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.
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 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.
∵ 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.
⇒ 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)
Since AD = BC [ ∵ ABCD is a rhombus]
∴ ½ AD = ½ BC
⇒ RA = SB
⇒ RA = OQ ….…(ii)
[
∵ PQ is drawn parallel to AD and AD = BC]
We have, AB = AD [Sides of
rhombus are equal]
⇒ ½ AB = ½ AD
⇒ 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.
⇒ AE = AD
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.
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
⇒ arc BAD = arc BCD
⇒ 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
Similarly, AD || BC
∴ 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.
∴ ∠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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