*NCERT Solutions For Class 9 Math Chapter – 10 Exercise – 10.6*

**Q1. Prove that the line of centers of two intersecting circles subtends equal angels at wo points of intersection.**

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**Solution:-**

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∆OAO’ and ∆OBO’

OA = OB (Radii of circle with center O)

O’A = O’B (Radii of circle with center O’)

OO’ = OO’ (Common)

Therefore by SSS criterion ∆OAO’ and ∆OBO’ are congruent to each other.

By CPCT ∟OAO’ = ∟OBO’

Hence it is proved that the line of centers of two intersecting circles subtends equal angels at it the two points of intersection.

**Q2. 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 of its center If the distance parallel between AB and CD is 6 cm Find the radius of the circle.**

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**Solution:-**

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AB = 5 cm

CD = 11 cm

MB = 2.5 cm

ND = 5.5 cm

Let OM = x cm and ON = 6 – x cm

∆OMB

By Pythagoras theorem ,

OM^{2} + MB^{2} = OB^{2}

X^{2} + 25^{2} = OB^{2}

X^{2} + 6.25 = OB^{2} (I)

∆OND

By Pythagoras theorem

ON^{2} + ND^{2} = OD^{2}

- X)
^{2}+ 5.5^{2}= OD^{2}

36 + x^{2} – 12x + 30.25 = OD^{2}

X^{2}_{ }– 12x + 66.25 = OD^{2}

OB and OD are the radii of the circle Therefore

OB = OD.

Equation (I) and (ii) we get,

X^{2} + 6.25 = x^{2} – 12x + 66.25

12x = 60

X = 5

Substituting the value of x in (1) or (2) we get the radius of circle = 5.59 cm.

**Q3. The length of two parallel chords of a circle are 6 cm and 8 cm If the smaller chord is at distance 4 cm from the center What is the distance of the other chord from the center?**

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**Solution:-**

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AB = 6 cm

CD = 8 cm

MB = 3 cm

ND = 4 cm

Given OM = 4 cm and let ON = x cm

∆OMB

OM^{2} + MB^{2} = OB^{2}

4^{2} + 3^{2} = OB^{2}

OB^{2} = 25

OB = 5 cm

OB and OD are the radii of the circle. Therefore OD = OB = 5 cm

Consider ∆OND

ON^{2} + ND^{2} = OD^{2}

X^{2} + 4^{2 } = 5^{2}

X = 3

The distance of the chord CD from the center is 3 cm

**Q4. Let the vertex of an angel ABC be located outside a circle and let the sides of the angel intersect equal chord AD and CE with the circle Prove that ****∟****ABC is equal to half the difference of the angels subtended b the chords AC and DE at the center.**

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**Solution:-**

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∆AOD and ∆COE,

OA = OC

OD = OE

AD = CE

∆AOD ∆COE (SSS congruence Rule)

∟OAD = ∟OCE [By CPCT]

∟ODA = ∟OEC [By CPCT]

∟OAD = ∟ODA (As OA = OD)

∟OAD = ∟OCE = ∟ODA = ∟OEC

Let ∟OAD = ∟OCE = ∟ODA = ∟OEC = X

In ∆OAC ,

OA = OC

∟OCA = ∟OAC

∆ODE,

OD = OE

∟OED = ∟ODE

ADEC is a cyclic quadrilateral

∟CAD + ∟DEC = 180^{0}

X + A + X + Y = 180^{0}

2X + A + Y = 180^{0}

Y = 180^{0} – 2x – a

∟DOE = 180^{0} – 2Y

∟AOC = 180^{0} – 2a

∟DOE – ∟AOC = 2A – 2Y = 2A – 2 (180^{0 } – 2X – A)

= 4a + 4x – 360^{0}

∟BAC + ∟CAD = 180^{0} (Linear pair)

∟BAC = 180^{0} – ∟CAD = 180^{0} – (a+x)

∟ACB = 180^{0} – (a+x)

In ∆ABC,

∟ABC + ∟BAC + ∟ACB = 180^{0}

∟ABC = 180^{0} – ∟BAC – ∟ACB

= 180^{0} – (180^{0} – a – x) – (180 – a – x)

= 2a + 2x – 180^{0}

= 1/2[4a + 4x – 360^{0}]

Using Equation (5)

∟ABC = 1/2(∟DOE – ∟AOC)

**Q5. Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of it diagonals.**

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**Solution:-**

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∟COD = 90^{0}

Also, in rhombus, the diagonal intersect each other at 90^{0}.

