NCERT Solutions for Class 9 Math Chapter 10 - Circles

Answer:

interior

Answer:

exterior

Answer:

diameter

Answer:

semicircle

Answer:

chord

Answer:

three

Answer:

TRUE

Answer:

FALSE
Correct statement: A circle has infinite number of equal chords.

Answer:

FALSE
Correct statement: If a circle is divided into three equal arcs, each is a minor arc.

Answer:

TRUE

Answer:

FALSE
Correct statement: Segment of a circle is the region between the chord and its corresponding arc.

Answer:

TRUE

Answer:

Two circles are said to be congruent if and only if one of them can be superposed on the other so as to cover it exactly.
Let C(O, r) and C(O′, s) be two circles. Let us imagine that the circle C(O′, s) is superposed on C(O, r) so that O′ coincide with O. Then it can easily be seen that C(O′, s) will cover C(O, r) completely if and only if r = s.
Hence, we can say that two circles are congruent, if and only if, they have equal radii. Now using this concept we have to prove that equal chords of congruent circles subtended equal angles at the centre which is proved as follows:
Given: PQ and RS are equal chords of congruent circles

C(O, r) and C′(O′ r).
To prove: ∠POQ = ∠RO′S
Proof: In triangles POQ and RO′S (See fig.)
OP = OQ = O′R = O′S
= r (Radius)
PQ = RS (Given)
$\therefore$ $∆$POQ $\cong$ $∆$RO′S
(By SSS congruency)
$\therefore$ ∠POQ = ∠RO′S
(Congruent part of congruent triangles)

Answer:

Given: Two chords PQ and RS of congruent circles C(O, r) and C′(O′, r) such that.
∠POQ = ∠RO′S
To prove: PQ = RS
Proof: In triangle POQ and RO′S (see fig.)

OP = O′Q = O′R = O′S = r (Radius)
∠POQ = ∠RO′S (Given)
$\therefore$ $∆$POQ $\cong$ $∆$RO′S
(By SAS congruency)
$\therefore$ PQ = RS
(Congruent parts of congruent triangles)

Answer:

From the fig. we observe that when different pair of circles are drawn; each pair have two points (say A and B) in common.
Maximum number of common points are two. Suppose that two circles C(O, r) and C(O′, s) intersect each other in three points, say A, B and C.
Then A, B and C are non-collinear points.

We know that:
There is one and only one circle passing through three non-collinear points.
Therefore a unique circle passes through A, B and C.
$\Rightarrow$ O′ coincides with O and s = r
A contradiction to the fact that
C(O′, s) $\ne$ C(O, r)
$\therefore$ Our supposition is wrong.
Hence, two different circles cannot intersect each other at more than two points.

Answer:

Steps of construction:
(i) Take any three points A, B and C on the circle.
(ii) Join AB and BC.
(iii) Draw perpendicular bisector say LM on AB.

(iv) Draw perpendicular bisector PQ on BC.
(v) Let LM and PQ intersect at the point O.
Then O is the centre of the circle.
Verification:
O lies on the perpendicular bisector of AB.
$\therefore$ OA = OB ... (i)
O lies on the perpendicular bisector of BC.
$\therefore$ OB = OC ... (ii)
From (i) and (ii) we observe that
OA = OB = OC = r (say)
Three non collinear points A, B and C are at equal distance (r) from the point O inside the circle.
Hence, O is the centre of the circle.

