Important Questions Class 10 Maths Chapter 10

Important Questions Class 10 Mathematics Chapter 10 – Circles

Mathematics is based on proper conceptual understanding and a strong grasp of the topics. The subject cannot be memorised and needs a detailed approach to every problem. The Important Questions Class 10 Mathematics Chapter 10 helps students to develop the reasoning and thinking abilities required in this fast-paced world. Hence, it is one of the crucial subjects in the primary and secondary syllabus.

The chapter is all about circles, and is associated with theorems and laws. The important topics covered in the Mathematics Class 10 Chapter 10  Important Questions include tangent to the Circle and the Number of Tangents from a point on the circle and all the theories and assumptions related to it. 

In the  Important Questions Class 10 Mathematics Chapter 10, we have covered every topic in detail with a proper analysis of each concept. It has certain examples which will help students to connect to their Mathematical studies in the future directly. Students will be able to apply these formulas as required in the problems once they refer to them. It includes everything you might be searching and looking up for during your preparation.

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Important Questions Class 10 Mathematics Chapter 10 – With Solutions

The following questions and their solutions are included in the Class 10 Mathematics Chapter 10  Important Questions. 

Question 1. The given radii of the two concentric circles are 4 cm, and 5 cm, the length of each chord of one circle that is tangent to the other circle is:

(A) 3 cm 

(B) 6 cm

(C) 9 cm 

(D) 1 cm

Answer 1:

As per the question,

OA = 4cm, OB = 5cm

Also, OA ⊥ BC

Hence, OB2 = OA2 + AB2

52 = 42 + AB2

 AB = √(25 – 16) = 3cm

BC = 2AB = 2 × 3cm = 6cm

 

Question 2. In Fig.  , if AOB = 125°, COD is equal to

(A) 62.5° (B) 45° 

(C) 35° (D) 55° 

Answer 2:

ABCD is the quadrilateral circumscribing of the circle

We have the opposite sides of the quadrilateral circumscribing the circle and subtending supplementary angles at the circle’s centre.

So, we get,

∠AOB + ∠COD = 180°

125° + ∠COD = 180°

∠COD = 55°

 

Question 3. In the given Fig. AB is the chord of the circle, and AOC is the diameter so that ACB = 50°. When AT is the tangent to the circle at the point A, then BAT is the same as

(A) 65° (B) 60°

(C) 50° (D) 40°

Answer 3:

As per the question,

A circle with the centre O, the diameter AC and ∠ACB = 50°

AT is the tangent of the circle at the point A

As the angle in the semicircle is a right angle

∠CBA = 90°

By the angle sum property of the triangle,

∠ACB + ∠CAB + ∠CBA = 180°

50° + ∠CAB + 90° = 180°

∠CAB = 40° … (1)

As the tangent to at any point on the circle is perpendicular to the radius passing through point of contact,

We have,

OA ⏊ AT

∠OAT = 90°

∠OAT + ∠BAT = 90°

∠CAT + ∠BAT = 90°

40° + ∠BAT = 90°  [from the equation (1)]

∠BAT = 50°

 

Question 4. From a point P which is at a distance of 13 cm from the centre O of the circle of radius 5 cm, the pair of the tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

(A) 60 cm2 (B) 65 cm2

(C) 30 cm2 (D) 32.5 cm2

Answer 4:

Construction: Draw the circle of radius 5 cm with centre O.

Let P be the point at the distance of 13 cm from O.

Draw the pair of tangents, PQ and PR.

OQ = OR = radius = 5cm …equation (1)

Also, OP = 13 cm

We have a tangent at any point on the circle that is perpendicular to the radius passing through the point of the contact,

Thus, we have,

OQ ⏊ PQ and OR ⏊PR

△POQ and △POR are the right-angled triangles.

Using the Pythagoras Theorem in △PQO,

(Base) 2+ (Perpendicular)2= (Hypotenuse)2

(PQ.)2 + (OQ)2= (OP)2

(PQ.)2 + (5)2 = (13)2

(PQ.)2 + 25 = 169

(PQ.)2 = 144

PQ = 12 cm

Tangents through the external point to the circle are equal.

