NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

# NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 10 Circles Ex 10.6.

Ex 10.6 Class 9 Maths Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution:
Given:
Two circles with centres O and O’ which intersect each other at C and D.

To prove: OCO = ODO’
Construction: Join OC, OD, O’C and O’D
Proof: In ∆OCO’ and ∆ODO’, we have
OC = OD               (Radii of the same circle)
O’C = O’D             (Radii of the same circle)
OO’ = OO’             (Common)
∆OCO’ ODO’            (By SSS congruency rule)
Hence,
OCO = ODO   (By CPCT)

Ex 10.6 Class 9 Maths 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:
Given: A circle with centre O. AB = 5 cm and CD = 11 cm are two chords such that AB || CD and the perpendicular distance between AB and CD is 6 cm.

Let r cm be the radius of the circle.
Let us draw OP
AB and OQ CD such that PQ = 6 cm
Join OA and OC.
Let OQ = x cm
OP = (6 – x) cm

The perpendicular drawn from the centre of a circle to chord bisects the chord.

Ex 10.6 Class 9 Maths 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 1:
If both the parallel chords are on the same side of the centre.

Given: A circle with centre O. Two parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre.

Let r cm be the radius of the circle and draw OP AB and join OA and OC.
OP AB
P is the mid-point of AB.
AP = ½ AB = ½ (6 cm) = 3 cm
Similarly, CQ = ½ CD = ½ (8 cm) = 4 cm
Now in ∆OPA, we have OA2 = OP2 + AP2
r2 = 42 + 32
r2 = 16 + 9 = 25
r = 25 = 5 cm

Again, in ∆OCQ, we have OC2 = OQ2 + CQ2
r2 = OQ2 + 42
OQ2 = r2 – 42
OQ2 = 52 – 42 = 25 – 16 = 9         [ r = 5]
OQ = √9 = 3 cm
The distance of the other chord (CD) from the centre is 3 cm.
Case 2: If the two parallel chords are on either side of the centre.

In ∆POA, OA2 = OP2 + PA2
r2 = 42 + 32 = 52
r = 5
In ∆QOC, OC2 = CQ2 + OQ2
r2 = 42 + OQ2
OQ2 = r2 – 42 = 52 – 42 = 9
OQ = 3 cm

Ex 10.6 Class 9 Maths 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:
Given: ABC is such that when we produce arms BA and BC, they make two equal chords AD and CE.
To prove:
ABC = ½ [DOE – AOC]
Construction: Join AE.
Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles.
In ∆BAE, we have
DAE = ABC + AEC               ……(i)
The chord DE subtends
DOE at the centre and DAE in the remaining part of the circle.

ABC = ½ [(Angle subtended by the chord DE at the centre) – (Angle subtended by the chord AC at the centre)]
ABC = ½ [Difference of the angles subtended by the chords DE and AC at the centre]

Ex 10.6 Class 9 Maths 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: A rhombus ABCD such that its diagonals AC and BD intersect at O.
Taking AB as diameter, a circle is drawn. Let us draw PQ || DA and RS || AB, both are passing through O.
P, Q, R and S are the mid-points of DC, AB, AD and BC, respectively.
Q is the mid-point of AB.
AQ = QB                ….…(i)
ABCD is a rhombus]
RA = SB
RA = OQ                ….…(ii)

We have, AB = AD           [Sides of rhombus are equal]
AQ = AR                 ….…(iii)
From (i), (ii) and (iii), we have AQ = QB = OQ
i.e. A circle drawn with Q as centre, will pass through A, B and O.
Thus, the circle passes through the of intersection ‘O’ of the diagonals of rhombus ABCD.

Ex 10.6 Class 9 Maths 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: A parallelogram ABCD. We have a circle passing through A, B and C is drawn such that it intersects CD at E.

∴ ∠AEC + B = 180°                  …..…(i)
[Opposite angles of a cyclic quadrilateral are supplementary]

But ABCD is a parallelogram. [Given]
∴ ∠D = B                                   …..…(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
AEC + D = 180°                     …..…(iii)

But AEC + AED = 180°     [Linear pair]           …..…(iv)
From (iii) and (iv), we have
D = AED
i.e., The base angles of ADE are equal.
Opposite sides must be equal.

Ex 10.6 Class 9 Maths 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:
Given: A circle in which two chords AC and BD are such that they bisect each other. Let their point of intersection be O.
To prove: (i) AC and BD are diameters.
(ii) ABCD is a rectangle.
Construction: Join AB, BC, CD and DA.

Proof: (i) In ∆AOB and ∆COD, we have
AO = CO                          [O is the mid-point of AC]
BO = DO                          [O is the mid-point of BD]

AOB = COD                     [Vertically opposite angles]
∆AOB ∆COD                 [Using the SAS congruence rule]
AB = CD                       [C.P.C.T.]
arc AB = arc CD                   ….…(1)
Similarly, arc AD = arc BC       ….…(2)
Adding (1) and (2), we get
arc AB + arc AD = arc CD + arc BC
BD divides the circle into two equal parts.
BD is a diameter.
Similarly, AC is a diameter.

(ii) We have proved in part (i) that ∆AOB ∆COD
OAB = OCD                 [C.P.C.T.]
CAB = ACD
AB || DC

ABCD is a parallelogram.
Since, opposite angles of a parallelogram are equal.
DAB = DCB
But
DAB + DCB = 180°
[Sum of the opposite angles of a cyclic quadrilateral is 180°]
DAB = DCB = 90°

Thus, ABCD is a rectangle.

Ex 10.6 Class 9 Maths Question 8.
Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – ½ A, 90° – ½ B and 90° – ½ C.

Solution:
Given: A triangle ABC inscribed in a circle, such that bisectors of A, B and C intersect the circumcircle at D, E and F respectively.
Construction: Join DE, EF and FD.
Proof:

Angles in the same segment are equal.
FDA = FCA                 ….…(i)
EDA = EBA                    ….…(ii)
Adding (i) and (ii), we have
FDA + EDA = FCA + EBA
FDE = FCA + EBA

Ex 10.6 Class 9 Maths Question 9.
Two congruent circles intersect each other at points 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:
We have two congruent circles such that they intersect each other at points A and B. A line segment passing through A, meets the circles at P and Q.
Let us draw the common chord AB. Since angles subtended by equal chords in the congruent circles are equal.
APB = AQB
Now, in ∆PBQ, we have
AQB = APB
So, their opposite sides must be equal.
BP = BQ.

Ex 10.6 Class 9 Maths Question 10.
In any ∆ABC, if the angle bisector of A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.

Solution:
Given: ∆ABC with O as centre of its circumcircle. The perpendicular bisector of BC passes through O. Join OB and OC. Suppose it cuts circumcircle at P.

In order to prove that the perpendicular bisector of BC and bisector of angle A of ∆ABC intersect at P, it is sufficient to show that AP is bisector of A of ∆ABC.

Arc BC makes angle θ at the circumference.
BOC = 2θ
[Angle at centre is double the angle made by an arc at circumference]
Also, in ∆BOC, OB = OC and OP is perpendicular bisector of BC.
So,
BOP = COP = θ
Arc CP makes angle θ at O, so it will make angle θ/2 at circumference.
Arc BP makes angle θ at O, so it will make angle θ/2 at circumference.
Hence, AP is the angle bisector of
A of ∆ABC.

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