Answer:

Proof: C(O, r) and C(O′, s) be two circles intersecting A and B. To prove OO′ is the perpendicular bisector of the chord AB. Draw line segments OA, OB, O′A and O′B (See fig.)
In triangles OAO′ and OBO′,

OA = OB = r
O′A = O′B = s
and OO′ = OO′
$\therefore$ $∆$OAO′ $\cong$ $∆$OBO′ (By SSS congruency)
$\therefore$ ∠AOO′ = ∠BOO′
Let M be the point of intersection of AB and
OO′. Then in triangles AOM and BOM,
OA = OB
∠AOM = ∠BOM
($\because$ ∠AOO′ = ∠AOM and ∠BOO′ = ∠BOM)
and OM = OM
$\therefore$ $∆$AOM $\cong$ $∆$BOM (By SAS axiom)
$\therefore$ AM = MB ... (i)
and ∠AMO = ∠BMO ... (ii)
Now ∠AMO + ∠BMO = 180$°$
(Linear pair axiom)
$\Rightarrow$ ∠AMO + ∠AMO = 180$°$
$\Rightarrow$ 2∠AMO = 180$°$

$\Rightarrow$ ∠AMO = 90$°$
Also ∠BMO = 90$°$ ($\because$∠AMO = ∠BMO)
Now, we have
AM = MB
∠AMO = ∠BMO = 90$°$
It proves that line OO′ joining centres O and
O′ lies on the perpendicular bisector of common chord AB.

Answer:

Let two circles with centres O and O′ intersect each other at points A and B. On joining A and B; AB is the common chord.

Radius OA = 5 cm, Radius O′A = 3 cm, distance between their centres OO′ = 4 cm.
We observe that in triangle AOO′;
5² = 4² + 3²
$\Rightarrow$ 25 = 16 + 9
$\Rightarrow$ 25 = 25
Pythagoras result, holds in $∆$AO′O.
Hence, $∆$AO′O is rt. ∠d triangle, rt. angle at O′.
As we know that perpendicular drawn from the centre of the circle bisects the chord.
Hence, O′ is the mid-point of the chord AB.
Also O′ is the centre of the circle II.
Therefore, length of chord AB = Diameter of circle II
$\therefore$ Length of chord AB = 2 × 3 cm = 6 cm.
ALITER

Let two circles with centre O and O′ intersect
each other at points A and B. Let common chord
AB intersects OO′ at point C.
Let OC = x cm
$\therefore$ O′C = 4 – x cm
As we know that line joining the centres of two circles is perpendicular bisector of the common chord of circles.
$\therefore$ In rt. ∠d $∆$OCA,
AC² + OC² = OA²
[Using Pythagoras Theorem]
$\Rightarrow$ AC² + x² = 52
$\Rightarrow$ AC² = 25 – x² ...(i)
Similarly in $∆$ACO′,
AC² + O′C² = AO′²
$\Rightarrow$ AC² + (4 – x)² = 32
$\Rightarrow$ AC² = 9 – (4 – x)² ...(ii)
From (i) and (ii), we get
25 – x² = 9 – (4 – x)²
$\Rightarrow$ 25 – x² = 9 – (16 + x² – 8x)
$\Rightarrow$ 25 – x² = 9 – 16 – x² + 8x
$\Rightarrow$ – 8x = 9 – 16 – 25 – x² + x²
$\Rightarrow$ – 8x = – 32
$\Rightarrow$ x = 4
$\therefore$ CO′ = 4 – x
$\Rightarrow$ CO′ = 4 – 4
$\Rightarrow$ CO′ = 0
It means that O′coincides with C.
$\therefore$ AC = Radius, AO′ = 3 cm.
Length of chord AB = Diameter of circle with
centre O′
$\therefore$ Length of chord AB = 2 × AO′
= 2 × AC
= 2 × 3 = 6 cm.

Answer:

Let AB and CD are two equal chords of a circle of centre O intersecting each other at point E within the circle.
We have to prove that:
(a) AE = CE
(b) BE = DE.
Construction: Draw OM ⊥ AB, ON ⊥ CD also join OE.