∟AOB = ∟B0C = ∟COD = ∟DOA = 90^{0}

O has to lie on the circle.

**Q6.****ABCD is a parallelogram the circle through A , B and C intersect CD at E. Prove that AE = AD.**

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**Solution:**

Here, ABCE is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of the opposite angles is 180°.

So, ∠AEC+∠CBA = 180°

As ∠AEC and ∠AED are linear pair,

∠AEC+∠AED = 180°

Or, ∠AED = ∠CBA … (1)

We know in a parallelogram; opposite angles are equal.

So, ∠ADE = ∠CBA … (2)

Now, from equations (1) and (2) we get,

∠AED = ∠ADE

Now, AD and AE are angles opposite to equal sides of a triangle,

∴ AD = AE (proved).

**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:**

Here chords AB and CD intersect each other at O.

Consider ΔAOB and ΔCOD,

∠AOB = ∠COD (They are vertically opposite angles)

OB = OD (Given in the question)

OA = OC (Given in the question)

So, by SAS congruency, ΔAOB ≅ ΔCOD

Also, AB = CD (By CPCT)

Similarly, ΔAOD ≅ ΔCOB

Or, AD = CB (By CPCT)

In quadrilateral ACBD, opposite sides are equal.

So, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal.

So, ∠A = ∠C

Also, as ABCD is a cyclic quadrilateral,

∠A+∠C = 180°

⇒∠A+∠A = 180°

Or, ∠A = 90°

As ACBD is a parallelogram and one of its interior angles is 90°, so, it is a rectangle.

∠A is the angle subtended by chord BD. And as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.

**8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°–(½)A, 90°–(½)B and 90°–(½)C.**

**Solution:**

Consider the following diagram

Here, ABC is inscribed in a circle with center O and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F respectively.

Now, join DE, EF and FD

As angles in the same segment are equal, so,

∠FDA = ∠FCA ————-(i)

∠FDA = ∠EBA ————-(i)

By adding equations (i) and (ii) we get,

∠FDA+∠EDA = ∠FCA+∠EBA

Or, ∠FDE = ∠FCA+∠EBA = (½)∠C+(½)∠B

We know, ∠A +∠B+∠C = 180°

So, ∠FDE = (½)[∠C+∠B] = (½)[180°-∠A]

∠FDE = [90-(∠A/2)]

In a similar way,

∠FED = [90° -(∠B/2)] °

And,

∠EFD = [90° -(∠C/2)] °

**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:**

The diagram will be

Here, ∠APB = ∠AQB (as AB is the common chord in both the congruent circles.)

Now, consider ΔBPQ,

∠APB = ∠AQB

So, the angles opposite to equal sides of a triangle.

∴ BQ = BP

**10. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.**

**Solution:**

Consider this diagram

Here, join BE and CE.

Now, since AE is the bisector of ∠BAC,

∠BAE = ∠CAE

Also,

∴arc BE = arc EC

This implies, chord BE = chord EC

Now, consider triangles ΔBDE and ΔCDE,

DE = DE (It is the common side)

BD = CD (It is given in the question)

BE = CE (Already proved)

So, by SSS congruency, ΔBDE ΔCDE.

Thus, ∴∠BDE = ∠CDE

We know, ∠BDE = ∠CDE = 180°

Or, ∠BDE = ∠CDE = 90°

∴ DE ⊥ BC (hence proved).

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