Hence,

PQ = PR = 12 cm … (2)

Thus, Area of the quadrilateral PQRS, A = area of △POQ + area of △POR

Area of the right angled triangle = ½ x base x perpendicular

A = (½ x OQ x PQ) + (½ x OR x PR)

A = (½ x 5 x 12) + (½ x 5 x 12)

A = 30 + 30 = 60 cm2

 

Question 5. At one end, A of the diameter AB of the circle of radius 5 cm, through which tangent XAY is drawn to the circle. The length of the chord CD is parallel to XY as well as at the distance of 8 cm from the point A, is

(A) 4 cm (B) 5 cm

(C) 6 cm (D) 8 cm

Answer 5:

From the question,

Radius of the circle, AO=OC = 5cm

AM=8CM

AM=OM+AO

OM =AM-AO

Putting these values in the equation,

OM= (8-5) =3CM

OM is perpendicular to the chord CD.

In the ∆OCM <OMC=90°

By the Pythagoras theorem,

OC²=OM²+MC²

Hence,

CD= 2 ×CM = 8 cm

 

Question 6: How many tangents can be drawn from the external point towards the circle?

Answer 6: Two tangents can be drawn from the external point towards the circle.

 

Question 7: Given: A triangle OAB that is an isosceles triangle and AB is the tangent to the circle with the centre O. Find out the measure of ∠OAB.

Answer 7: The measure of the angle ∠OAB for the given isosceles triangle OAB will be 45 degrees.

 

Question 8: What should be the angle between the two tangents that are drawn at the end of the two radii and are inclined at the angle of 45 degrees?

Answer 8: The angle between them should be 135 degrees.

 

Question 9: Given that the right triangle QPR  is the right-angled at Q. QR = 12 cm, PQ = 5 cm. The radius for the circle which is inside a triangle QPR will be?

Answer 9: The radius of a circle will be 2 cm.

 

Question 10: Define Tangent and Secant.

Answer 10: A tangent is a line that meets the circle only at one point.

A secant is a line that meets the circle at two points while intersecting it. These two points are always distinct.

 

Question 11: What is a circle?

Answer 11: The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle.

 

Question 12. When the chord AB makes an angle of 60° at the circle’s centre, the angle between the tangents on A and B is at 60°.

Answer 12:

False

Explanation:

For example,

Consider the given figure for which we have a circle with centre O and AB as the chord with ∠AOB = 60°

As the tangent to any point on the circle is perpendicular to the radius through the point of the contact,

We have,

OA ⏊ AC and OB ⏊ CB

∠OBC = ∠OAC = 90° … eq(1)

Using the angle addition property of the quadrilateral in Quadrilateral AOBC,

We have,

∠OBC + ∠OAC + ∠AOB + ∠ACB = 360°

90° + 90° + 60° + ∠ACB = 360°

∠ACB = 120°

So, the angle between the two tangents is 120°.

Thus we can conclude that the given sentence is false.

 

Question 13. The length of the tangent for the external point on the circle is always greater than the circle’s radius.

Answer 13:

False

Explanation:

The length of the tangent from the external point P on the circle might be or not greater than the circle’s radius.

 

Question 14. How many tangents could a circle have?

Answer 14:

There can be infinite tangents to the circle. A circle is made up of the infinite points that are at an equal distance from the point. As there are infinite points at the circumference of the circle, infinite tangents could be drawn from them.

 

Question 15. Fill in the blanks:

(i) A tangent to the circle cuts it in …………… point(s).

(ii) A line intersecting the circle in two points is called the ………….

(iii) A circle could have …………… parallel tangents at the most.

(iv) The common point of the tangent to the circle and the circle is called …………

Answer 15:

(i) A tangent to the circle cuts it into one point(s).

(ii) A line intersecting the circle in two points is called a secant.

(iii) A circle could have two parallel tangents at the most.

(iv) The common point of the tangent to the circle and the circle is called the point of contact.

 

Question 16. A tangent PQ at the point P of the circle of radius 5 cm meets the line passing through the centre O at the point Q such that OQ = 12 cm. The Length PQ is :

(A) 12 cm

(B) 13 cm

(C) 8.5 cm

(D) √119 cm

Answer 16: (D) √119 cm

In the given figure, the line which is drawn from the centre of the given circle to the tangent PQ is perpendicular to the line PQ.