Proof: In rt.∠d $∆$OME and $∆$ONE,
∠OME = ∠ONE (Each 90$°$)
OM = ON
[Equal chords are equidistant from the centre]
hyp. OE = hyp. OE (Common)
$\therefore$ $∆$OME $\cong$ $∆$ONE
[By RHS congruency]
$\therefore$ ME = NE
(Congruent parts of congruent triangles) ...(i)
Now; O is the centre of circle and
OM ⊥ AB

But AB = CD (Given)
From (ii) and (iii), we get
AM = NC ...(iv)
Also MB = DN ...(v)
Adding (i) and (iv), we get
AM + ME = NC + NE
$\Rightarrow$ AE = CE Proved part (a)
Now, AB = CD (Given)
AE = CE (Proved above)
AB – AE = CD – CE
$\Rightarrow$ BE = DE Proved part (b)

Answer:

AB and CD be two equal chords of a circle with centre O intersecting each other with in the circle at point E. OE is joined. We have to prove that
∠OEM = ∠OEN.
Construction: Draw OM ⊥ AB,
ON ⊥ CD.

Proof: In rt. ∠d $∆$OME and $∆$ONE,
∠OME = ∠ONE (Each 90$°$)
OM = ON
(Equal chords are equidistant from the centre)
Hyp. OE = Hyp. OE (Common)
$\therefore$ $∆$OME $\cong$ $∆$ONE
(By RHS congruency)
$\therefore$ ∠OEM = ∠OEN
(Congruent parts of congruent triangles)

Answer:

Line l intersects two concentric circles with centre O at points A, B, C and D.
We have to prove that
AB = CD

Construction: Draw OL ⊥ l.
Proof: AD is a chord of outer circle and OL ⊥ AD.
$\therefore$ AL = LD
[$\because$ Perpendicular drawn from the centre bisects the chord] ...(i)
Now, BC is a chord of inner circle and OL ⊥ BC.
$\therefore$ BL = LC
[Same reason as above] ...(ii)
Subtracting (ii) from (i), we get
AL – BL = LD – LC
$\Rightarrow$ AB = CD (Proved

Answer:

Let position of Reshma, Salma and Mandip is denoted by points A, B and C.
Given that the distance between Reshma and Salma is 6 m and distance between Salma and Mandip is 6 m. It means:
AB = BC = 6 m
$\therefore$ Centre lies on the bisector of ∠BAC.
Let M be the point of intersection of BC and OA.
Again, since AB = BC
and AM bisects ∠CAB
$\therefore$ AM ⊥ CB and M is the mid-point of CB.
Let OM = x
then MA = 5 – x
Now, from right-angled $∆$OMB,
OB² = OM² + MB²
5² = x² + MB² ...(1)

$\Rightarrow$ MB = 4.8 m
$\therefore$ BC = 2MB = 2 × 4.8 = 9.6 m
Hence, distance between Reshma and Mandip is 9.6 m.

Answer:

Let position of three boys Ankur, Syed and David is denoted by points A, B and C.
Three points are at equal distances.
$\therefore$ AB = BC = AC = a m (say)

Equal sides of equilateral triangles are as equal chords and perpendicular distances of equal chords of a circle are equidistant from the centre.
$\therefore$ OD = OE = OF = x m (say)
Join OA, OB and OC.
Now, we have three congruent triangles,
$∆$OAB, $∆$OBC and $∆$AOC
$\therefore$ area ($∆$AOB) = area ($∆$BOC)
= area ($∆$AOC) ...(i)
Now area of equilateral
triangle ABC of side a = ar ($∆$AOB)
+ ar ($∆$BOC) + ar ($∆$AOC) ...(ii)
$\Rightarrow$ Area ($∆$ABC) = 3Area ($∆$BOC)
[Using (i) in (ii)]

Answer:

Answer:

Answer:

Answer:

In $∆$ABC ;
$\Rightarrow$ ∠BAC + 69$°$ + 31$°$ = 180$°$
$\Rightarrow$ ∠BAC = 180$°$ – 69$°$ – 31$°$
$\Rightarrow$ ∠BAC = 80$°$ ... (i)
A and D are the points in the same segment of circle
Therefore ∠BDC = ∠BAC
[$\because$ Angles subtended by the same arc at any points in the alternate segment of a circle are equal.]
$\Rightarrow$ ∠BDC = 80$°$ [Using (i)]

Answer:

