) ...(i)(Angles in the same segment)
In circle II,
\(\angle{QCD}\) = \(\angle{QBD}\) ...(ii)(Angles in the same segment)
Also, \(\angle{ABP}\) = \(\angle{QBD}\) ...(Vertically opposite angles)
From Equation (i) and (ii), we get,
\(\angle{ACP}\) = \(\angle{QCD}\)
Hence, proved.
Answer :
Given: Two circles are drawn with sides AC and AB of \(\triangle{ABC}\) as diameters. Both circles intersect each other at D.
To prove: D lies on BC.
Construction: Join AD.
Proof: Since, AC and AB are the diameters of the two circles,
\(\angle{ADB}\) = \(90^\circ\) ...(i)(Angles in a semi-circle)
And \(\angle{ADC}\) = \(90^\circ\) ...(ii)(Angles in a semi-circle)
On adding Equation (i) and (ii), we get,
\(\angle{ADB}\) + \(\angle{ADC}\) = \(90^\circ\) + \(90^\circ\) = \(180^\circ\)
Hence, we can say that, BCD is a straight line.
So, D lies on BC.
Answer :
Draw a circle with AC as diameter passing through B and D. Join BD.
\(\triangle{ADC}\) and \(\triangle{ABC}\) are right angled triangles with common hypotenuse.
\(\angle{CBD}\) = \(\angle{CAD}\) ...(Since, angles in the same segment are equal)
Hence, proved.
Answer :
Given: PQRS is a parallelogram inscribed in a circle.
To prove: PQRS is a rectangle.
Proof: Since, PQRS is a cyclic quadrilateral.
Thus, \(\angle{P}\) + \(\angle{R}\) = \(180^\circ\) ...(i)(Since, Sum of opposite angles in a cyclic quadrilateral is \(180^\circ\))
But, \(\angle{P}\) = \(\angle{R}\) ...(ii)(Since, in a parallelogram, opposite angles are equal)
from eq. (i) and (ii), we get,
\(\angle{P}\) = \(\angle{R}\) = \(90^\circ\)
Similarly, \(\angle{Q}\) = \(\angle{S}\) = \(90^\circ\)
Thus, Each angle of PQRS is \(90^\circ\).
Hence, it is proved that PQRS is a rectangle.