Question - 13 : - In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC. [NCERT Exemplar]

Answer - 13 : -

Let O be the centre of the given circle. Suppose, the tangent at P meets BC at Q.

Join BP.

To prove : BQ = QC

[angles in alternate segment]

Proof : ∠ABC = 90°

[tangent at any point of circle is perpendicular to radius through the point of contact]

In ∆ABC, ∠1 + ∠5 = 90°

[angle sum property, ∠ABC = 90°]

∠3 = ∠1

[angle between tangent and the chord equals angle made by the chord in alternate segment]

∠3 + ∠5 = 90° ……..(i)

Also, ∠APB = 90° [angle in semi-circle]

∠3 + ∠4 = 90° …….(ii)

[∠APB + ∠BPC = 180°, linear pair]

From Eqs. (i) and (ii), we get

∠3 + ∠5 = ∠3 + ∠4

∠5 = ∠4

=> PQ = QC

[sides opposite to equal angles are equal]

Also, QP = QB

[tangents drawn from an internal point to a circle are equal]

=> QB = QC

Hence proved.

Question - 14 : - From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ∆PCD.

Answer - 14 : -

PA and PB are the tangents drawn from a point P out side the circle with centre O

CD is another tangents to the circle at point E which intersects PA and PB at C and D respectively

PA = 14 cm

PA and PB are the tangents to the circle from P

PA = PB = 14 cm

Now CA and CE are the tangents from C

CA = CE ….(i)

Similarly DB and DE are the tangents from D

DB = DE ….(ii)

Now perimeter of ∆PCD

= PC + PD + CD

= PC + PD + CE + DE

= PC + CE + PD + DE

= PC + CA + PD = DB {From (i) and (ii)}

= PA + PB

= 14 + 14

= 28 cm

Question - 15 : - In the figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle. [CBSE 2002]

Answer - 15 : -

In right ∆ABC, ∠B = 90°

BC = 6 cm, AB = 8 cm

Let r be the radius of incircle whose centre is O and touches the sides A B, BC and CA at P, Q and R respectively

AP and AR are the tangents to the circle AP = AR

Similarly CR = CQ and BQ = BP

OP and OQ are radii of the circle

OP ⊥ AB and OQ ⊥ BC and ∠B = 90° (given)

BPOQ is a square

BP = BQ = r

AR = AP = AB – BD = 8 – r

and CR = CQ = BC – BQ = 6 – r

But AC² = AB² + BC² (Pythagoras Theorem)

= (8)² + (6)² = 64 + 36 = 100 = (10)²

AC = 10 cm

=> AR + CR = 10

=> 8 – r + 6 – r = 10

=> 14 – 2r = 10

=> 2r = 14 – 10 = 4

=> r = 2

Radius of the incircle = 2 cm

Question - 16 : - Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. [NCERT Exemplar]

Answer - 16 : -

Let mid-point of an arc AMB be M and TMT’ be the tangent to the circle.

Join AB, AM and MB.

Since, arc AM = arc MB

=> Chord AM = Chord MB

In ∆AMB, AM = MB

=> ∠MAB = ∠MBA ……(i)

[equal sides corresponding to the equal angle]

Since, TMT’ is a tangent line.

∠AMT = ∠MBA

[angle in alternate segment are equal]

∠AMT = ∠MAB [from Eq. (i)]

But ∠AMT and ∠MAB are alternate angles, which is possible only when AB || TMT’

Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Hence proved

Question - 17 : - From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle show that ∆APB is equilateral.

Answer - 17 : -

Given : From a point P outside the circle with centre O, PA and PB are the tangerts to the circle such that OP is diameter.

AB is joined.

To prove: APB is an equilateral triangle

Const : Join OP, AQ, OA

Proof : OP = 2r

=> OQ + QP = 2r

=> OQ = QP = r (OQ = r)

Now in right ∆OAP,

OP is its hypotenuse and Q is its mid point

OA = AQ = OQ

(mid-point of hypotenuse of a right triangle is equidistances from its vertices)

∆OAQ is equilateral triangle ∠AOQ = 60°

Now in right ∆OAP,

∠APO = 90° – 60° = 30°

=> ∠APB = 2 ∠APO = 2 x 30° = 60°

But PA = PB (Tangents from P to the circle)

=> ∠PAB = ∠PBA = 60°

Hence ∆APB is an equilateral triangle.

Question - 18 : - Two tangents segments PA and PB are drawn to a circle with centre O such that ∠APB = 120°. Prove that OP = 2 AP. [CBSE 2014]

Answer - 18 : -

Given : From a point P. Out side the circle with centre O, PA and PB are tangerts drawn and ∠APB = 120°

OP is joined To prove : OP = 2 AP

Const: Take mid point M of OP and join AM, join also OA and OB.

Proof : In right ∆OAP,

∠OPA = 1/2 ∠APB = 1/2 x 120° = 60°

∠AOP = 90° – 60° = 30°

M is mid point of hypotenuse OP of ∆OAP

MO = MA = MP

∠OAM = ∠AOM = 30° and ∠PAM = 90° – 30° = 60°

∆AMP is an equilateral triangle

MA = MP = AP

But M is mid point of OP

OP = 2 MP = 2 AP

Hence proved.

Question - 19 : - If ∆ABC is isosceles with AB = AC and C (0, r) is the incircle of the ∆ABC touching BC at L. Prove that L bisects BC.

Answer - 19 : -

Given: In ∆ABC, AB = AC and a circle with centre O and radius r touches the side BC of ∆ABC at L.

To prove : L is mid point of BC.

Proof : AM and AN are the tangents to the circle from A

AM = AN

But AB = AC (given)

AB – AN = AC – AM

BN = CM

Now BL and BN are the tangents from B

BL = BN

Similarly CL and CM are tangents

CL = CM

But BM = CM (proved)

BL = CL

L is mid point of BC.

Question - 20 : - AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD. [NCERT Exemplar]

Answer - 20 : -

To prove, BC = BD

Join BC and OC.

Given, ∠BAC = 30°

=> ∠BCD = 30°

[angle between tangent and chord is equal to angle made by chord in the alternate segment]

∠ACD = ∠ACO + ∠OCD

∠ACD = 30° + 90° = 120°

[OC ⊥ CD and OA = OC = radius => ∠OAC = ∠OCA = 30°]

In ∆ACD,

∠CAD + ∠ACD + ∠ADC = 180°

[since, sum of all interior angles of a triangle is 180°]

=> 30° + 120° + ∠ADC = 180°

=> ∠ADC = 180° – (30° + 120°) = 30°

Now, in ∆BCD,

∠BCD = ∠BDC = 30°

=> BC = BD

[since, sides opposite to equal angles are equal]

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