According to fig.,
∠CED + ∠BEC = 180$°$ (Linear pair)
$\Rightarrow$ ∠CED + 130$°$ = 180$°$
$\Rightarrow$ ∠CED = 180$°$ – 130$°$
$\Rightarrow$ ∠CED = 50$°$ ... (i)
∠AEB = ∠CED
(Vertically opp. angles)
$\Rightarrow$ ∠AEB = 50$°$ [Using (i)]
Now,
∠ABD = ∠ACD
[Angles subtended by the arc AD in the same segment of a circle are equal]
$\Rightarrow$ ∠ABD = 20$°$
[$\because$ ∠ACD = 20$°$ (Given)]
Now in $∆$AEB;
∠BAE + ∠ABE + ∠AEB = 180$°$
[Angle sum property of a triangle]
$\Rightarrow$ ∠BAE + 20$°$ + 50$°$ = 180$°$
$\Rightarrow$ ∠BAE = 180$°$ – 20$°$ – 50$°$
$\Rightarrow$ ∠BAE = 110$°$
or ∠BAE = 110$°$

Answer:

∠BDC = ∠BAC [Angles of the same segment]
$\Rightarrow$ ∠BDC = 30$°$
($\because$ ∠BAC = 30$°$)
In $∆$BCD;
∠BCD + ∠DBC + ∠BDC = 180$°$
$\Rightarrow$ ∠BCD + 70$°$ + 30$°$ = 180$°$
[$\because$∠DBC = 70$°$]
$\Rightarrow$ ∠BCD = 180$°$ – 70$°$ – 30$°$
$\Rightarrow$ ∠BCD = 80$°$ ... (i)
If AB = BC
Then in $∆$ABC;
∠ACB = ∠BAC
(Angles opp. to equal sides of a triangle are equal)
$\Rightarrow$ ∠ACB = 30$°$ ... (ii)
Now,
∠BCD = ∠ACB + ∠ACD
$\Rightarrow$ 80$°$ = 30$°$ + ∠ACD
[Using (i) and (ii)]
$\Rightarrow$ 80$°$ – 30$°$ = ∠ACD
$\Rightarrow$ 50$°$ = ∠ACD
or ∠ACD = 50$°$
or ∠ECD = 50$°$

Answer:

AC is the diameter
$\therefore$ ∠B = ∠D = 90$°$ ... (1)
(Angle is a semicircle is right angle)
Similarly, BD is the diameter

$\Rightarrow$ AD = BC
(Chords opposite to equal arcs) ... (3)
Similarly, AB = DC ... (4)
From (1), (2), (3) and (4) we observe that each angle of the quadrilateral is 90$°$ and opposite sides are equal.
Hence, ABCD is a rectangle.

Answer:

Given: A trapezium ABCD in which AB CD and AD = BC.

To prove: The points A, B, C and D are concyclic, (i.e., trapezium ABCD is cyclic).
Construction: Draw DE II CB.
Proof: DE II CB and EB II DC.
$\therefore$ EBCD is a IIgm
$\therefore$ DE = CB and ∠DEB = ∠DCB.
$\because$ Opposite ∠s of a IIgm are equal.
Now, $\because$ AD = BC and BC = DE
$\therefore$ DA = DE $\Rightarrow$ ∠DAE = ∠DEA.
[$\because$ Angles opposite to equal sides in a
triangle are equal]
But ∠DEA + ∠DEB = 180$°$
... (Linear pair)
$\Rightarrow$ ∠DAE + ∠DCB = 180$°$
[$\because$ ∠DEA = ∠DAE & ∠DEB = ∠DCB
(Proved above)]
$\Rightarrow$ ∠DAB + ∠DCB = 180$°$ ....(1)
$\Rightarrow$ ∠A + ∠C = 180$°$
Hence, ABCD is a cyclic trapezium.
[$\because$ Opposite angles of a cyclic quadrilateral are supplementary, as in Result (1)].

