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Exercise 10.6 - Chapter 10 Circles class 9 ncert solutions Maths - SaraNextGen [2024]


Question 1:

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

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_2642de33.jpg

Let two circles having their centres as O and https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  intersect each other at point A and B respectively. Let us join Ohttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif .

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m3cf42d26.jpg

In ΔAOhttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  and BOhttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif ,

OA = OB (Radius of circle 1)

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif A = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  B (Radius of circle 2)

Ohttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  = Ohttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  (Common)

ΔAOhttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  ≅ ΔBOhttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  (By SSS congruence rule)

∠OAhttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  = ∠OBhttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2581/Chapter%2010_html_m6b2683df.gif  (By CPCT)

Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

 

Question 2:

Two chords AB and CD of lengths 5 cm 11cm 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.

Answer:

Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2584/Chapter%2010_html_m1e2f9fc7.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2584/Chapter%2010_html_me2bd84e.gif  (Perpendicular from the centre bisects the chord)

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2584/Chapter%2010_html_79fa0f8c.gif

Let ON be x. Therefore, OM will be 6− x.

In ΔMOB,

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2584/Chapter%2010_html_m54540597.gif

In ΔNOD,

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2584/Chapter%2010_html_m67968e8a.gif

We have OB = OD (Radii of the same circle)

Therefore, from equation (1) and (2),

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2584/Chapter%2010_html_m1148a5db.gif

From equation (2),

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2584/Chapter%2010_html_m2042036b.gif

Therefore, the radius of the circle ishttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2584/Chapter%2010_html_4638faa9.gif cm.

 

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?

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2599/Chapter%2010_html_4e266f.jpg

Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.

Distance of smaller chord AB from the centre of the circle = 4 cm

OM = 4 cm

MB = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2599/Chapter%2010_html_7da1dcc8.gif

In ΔOMB,

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2599/Chapter%2010_html_777b1314.gif

In ΔOND,

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2599/Chapter%2010_html_4f5219c4.gif

Therefore, the distance of the bigger chord from the centre is 3 cm.

 

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.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2600/Chapter%2010_html_m3c528580.jpg

In ΔAOD and ΔCOE,

OA = OC (Radii of the same circle)

OD = OE (Radii of the same circle)

AD = CE (Given)

∴ ΔAOD ≅ ΔCOE (SSS congruence rule)

∠OAD = ∠OCE (By CPCT) … (1)

∠ODA = ∠OEC (By CPCT) … (2)

Also,

∠OAD = ∠ODA (As OA = OD) … (3)

From equations (1), (2), and (3), we obtain

∠OAD = ∠OCE = ∠ODA = ∠OEC

Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x

In Δ OAC,

OA = OC

∴ ∠OCA = ∠OAC (Let a)

In Δ ODE,

OD = OE

∠OED = ∠ODE (Let y)

ADEC is a cyclic quadrilateral.

∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)

x + a + x + y = 180°

2x + a + y = 180°

y = 180º − 2x − a … (4)

However, ∠DOE = 180º − 2y

And, ∠AOC = 180º − 2a

∠DOE − ∠AOC = 2a − 2= 2− 2 (180º − 2x − a)

= 4a + 4x − 360° … (5)

∠BAC + ∠CAD = 180º (Linear pair)

⇒ ∠BAC = 180º − ∠CAD = 180º − (a + x)

Similarly, ∠ACB = 180º − (a + x)

In ΔABC,

∠ABC + ∠BAC + ∠ACB = 180º (Angle sum property of a triangle)

∠ABC = 180º − ∠BAC − ∠ACB

= 180º − (180º − a − x) − (180º − a −x)

= 2a + 2− 180º

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2600/Chapter%2010_html_eeecab0.gif [4a + 4x − 360°]

∠ABC = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2600/Chapter%2010_html_m5a4d85ce.gif [∠DOE − ∠ AOC] [Using equation (5)]

 

Question 5:

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

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2602/Chapter%2010_html_1ff1bfa9.jpg

Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.

∴ ∠COD = 90°

Also, in rhombus, the diagonals intersect each other at 90°.

∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Clearly, point O has to lie on the circle.

 

Question 6:

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

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2606/Chapter%2010_html_c087b35.jpg

It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.

∠AEC + ∠CBA = 180°

∠AEC + ∠AED = 180° (Linear pair)

∠AED = ∠CBA … (1)

For a parallelogram, opposite angles are equal.

∠ADE = ∠CBA … (2)

From (1) and (2),

∠AED = ∠ADE

AD = AE (Angles opposite to equal sides of a triangle)

 

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.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2610/Chapter%2010_html_3048653e.jpg

Let two chords AB and CD are intersecting each other at point O.

In ΔAOB and ΔCOD,

OA = OC (Given)

OB = OD (Given)

∠AOB = ∠COD (Vertically opposite angles)

ΔAOB ≅ ΔCOD (SAS congruence rule)

AB = CD (By CPCT)

Similarly, it can be proved that ΔAOD ≅ ΔCOB

∴ AD = CB (By CPCT)

Since in quadrilateral ACBD, opposite sides are equal in length, ACBD is a parallelogram.

We know that opposite angles of a parallelogram are equal.

∴ ∠A = ∠C

However, ∠A + ∠C = 180° (ABCD is a cyclic quadrilateral)

⇒ ∠A + ∠A = 180°

⇒ 2 ∠A = 180°

⇒ ∠A = 90°

As ACBD is a parallelogram and one of its interior angles is 90°, therefore, 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.

 

Question 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° https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2611/Chapter%2010_html_21fe8425.gif .

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2611/Chapter%2010_html_m2ecc2dce.jpg

It is given that BE is the bisector of ∠B.

∴ ∠ABE = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2611/Chapter%2010_html_177ed28a.gif

However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)

⇒ ∠ADE = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2611/Chapter%2010_html_177ed28a.gif

Similarly, ∠ACF = ∠ADF = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2611/Chapter%2010_html_m7a78a323.gif (Angle in the same segment for chord AF)

∠D = ∠ADE + ∠ADF

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2611/Chapter%2010_html_4e39744f.gif

Similarly, it can be proved that

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2611/Chapter%2010_html_5d49ab0c.gif

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.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2616/Chapter%2010_html_m962b15a.jpg

AB is the common chord in both the congruent circles.

∴ ∠APB = ∠AQB

In ΔBPQ,

∠APB = ∠AQB

∴ BQ = BP (Angles opposite to equal sides of a triangle)

Question 10:

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

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2619/Chapter%2010_html_m78d740da.jpg

Let perpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of side BC intersect it at E.

Perpendicular bisector of side BC will pass through circumcentre O of the circle. ∠BOC and ∠BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively. We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∠BOC = 2 ∠BAC = 2 ∠A … (1)

In ΔBOE and ΔCOE,

OE = OE (Common)

OB = OC (Radii of same circle)

∠OEB = ∠OEC (Each 90° as OD ⊥ BC)

∴ ΔBOE ≅ ∠COE (RHS congruence rule)

∠BOE = ∠COE (By CPCT) … (2)

However, ∠BOE + ∠COE = ∠BOC

⇒ ∠BOE +∠BOE = 2 ∠A [Using equations (1) and (2)]

⇒ 2 ∠BOE = 2 ∠A

⇒ ∠BOE = ∠A

∴ ∠BOE = ∠COE = ∠A

The perpendicular bisector of side BC and angle bisector of ∠A meet at point D.

∴ ∠BOD = ∠BOE = ∠A … (3)

Since AD is the bisector of angle ∠A,

∠BAD = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/107/2619/Chapter%2010_html_77164196.gif

⇒ 2 ∠BAD = ∠A … (4)

From equations (3) and (4), we obtain

∠BOD = 2 ∠BAD

This can be possible only when point BD will be a chord of the circle. For this, the point D lies on the circum circle.

Therefore, the perpendicular bisector of side BC and the angle bisector of ∠A meet on the circum circle of triangle ABC.

Also Read : Exercise-11.1-Chapter-11-Constructions-class-9-ncert-solutions-Maths

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