RBSE Solutions for Class 10 Maths Chapter 13 Circle and Tangent Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 13 Circle and Tangent Additional Questions.

## Rajasthan Board RBSE Class 10 Maths Chapter 13 Circle and Tangent Additional Questions

**Multiple Choice Questions :**

Question 1.

A point Q is 13 cm away from the center (RBSESolutions.com) of a circle and the length PQ of the tangent drawn from P to the circle is 12 cm, then radius of circle is (in cm) [CBSE 2012]

(A) 25

(B) \(\sqrt { 313 }\)

(C) 5

(D) 1

Solution :

PQ is the tangent on circle drawn from external point Q and OP is radius of circle.

So, ∠OPQ = 90° (By theorem 13.1)

From right angled OPQ

OQ^{2} = PQ^{2} + OP^{2} (By Pythagoras theorem)

⇒ 13^{2} = 12^{2} + OP^{2}

⇒ OP^{2} = 13^{2} – 12^{2}

⇒ OP^{2} = (13 + 12) (13 – 12)

⇒ OP^{2} = 25

⇒ OP = \(\sqrt { 25 }\) = 5 cm

Hence, correct choice is (C).

Question 2.

In the figure AP, AQ and BC are the (RBSESolutions.com) tangent of a circle. If AB = 5 cm,AC = 6 cm and BC = 4 cm, then length of AP is (in cm). [CBSE 2012]

(A) 7.5

(B) 15

(C) 10

(D) 9

Solution :

Since we know that length of tangents drawn on cine from an external point are equal

So, BP = BD (Tangents on circle from point B) ……(i)

CQ = CD (Tangents on circle from point C) ……(ii)

AP = AQ (Tangents (RBSESolutions.com) on circle from point A) ……(iii)

AP = AB + BP ⇒ AP = 5 + BD ……(iv)

AQ = AC + CQ ⇒ AQ = 6 + CD ……(v)

Adding equation (iv) and (v)

AP + AQ = 5 + BD + 6 + CD

⇒ AP + AP = 11 + BD + CD [By using equation (iii)]

⇒ 2 AP = 11 + BC

⇒ 2 AP = 11 + 4

⇒ 2 AP = 15 ⇒ AP = 7.5 cm

Hence, correct choice is (A).

Question 3.

In the figure PA and PB are tangents (RBSESolutions.com) to the circle C(O, r) from an external point P. If angle between in these tangents is 70°, then ∠POA is equal to.

(A) 60°

(B) 110°

(C) 55°

(D) 90°.

Solution :

PA and PB are tangents to a circle from point P and OA and OB are radius of circle. So AP ⊥ OA and PB ⊥ OB.

∴ ∠OAP = 90°

and ∠OBP = 90°

Now ∠AOB + ∠APB = 180°

⇒ ∠AOB + 70° = 180°

⇒ ∠AOB = 180° – 70°

∴ ∠AOB = 110°

∵ line OP is bisect of ∠AOB

∴ ∠POA = \(\frac { \angle AOB }{ 2 } \) = \(\frac { { 110 }^{ \circ } }{ 2 } \) = 55°

∴ ∠POA = 55°

Hence, correct choice is (C).

Question 4.

If four sides of rhombus ABCD touch (RBSESolutions.com) a circle, then

(A) AC + AD = BD + CD

(B) AB + CD = BC + AD

(C) AB + CD = AC + BC

(D) AC + AD = BC + DB.

Solution :

Correct choice is (B).

Question 5.

Two circle intersects each other at …… point.

(A) One

(B) To

(C) Three

(D) None of these

Solution :

Correct choice is (B).

Question 6.

In the figure, AT is a tangent to a (RBSESolutions.com) circle with center O. If OT = 4 cm and ∠OTA = 30°, then AT is equals to.

(A) 4 cm

(B) 2 cm

(C) 2\(\sqrt { 3 }\) cm

(D) 4\(\sqrt { 3 }\) cm [NCERT Exemplar Problem]

Solution :

Join OA. since AT is the tangent on the circle an OA is the radius of the circle. and OA is the radius of the circle.

So, AT ⊥ OA i.e., ∠OAT = 90°

From the right angled ΔOAT

∠OTA = 30° and OT = 4 cm

cos 30° = \(\frac { AT }{ OT }\)

⇒ \(\frac { \sqrt { 3 } }{ 2 }\) = \(\frac { AT }{ 4 }\)

⇒ AT = \(\frac { 4\sqrt { 3 } }{ 2 } \)

= 2\(\sqrt { 3 }\)

Hence, correct choice is (C).