Answer:

Arc BC of circle I subtends ∠1 and ∠2 in the
same segment of circle.
$\therefore$ ∠1 = ∠2
[Angles subtended by the same arc in the alternate segment of a circle are equal]
Arc BC of circle II subtends ∠3 and ∠4 in the
same segment of a circle.
$\therefore$ ∠3 = ∠4
[same reason as above]
Now in $∆$ACD,
∠A + ∠C + ∠D = 180$°$
[Angle sum property of a triangle]
$\Rightarrow$ ∠1 + ∠5 + ∠6 + ∠3 = 180$°$ ... (i)
In $∆$PCQ;
∠P + ∠C + ∠Q = 180$°$
[Angle sum property of a triangle]
$\Rightarrow$ ∠2 + ∠5 + ∠7 + ∠4 = 180$°$ ... (ii)
From (i) and (ii),
∠1 + ∠5 + ∠6 + ∠3 = ∠2 + ∠5 + ∠7 + ∠4
... (iii)
But ∠1 = ∠2 and ∠3 = ∠4
(Proved above)
$\therefore$ From (iii), we get
∠1 + ∠5 + ∠6 + ∠3 = ∠1 + ∠5 + ∠7 + ∠3
$\Rightarrow$ ∠6 = ∠7
or ∠ACP = ∠QCD [Hence proved]

Answer:

Given: Two circles intersect each other at points A and B. AP and AQ be their respective diameters.

To prove: Point B lies on the third side PQ.
Construction: Join A and B.
Proof: AP is a diameter.
$\therefore$ ∠1 = 90$°$ (Angle in a semicircle)
Also AQ is a diameter.
$\therefore$ ∠2 = 90$°$ (Angle in a semicircle)
∠1 + ∠2 = 90$°$ + 90$°$
$\Rightarrow$ ∠PBQ = 180$°$
$\Rightarrow$ PBQ is a line.
Thus, B (the point of intersection of these circles) lies on the third side i.e., on PQ.

Answer:

We have ABC and ADC two right triangles rt.
angled at B and D respectively.
$\therefore$ ∠ABC = ∠ADC (Each 90$°$)
If we draw a circle with AC (the common hypotenuse) as diameter, this circle will definitely passes through points B and D.
[Because B and D are the points in the alternate segment of an arc AC.]

Answer:

Let ABCD be a cyclic parallelogram. To prove that it is a rectangle, it is sufficient to show that one of the angles of parallelogram ABCD is a right angle.

Now ABCD is a parallelogram.
$\Rightarrow$ ∠B = ∠D ... (i)
[$\because$ Opposite angles of a gm are equal]
Also, ABCD is a cyclic quadrilateral.
$\Rightarrow$ ∠B + ∠D = 180$°$ ... (ii)
From (i) and (ii), we get:
∠B + ∠B = 180$°$
$\Rightarrow$ 2∠B = 180$°$
$\Rightarrow$ ∠B = 90$°$
Thus, ∠B = ∠D = 90$°$
Hence, ABCD is a rectangle.

Answer:

Let two circles with respective centres A and B intersect each other at points C and D.
We have to prove ∠ACB = ∠ADB

Proof: In $∆$ABC and $∆$ABD,
AC = AD (each radii)
BC = BD (each radii)
AB = AB (common)
$\therefore$ $∆$ABC $\cong$ $∆$ABD (By SSS of congruency)
$\Rightarrow$ ∠ACB = ∠ADB
(Congruent parts of congruent triangles)

Answer:

Let O be the centre of circle.
Join OA and OC.
Since perpendicular from the centre of the circle to the chord bisects the chord.

Answer:

Let AB = 6 cm and CD = 8 cm are the chords of circle with centre O.
Join OA and OC.
Since perpendicular from the centre of the circle to the chord bisects the chord.

