Question - 2 : -
Two chords AB and CD of lengths 5 cm and 11 cm respectively of acircle are parallel to each other and are on opposite sides of its centre. Ifthe distance between AB and CD is 6 , find the radius of the circle.

Answer - 2 : -

Here, OM ⊥ AB and ON ⊥ CD. is drawn and OB and OD are joined.

We know that ABbisects BM as the perpendicular from the centre bisects chord.

Since AB = 5 so,

BM = AB/2

Similarly, ND = CD/2 =11/2

Now, let ON be x.

So, OM = 6−x.

Consider ΔMOB,

OB² =OM²+MB²

Or,

Consider ΔNOD,

OD²=ON²+ ND²

We know, OB = OD(radii)

From equation 1 andequation 2 we get

Now, from equation (2)we have,

OD²= 1²+(121/4)

Or OD = (5/2)×√5 cm

Question - 3 : -
The lengths of two parallel chords of a circle are 6 cm and 8 cm. Ifthe smaller chord is at a distance 4 cm from the centre, what is the distanceof the other chord from the centre?

Answer - 3 : -

Consider the followingdiagram

Here AB and CD are 2parallel chords. Now, join OB and OD.

Distance of smallerchord AB from the centre of the circle = 4 cm

So, OM = 4 cm

MB = AB/2 = 3 cm

Consider ΔOMB

OB² =OM²+MB²

Or, OB = 5cm

Now, consider ΔOND,

OB = OD = 5 (sincethey are the radii)

ND = CD/2 = 4 cm

Now, OD²=ON²+ND²

Or, ON = 3 cm.

Question - 4 : -
4. Let the vertex of an angle ABC be located outside a circle and let thesides of the angle intersect equal chords AD and CE with the circle. Prove that∠ABC is equal to half the difference ofthe angles subtended by the chords AC and DE at the centre.

Answer - 4 : -

Consider the diagram

In ΔAOD andΔCOE,

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 aresupplementary)

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 − 2y = 2a − 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 +2x −180º

= [4a + 4x − 360°]

∠ABC =

[∠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 - 5 : -

To prove: A circle drawn with Qas centre, will pass through A, B and O (i.e. QA = QB = QO)

Since all sides of arhombus are equal,

AB = DC

Now, multiply (½) onboth sides

(½)AB = (½)DC

So, AQ = DP

BQ = DP

Since Q is themidpoint of AB,

AQ= BQ

Similarly,

RA = SB

Again, as PQ is drawnparallel to AD,

RA = QO

Now, as AQ = BQ and RA= QO we get,

QA = QB = QO (henceproved).

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 - 6 : -

Here, ABCE is a cyclicquadrilateral. In a cyclic quadrilateral, the sum of the opposite angles is180°.

So, ∠AEC+∠CBA = 180°

As ∠AEC and ∠AED are linear pair,

∠AEC+∠AED = 180°

Or, ∠AED = ∠CBA … (1)

We know in aparallelogram; opposite angles are equal.

So, ∠ADE = ∠CBA … (2)

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

∠AED = ∠ADE

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

∴ AD = AE (proved).

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 - 7 : -

Here chords AB and CDintersect each other at O.

Consider ΔAOB andΔCOD,

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

OB = OD (Given in thequestion)

OA = OC (Given in thequestion)

So, by SAS congruency,ΔAOB ≅ ΔCOD

Also, AB = CD (ByCPCT)

Similarly, ΔAOD ≅ ΔCOB

Or, AD = CB (By CPCT)

In quadrilateral ACBD,opposite sides are equal.

So, ACBD is aparallelogram.

We know that oppositeangles of a parallelogram are equal.

So, ∠A = ∠C

Also, as ABCD is acyclic quadrilateral,

∠A+∠C = 180°

⇒∠A+∠A = 180°

Or, ∠A = 90°

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

∠A is the angle subtended bychord BD. And as ∠A = 90°, therefore, BDshould be the diameter of the circle. Similarly, AC is the diameter of thecircle.

Question - 8 : -
Bisectors of angles A, B and C of a triangle ABC intersect itscircumcircle at D, E and F respectively. Prove that the angles of the triangleDEF are 90°–(½)A, 90°–(½)B and 90°–(½)C.

Answer - 8 : -

Consider the followingdiagram

Here, ABC is inscribedin a circle with center O and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and Frespectively.

Now, join DE, EF andFD

As angles in the samesegment 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)] °

Question - 9 : -
Two congruent circles intersect each other at points A and B. ThroughA any line segment PAQ is drawn so that P, Q lie on the two circles. Prove thatBP = BQ.

Answer - 9 : -

The diagram will be

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

Now, consider ΔBPQ,

∠APB = ∠AQB

So, the anglesopposite to equal sides of a triangle.

∴ BQ = BP

Question - 10 : -
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersecton the circumcircle of the triangle ABC.

Answer - 10 : -

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

Perpendicular bisector of side BC will pass throughcircumcentre O of the circle. ∠BOC and ∠BAC are theangles subtended by arc BC at the centre and a point A on the remaining part ofthe circle respectively. We also know that the angle subtended by an arc at thecentre is double the angle subtended by it at any point on the remaining partof 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 (RHScongruence rule)

∠BOE = ∠COE (By CPCT) … (2)

However, ∠BOE + ∠COE = ∠BOC

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

⇒ 2 ∠BOE = 2 ∠A

⇒ ∠BOE = ∠A

∴ ∠BOE = ∠COE = ∠A

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

∴ ∠BOD = ∠BOE = ∠A … (3)

Since AD is the bisector of angle ∠A,

∠BAD =

⇒ 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 ofthe 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.

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