Question 7.

In the following figure BC is diameter (RBSESolutions.com) of circle. AS is the tangent of circle at point A. If ∠ABC = 38°, then find ∠BAS.

(A) 52°

(B) 50°

(C) 62°

(D) 75°

Solution :

∵ BC is the diameter of circle.

∴ In ∆ABC,

∠BAC = 90°

∵ ∠ABC = 38° (given)

∵ ∠ACB = 180° – (∠ABC + ∠BAC)

= 180° – (38° + 90°)

or ∠ACB = 52°

Now, ∵ ∠BAS = ∠ACB

∴ ∠BAS = 52°

Hence, correct choice is (A).

Question 8.

According to following (RBSESolutions.com) figure if ∠PQR = 50°, then value of ∠QSR will be

(A) 90°

(B) 50°

(C) 40°

(D) 30°

Solution :

∠QSR is the angle made in (RBSESolutions.com) alternate circle sector of ∠PQR.

∴ ∠PQR = ∠QSR

= 50°

So, correct choice is (B).

Question 9.

In the figure, the value of ∠PQR is

(A) 30°

(B) 50°

(C) 60°

(D) 90°

Solution :

We know that, angle made in

semicircle in right angle.

∴ ∠PQR = 90°

So, correct choice is (D).

Question 10.

In the figure OP = 5 cm and radius (RBSESolutions.com) of circle = 3 cm, then find the length of tangent OPT drawn from point P circle.

(A) 3 cm

(B) 4 cm

(C) 5 cm

(D) 6 cm

Solution :

∵ ∠OTP = 90°

From right angled ∆OTP

OP^{2} = OT^{2} + PT^{2}

PT^{2} = OP^{2} – OT^{2}

= 5^{2} – 3^{2}

= 25 – 9 = 16

PT = \(\sqrt { 16 }\)

PT = 4 cm

So, correct choice is (B).

**Short/Long Answer Type Questions**

Question 1.

Answer the (RBSESolutions.com) following in ‘Yes’ and ‘No’.

(i) The length of two tangents of a circle is always same,

(ii) Scant drawn on any circle intersect two points on the circle.

(iii) In the circumscribed circle of an isoscale triangle ABC, the tangents on point A are such as AB = AC, which is parallel to BC.

(iv) Tangent and common points of a circle are called tangent points.

(v) The tangent on a point of a circle is parallel to radius passing thought the point of contact.

Solution :

(i) Yes

(ii) Yes

(iii) Yes

(iv) Yes

(v) No.

Question 2.

Fill in the blanks:

(i) Tangent at any point of circle is (RBSESolutions.com) perpendicular to …… passing through point of contact.

(ii) There are exactly ……. tangents to a circle through a point lying outside the circle.

(iii) Tangent and common point of circle are called ………… .

(iv) There is no common point on line and circle. In such condition line in respect to circle is …….

Solution :

(i) Radius

(ii) Two

(iii) Point of Contact

(iv) Non-intersection line.

Question 3.

How many tangents of circle can have?

Solution :

A tangent can be drawn from each point of the circumference of a circle as there are infinite points on the circumference of a circle. Hence, infinite tangents can be (RBSESolutions.com) drawn on circle.

Question 4.

Draw a tangent at any point of circle with radius 3 cm.

Solution :

Draw a tangent with the circle of radius 3 cm.

Construction : Let any point O. Draw a circle with radius r and take O as center. Mark a point P at the circumference of the circle and join O to P. OP is the radius of circle which is perpendicular to AB.

Note : Tangent drawn from only point of (RBSESolutions.com) the circle is perpendicular to the line which joins the centre.

OP ⊥ AB.

Question 5.

Two tangents RA and RB are drawn to a circle with center O from an external point R. If angle between tangents O and ∠AOB = 140° then find the value of angle θ. [Higher secondary board Raj. 2015]

Solution :

Given :

∠AOB = 140°

θ + 90° + 90° + 140° = 360°

θ + 320° = 360°

θ = 360° – 320° = 40°.

Question 6.

In the following figure, show that angle (RBSESolutions.com) which a chord of circle makes inalternate segment.