Thus, OP ⊥ PQ

By using the Pythagoras theorem in the triangle ΔOPQ we have,

OQ2 = OP2+PQ2

(12)2 = 52+PQ2

PQ2 = 144-25

PQ2 = 119

PQ = √119 cm

Hence, option D, which is √119 cm, is the length of PQ.

 

Question 17. Draw the circle and two lines parallel to the given line so that one is the tangent and the

other the secant to the circle.

Answer 17:

In the given figure, XY as well as AB are two parallel lines. The line segment AB is tangent at the point C, whereas the line segment XY is the secant.

 

Question 18. From the point Q, the length of the tangent to the circle is 24 cm, as well as the distance having Q from the centre is 25 cm. The radius for the circle is

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

Answer 18: (A) 7 cm

First, draw the perpendicular from the centre O of the triangle to the point P on the circle that is touching the tangent. This line would be perpendicular to the tangent of the circle.

Thus, OP is the perpendicular to PQ, that is, OP ⊥ PQ

From the given figure, it is also seen that △OPQ is the right-angled triangle.

It is also given that

OQ = 25 cm and PQ = 24 cm

By using the Pythagoras theorem in the△OPQ,

OQ2 = OP2 +PQ2

(25)2 = OP2+(24)2

OP2 = 625-576

OP2 = 49

OP = 7 cm

Thus, option A, which is 7 cm, is the circle’s radius.

 

Question 19. In the Fig., when TP and TQ are the two tangents to the circle with centre O so that ∠POQ. = 110°, the ∠PTQ is equal to

(A) 60°

(B) 70°

(C) 80°

(D) 90°

Answer 19: (B) 70°

From the given question, we have that OP is the radius of the circle to the tangent line PT, and OQ is the radius to the tangent TQ.

Thus, OP ⊥ PT and TQ. ⊥ OQ

∠OPT = ∠OQT = 90°

For the quadrilateral POQT, we know that the sum for the interior angles is 360°

Thus,  ∠PTQ+∠POQ+∠OPT+∠OQT = 360°

By putting the respective values, we have

∠PTQ +90°+110°+90° = 360°

∠PTQ = 70°

Thus, ∠PTQ is 70° which is option B.

 

Question 20. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠ POA is equal to

(A) 50°

(B) 60°

(C) 70°

(D) 80°

Answer 20: (A) 50°

Firstly, draw the diagram according to the given statement.

Then, in the above given diagram, OA is the radius to tangent PA, and OB is the radius to tangent PB.

thus, OA is perpendicular to PA, and OB is perpendicular to PB, i.e. OA ⊥ PA and OB ⊥ PB

So, ∠OBP = ∠OAP = 90°

Now, in the quadrilateral A

.O.B.P.,

The addition of all the interior angles will be 360°

So, ∠AOB+∠OAP+∠OBP+∠APB = 360°

Putting their values, we get,

∠AOB + 260° = 360°

∠AOB = 100°

then, consider the triangles △OPB and △OPA. Here,

AP = BP (Since the tangents from the point are always equal)

OA = OB (Which are the radii for the circle)

OP = OP (It is the common side)

then, we can say that triangles OPB and OPA are similar using SSS congruency.

therefore, △OPB ≅ △OPA

So, ∠POB = ∠POA

∠AOB = ∠POA+∠POB

2 (∠POA) = ∠AOB

By putting the respective values, we observe,

=>∠POA = 100°/2 = 50°

As the angle ∠POA is 50°, option A is the correct option.

 

Question 21. Prove us that the tangents drawn at the ends of the diameter of a circle are parallel.

Answer 21:

Firstly, draw the circle and connect two points, A and B, such that AB becomes the diameter of the circle. then, draw two tangents, PQ and RS, at points A and B, respectively.

then, both radii, i.e. AO and OB, are perpendicular to the tangents.

thus, OB is perpendicular to RS and OA perpendicular to PQ

thus, ∠OAP = ∠OAQ = ∠OBR = ∠OBS = 90°

From the above given figure, angles OBR and OAQ are alternate interior angles.

and, ∠OBR = ∠OAQ and ∠OBS = ∠OAP (Since they are also alternate interior angles)

thus, it can be said that line PQ and the line RS will be parallel to each other. (Hence Proved).

 

Question 22. Proves that the perpendicular at the point of contact to the tangent to the circle passes through the centre.

Answer 22:

Let’s assume O is the centre of the given circle.

