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# NCERT Solutions for Class 9 Maths Exercise 10.6

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NCERT solutions for Class 9 Maths Circles

## NCERT Solutions for Class 9 Mathematics Circles

###### . Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Ans. Let two circles with respective centers A and B intersect each other at points C and D.

We have to prove ACB = ADB

Proof: In triangles ABC and ABD,

BC = BD =

AB = AB [Common]

ABC ABD

[SSS rule of congruency]

NCERT Solutions for Class 9 Maths Exercise 10.6

###### 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 he radius of the circle.

Ans. Let O be the centre of the circle. Join OA and OC.

Since perpendicular from the centre of the circle to the chord bisects the chord.

AE = EB = AB = x 5 = cm

And CF = FD = CD = x 11 = cm

Let OE =

OF =

Let radius of the circle be

In right angled triangle AEO,

AO2 = AE2 + OE2

[Using Pythagoras theorem]

….(i)

Again In right angled triangle CFO,

OC2 = CF2 + OF2

###### [Using Pythagoras theorem]

….(ii)

Equating eq. (i) and (ii),

Now from eq. (i),

cm

Hence radius of the circle is cm.

NCERT Solutions for Class 9 Maths Exercise 10.6

###### 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord form the centre?

Ans. 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.

AE = EB = AB = x 6 = 3 cm

And CF = FD = CD = x 8 = 4 cm

Perpendicular distance of chord AB from the centre O is OE.

OE = 4 cm

Now in right angled triangle AOE,

###### OA2 = AE2 + OE2 [Using Pythagoras theorem]

= 32 + 42

= 9 + 16 = 25

= 5 cm

Perpendicular distance of chord CD from the center O is OF.

Now in right angled triangle OFC,

OC2 = CF2 + OF2 [Using Pythagoras theorem]

= 42 + OF2

= 16 + OF2

OF2 = 16

OF = 3cm

Hence distance of other chord from the centre is 3 cm.

NCERT Solutions for Class 9 Maths Exercise 10.6

###### 4. Let vertex of an angle ABC be located outside a circle and let the sides of the angle intersect 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.

Ans. Vertex B of ABC is located outside the circle with centre O.

Side AB intersects chord CE at point E and side BC intersects chord AD at point D with the circle.

We have to prove that

ABC = [AOC – DOE]

Join OA, OC, OE and OD.

Now AOC = 2AEC

[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 the circle]

AOC = AEC …(i)

Similarly DOE = DCE ….(ii)

Subtracting eq. (ii) from eq. (i),

[AOC – DOE] = AEC – DCE ….(iii)

[Angles in same segment in circle] ….(iv)

Also DCE = DAE

[Angles in same segment in circle] ….(v)

Using eq. (iv) and (v) in eq. (iii),

[AOC – DOE]

= DAE + ABD – DAE

[AOC – DOE] = ABD

Or [AOC – DOE] = ABC

Hence proved.

###### 5. Prove that the circle drawn with any drawn with any side of a rhombus as a diameter, passes through the point of intersection of its diagonals.

Ans. 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.

AOB =

And if we draw a circle with side AB as diameter, it will definitely pass through point O (the point intersection of diagonals) because then AOB = will be the angle in a semi-circle.

NCERT Solutions for Class 9 Maths Exercise 10.6

###### 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced it necessary) at E. Prove that AE = AD.

Ans. In figure (a),

ABCD is a parallelogram.

1 = 3 ….(i)

1 + 6 = ….(ii)

And 5 + 6 = ….(iii)

[Linear pair]

From eq. (ii) and (iii), 1 = 5 ….(iv)

Now, from eq. (i) and (iv),

3 = 5

AE = AD [Sides opposite to equal angles are equal]

(a) (b)

In figure (b),

ABCD is a parallelogram.

1 = 3 and 2 = 4

Also ABCD and BC meets them.

1 + 2 = ….(i)

And ADBC and EC meets them.

5 = 2 ….(ii) [Corresponding angles]

Since ABCE is a cyclic quadrilateral.

1 + 6 = ….(iii)

From eq. (i) and (iii),

1 + 2 = 1 + 6

2 = 6

But from eq. (ii), 2 = 5

5 = 6

Now in triangle AED,

5 = 6

AE = AD [Sides opposite to equal angles]

Hence in both the cases, AE = AD

NCERT Solutions for Class 9 Maths Exercise 10.6

###### 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.

Ans. Given: AC and BD of a circle bisect each other at O.

Then OA = OC and OB = OD

To prove: (i) AC and BD are the diameters. In other words, O is the centre of the circle.

(ii) ABCD is a rectangle.

Proof: (i) In triangles AOD and BOC,

AO = OC [given]

AOD = BOC [Vertically opp.]

OD = OB [given]

AODCOB [SAS congruency]

Similarly AOBCOD

AB = CD

[Arcs opposite to equal chords]

AC = BD [Chords opposites to equal arcs]

###### AC and BD are the diameters as only diameters can bisects each other as the chords of the circle.

(ii) Ac is the diameter. [Proved in (i)]

B = D = ….(i) [Angle in semi-circle]

Similarly BD is the diameter.

A = C = …(ii) [Angle in semi-circle]

Now diameters AC = BD

[Arcs opposite to equal chords]

AD = BC [Chords corresponding to the equal arcs] ….(iii)

Similarly AB = DC ….(iv)

From eq. (i), (ii), (iii) and (iv), we observe that each angle of the quadrilateral is and opposite sides are equal.

Hence ABCD is a rectangle.

NCERT Solutions for Class 9 Maths Exercise 10.6

###### 8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that angles of the triangle are and respectively.

Ans. According to question, AD is bisector of A.

1 = 2 =

And BE is the bisector of B.

3 = 4 =

Also CF is the bisector of C.

5 = 6 =

###### Since the angles in the same segment of a circle are equal.

9 = 3 [angles subtended by ] ….(i)

And 8 = 5 [angles subtended by ] ….(ii)

9 + 8 = 3 + 5

D =

Similarly E =

And F =

In triangle DEF,

D + E + F =

D = (E + F )

D =

D =

D = [A + B + C = ]

D =

Similarly, we can prove that

E = and F =

NCERT Solutions for Class 9 Maths Exercise 10.6

###### 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.

Ans. 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.

Arc 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 triangle PBQ,

1 = 2 [proved]

Sides opposite to equal angles of a triangle are equal.

Then we have, BP = BQ

NCERT Solutions for Class 9 Maths Exercise 10.6

###### 10. In any triangle ABC, if the angle bisector of and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.

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

Angle bisector of A and the perpendicular bisector (say ) of its opposite side BC intersect each other at a point P.

To prove: Circumcircle of triangle ABC also passes through point P.

Proof: Since any point on the perpendicular bisector is equidistant from the end points of the corresponding side,

BP = PC ….(i)

Also we have 1 = 2 [AP is the bisector of A (given)] ….(ii)

From eq. (i) and (ii) we observe 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 congruent arcs and acts as parts of circumcircle. Hence point P lies on the circumcircle. In other words, points A, B, P and C are concyclic (proved).

## NCERT Solutions for Class 9 Maths Exercise 10.6

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