Solution :

In the figure XY is a tangent and PT is a chord. Hence ∠PAT and ∠PBT are angle made in alternate segment.

Question 7.

Prove that the tangents drawn at the ends of a diameter of a circle are parallel. [CBSE 2012]

Solution :

Given :

A circle with center O has (RBSESolutions.com) diameter AB. PQ and RS are the tangents at the points A and B.

To prove: PQ || RS.

Proof : PQ is the tangents to the circle at the point A. OA is radius

∴∠1 = 90°

Similarly RS ⊥ OB

∴∠2 = 90°

Now, ∠1 = ∠2

But this ¡s alternate angle of two (RBSESolutions.com) parallel lines, when a transversal cuts them.

∴ PQ || RS

Hence, the tangents drawn at the ends of a diameter of a circle are parallel.

Question 8.

In the figure, PA and PB are two tangents to the circle with center O. One point M is on the circle, then prove that

PL + LM = PN + MN.

Solution :

Given :

PA and PB are two tangent of the (RBSESolutions.com) circle. Line segment LN touches the circle at point M.

To prove:

PL + LM = PN + MN

Proof :

∴ Two tangents PA and PB drawn from point P.

So PA = PB

or PL + LA = PN + NB

Two tangents LA and LM are from point L

So LA = LM …(ii)

Similarly NB = NM .. .(iii)

From equation (i), (ii) and (iii) we get.

PL + LM = PN + NM.

Question 9.

In the figure, a circle touches all the (RBSESolutions.com) four sides of a quadrilateral ABCD. If AB = 8 cm BC = 6 cm and AD = 5 cm, then find CD.

Solution :

Let the circle touches all the sides of quadrilateral ABCD at the point P, Q, R and S.

We know that tangents to a circle (RBSESolutions.com) from an external point are equal.

AP = AS …..(i)

BP= BQ …(ii)

CR = CQ …..(iii)

DR = DS ……(iv)

Adding equation (i), (ii), (iii) and (iv)

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ AP + BP + CR + DR = AS + DS + BQ + CQ

⇒ AB + CD = AD + BC

⇒ 8 + CD = 5 + 6

⇒ 8 + CD = 11

⇒ CD = 11 – 8

∴ CD = 3 cm

Question 10.

In the figure, PQ is a chord of circle with (RBSESolutions.com) center O and PT is a tangent If ∠QPT = 60°, then find ∠PRQ. [CBSE 2015]

Solution : From figure,

OP ⊥ PT i.e. ∠OPT = 90°

∠OPQ = ∠OPT – ∠QPT

∠OPQ = 90° – 60°

= 30°

In ΔOPQ, (equal radii of circle)

∠OQP = ∠OPQ = 30°

(Opposite angles are equal to equal sides)

Now

∠OQP + ∠OPQ + ∠POQ = 180°

Sum of angles = 30° + 30° + ∠POQ = 180°

∠POQ = 180° – 60° = 120°

Exterior ∠POQ = 360° – 120° = 240°

We know that angle (RBSESolutions.com) subtended by centre of circle is tWice the angle subtended on the circumference of circle.

∠POQ = 2∠PRQ

⇒ 240° = 2∠PRQ

⇒ ∠PRQ = \(\frac { { 240 }^{ \circ } }{ 2 } \) = 120°

Hence ∠PRQ = 120°.

Question 11.

Prove that the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.

Solution :

Given :

A circle which has center O. PQ and PR are (RBSESolutions.com) tangents an the circle drawn from external point P of the circle.

To prove : ∠ROQ + ∠QPR = 180°.

Proof : OQ is radius and PQ is tangent touches at point Q of the circle from point P.

∠OQP = 90°

[∵ Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.]

Similarly ∠ORP = 90°

In quadrilateral ROQP

∠ROQ + ∠PRO + ∠OQP + ∠QPR = 360°

⇒ ∠ROQ + 90° + 90° + ∠QPR = 360°

⇒ ∠ROQ + ∠QPR = 360° – 180°

⇒ ∠ROQ + ∠QPR = 180°

So, ∠QPR, supplementary of ∠ROQ.

Question 12.

In the given figure, a circle with center O, is (RBSESolutions.com) Inscribed in quadrilateral ABCD such that it touches the sides AB, BC, CD and AD at point P, Q, R and S respectively. If radius of circle ¡s 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD, then find the length of CD. [CBSE 2013]

Solution :

DR and DS are the tangents from point D and OS and OR are the radius of circle.