Perpendicular distance of chord AB from the centre O is OE.
$\therefore$ OE = 4 cm
Now in rt. ∠d $∆$AOE
OA² = AE² + OE²
[Using Pythagoras result]
$\Rightarrow$ r² = 3² + 4²
$\Rightarrow$ r² = 9 + 16
$\Rightarrow$ r² = 25

Answer:

Join OA, OC, OE and OD.
Now, ∠AOC = 2∠AEC
[Angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any point in the alternate segment of circle.]

Answer:

Let ABCD be a rhombus in which diagonals AC and BD intersect each other at point O.

As we know that diagonals of a rhombus bisect and perpendicular to each other.
$\therefore$ ∠AOB = 90$°$
and if we draw a circle with side AB as diameter; it will definitely pass through point O (the point of intersection of diagonals) because then ∠AOB = 90$°$ will be the angle in a semicircle.

Answer:

In Fig. (a),
ABCD is a parallelogram
$\Rightarrow$ ∠1 = ∠3 ...(i)
(Opposite angles of a parallelogram)
ABCE is a cyclic quadrilateral
∠1 + ∠6 = 180$°$ ...(ii)
∠5 + ∠6 = 180$°$ (Linear pair) ...(iii)
From (ii) and (iii),
∠1 = ∠5 ...(iv)
Now, from (i) and (iv)
∠3 = ∠5
Now in $∆$AED,
∠3 = ∠5
$\Rightarrow$ AE = AD
($\because$ Sides opposite to equal angles in a triangle
are equal)
In fig. (b),
ABCD is a || gm.
$\Rightarrow$ ∠1 = ∠3
(Opposite angles of a ||gm)
∠2 = ∠4
Also AB CD and BC meets them
∠1 + ∠2 = 180$°$ ...(1)
And AD BC and EC meets them
∠5 = ∠2 (Corr. ∠s) ...(2)
ABCE is cyclic
$\therefore$ ∠1 + ∠6 = 180$°$ ...(3)
From (1) & (3), we get
∠1 + ∠2 = ∠1 + ∠6
$\Rightarrow$ ∠2 = ∠6
But from (2),
∠2 = ∠5
$\Rightarrow$ ∠5 = ∠6
Now in $∆$AED,
∠5 = ∠6
$\Rightarrow$ AE = AD (Q.E.D.)
Hence, in both the cases,
AE = AD.

Answer:

Let chords AC and BD of a circle bisect each other at O.
Then OA = OC and OB = OD.
We have to prove that (i) AC and BD are the
diameters. In other words, O is the centre of circle.

$\Rightarrow$ AD = BC
[Chords corresponding to the equal arcs]
...(3)
Similarly, AB = DC ...(4)
From (1), (2), (3) and (4) we observe that each angle of the quadrilateral is 90$°$ and opposite sides are equal.
Hence, ABCD is a rectangle.

Answer:

Answer:

Given: Two equal circles intersect in A and B.
A straight line through A meets the circles in P and Q.
To prove: BP = BQ
Construction: Join A and B.
Proof: AB is a common chord and the circles are equal.
$\therefore$ Arcs about the common chord are equal, i.e.,

Since equal arcs of two equal circles subtend equal angles at any point on the remaining part of the circle,
then, we have
∠1 = ∠2
In $∆$PBQ, we have
∠1 = ∠2 (Proved)
$\therefore$ Sides opposite to equal angles of a triangle are equal.
Then, we have
BP = BQ.

Answer:

Given that ABC is a triangle and a circle passes through its vertices.

Angle bisector of ∠A and the perpendicular bisector (say l) of its opposite side BC intersect each other at a point P (say).
We have to prove that circumcircle of $∆$ABC also passes through point P.
Proof: As we know that any point on the perpendicular bisector is equidistant from the end points of the corresponding side,
$\therefore$ BP = PC ...(i)
Also we have ∠1 = ∠2
[$\because$ AP is the bisector of ∠A (given)] ...(ii)
From (i) and (ii) we observed that equal line segments are subtending equal angles in the same segment (i.e., at point A) of circumcircle of $∆$ABC. Therefore BP and PC acts as chords of circumcircle of $∆$ABC and the corresponding