∴ AD ⊥ OS and DR ⊥ OR (by theorem 13.1)

AD ⊥ CD (given)

In (RBSESolutions.com) quadrilateral DROS

∠D + ∠R + ∠O + ∠S = 360°

⇒ 90° + 90° + ∠O + 90° = 360°

⇒ ∠O = 360° – 270° = 90°

Similarly in quadrilateral DROS

∠D = ∠R = ∠O = ∠S = 90°

and OS = OR [radii of a circle

So, DROS is a square

So, SD = DR = 10 cm (Tangents on circle from point D.)

∵ Tangents BP and BQ on circle from point B.

∴ BP = BQ = 27 cm

CQ = BC – BQ

⇒ CQ = 38 – 27 = 11 cm

∵ CR and EQ are the tangents of circle from point C.

∴ CR = CQ

⇒ CR = 11 cm

Now CD = CR + DR

⇒ CD = 11 + 10

⇒ CD = 21 cm

Question 13.

If two tangents are drawn to (RBSESolutions.com) a circle of radius 3 cm such that the angle between them is 60°, then find the length of each tangent. [NCERT Exemplar Problem]

Solution :

Let PQ and PR are the tangents on circle from point P. OQ and OR are the radius of circle.

So, PQ ⊥ OQ and PR ⊥ OR

In right angled ∆POQ and ∆POR,

∠OQP = ∠ORP (each 90°)

hypotenuse PO = hypotenuse PO (common side)

and QQ = OR(equal radii of circle)

∴ ∆POQ = ∆POR (by RHS congruence)

⇒ ∠QPO = ∠RPO (CPCT)

⇒ ∠QPO = ∠RPO

= \(\frac { { 60 }^{ \circ } }{ 2 } \) = 30°

In right angled ∆OQP

tan 30° = \(\frac { OQ }{ PQ }\)

⇒ \(\frac { 1 }{ \sqrt { 3 } } \) = \(\frac { 3 }{ PQ }\) ⇒ PQ = 3\(\sqrt { 3 }\)

Since PQ and PQ are the tangents (RBSESolutions.com) from point P. we know that tangents to a circle from an external points are equal.

So, PR = PQ

= 3\(\sqrt { 3 }\) cm

Question 14.

In the following figure, O is the center of circle. PAQ is the tangent of circle at point A. If ∠OBA = 32°, then find the value of x and y.

Solution :

In ∆OAB

∵ OA = OB (radii of a circle)

∴ ∠OBA = ∠OAB (Opposite (RBSESolutions.com) angle of the equal side of triangle)

∴ ∠OAB = 32°

⇒ ∠x = 32°

(∠OBA = 32°) …..(i)

Again, PAQ the tangent at point A of circle and OA is radius.

∴ OA ⊥ PQ

⇒ ∠OAQ = 90°

∴ ∠BAQ + ∠OAB = 90°

∴ ∠BAQ + 32° = 90° [∵ ∠OAB = 32°]

∠BAQ = 90° – 32° = 58°

∠BAQ = ∠ACB [angle made in alternate segment]

58° = ∠y

∠x = 32°, ∠y = 58°

Question 15.

In the figure, A circle is (RBSESolutions.com) inscribed in a ∆ABC touching AB, BC and CA at P, Q and R respectively. If the length of tangent AP, BQ and CR are 3 cm, 4 cm and 35 cm respectively. Find perimeter of ∆ABC.

Solution :

We know that the tangents (RBSESolutions.com) drawn from an external point to a circle are euqal.

∴ AP = AR

BP = BQ

CQ = CR

∴ AP = AR = 3 cm

BP = BQ = 4 cm

and CR = CQ = 3.5 cm

AB = AP + PB

= (3 + 4) cm = 7 cm

BC = BQ + QC

= (4 + 3.5) cm = 7.5 cm

CA = AR + CR

= (3 + 3.5) cm = 6.5 cm

Perimeter of ∆ABC = (7 + 7.5 + 6.5) cm

= 21 cm

We hope the given RBSE Solutions for Class 10 Maths Chapter 13 Circle and Tangent Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 13 Circle and Tangent Additional Questions, drop a comment below and we will get back to you at the earliest.

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