A tangent line PR has been drawn, touching the circle at point P.

Draw QP ⊥ RP at point P, so that point Q lies on the circle.

∠OPR = 90° (radius ⊥ tangent)

Also, ∠QPR = 90° (Given)

therefore ∠OPR = ∠QPR

then, the above case is possible only when the centre O lies on the line QP.

thus, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

 

Question 23. The length of the tangent for the point A at the distance of 5 cm from the centre of the circle of 4 cm. Find out the radius of the circle.

Answer 23:

Draw the diagram as shown below.

Where AB is the tangent that is drawn on the circle from point A.

thus, the radius OB will be perpendicular to AB, i.e. OB ⊥ AB

We get, OA = 5cm and AB = 4 cm

then, In △ABO,

OA2 =AB2+BO2 (Using Pythagoras theorem)

52 = 42+BO2

BO2 = 25-16

BO2 = 9

BO = 3

thus, the radius of the given circle, i.e. BO, is 3 cm.

 

Question 24. Two concentric circles are of the radii 5 cm and 3 cm. Find out the length of a chord of the larger circle which touches the smaller circle.

Answer 24:

Draw two concentric circles having the centre O. and then draw a chord AB in the larger circle that touches the smaller circle at a point P, as shown in the figure below.

From the above given diagram, AB is tangent to the smaller circle to point P.

therefore, OP ⊥ AB

Using Pythagoras theorem in triangle OPA,

OA2= AP2+OP2

52 = AP2+32

AP2 = 25-9

AP = 4

Now, as OP ⊥ AB,

hence the perpendicular from the centre of the circle bisects the chord. AP will be equal to PB.

thus, AB = 2AP = 2×4 = 8 cm

here, the length of the chord of the larger circle is 8 cm.

 

Question 25. A quadrilateral ABCD has been drawn to circumscribe a circle (see in Fig.  ). Prove that AB + CD = AD + BC

Answer 25:

The figure given is:

From the given figure we can conclude a few points, which are:

(i) DR = DS

(ii) BP = BQ

(iii) AP = AS

(iv) CR = CQ

hence they are tangents on the circle from points D, B, A, and C, respectively.

then, adding the LHS and RHS of the above equations we observe,

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

By the rearranging them, we observe,

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

By the simplifying,

AD+BC= CD+AB

 

Question 26. In Fig., XY and X′Y′ are two parallel tangent line to a circle with centre O and another tangent line AB with the point of contact C intersecting XY at A, and X′Y′ at the point B. Prove that ∠ AOB = 90°.

Answer 26:

From the figure given in our the textbook, join OC Now, the diagram will be as-

Now the triangles △OPA and △OCA are same using SSS congruency as:

(i) OP = OC it is the radii of the same circle

(ii) AO = AO these are the common side

(iii) AP = AC This is the tangents from the point A

So, △OPA ≅ △OCA

Similarly,

△OQB ≅ △OCB

thus,

∠POA = ∠COA … (Equation i)

And, ∠QOB = ∠COB … (Equation ii)

hence the line POQ is a straight line, it could be considered as a diameter of the circle.

thus, ∠POA +∠COA +∠COB +∠QOB = 180°

then, from equations (i) and equation (ii), we get,

2∠COA+2∠COB = 180°

∠COA+∠COB = 90°

∴∠AOB = 90°

 

Question 27. Prove that the angle formed between the two tangents drawn from the external point to the circle is supplementary to the angle subtended with the line segment joining to the points of contact at the centre.

Answer 27:

Firstly, draw a circle with centre O. Choose the external point P and draw two tangents PA as well as PB at point A and point B, respectively. then, join A and B to make AB in a way that subtends ∠AOB at the centre of the circle. The diagram is as follows:

From the above diagram, it is seen that the line segments OA and PA are perpendicular.

thus, ∠OAP = 90°

In a symmetrical way, the line segments OB ⊥ PB and so, the angle ∠OBP = 90°

then, in the quadrilateral OAPB,

therefore, ∠APB+∠OAP +∠PBO +∠BOA = 360° (since the sum of all interior angles will be 360°)

By putting the values, we get,

∠APB + 180° + ∠BOA = 360°

So, ∠APB + ∠BOA = 180° (Hence proved).

 

Question 28. Prove that the parallelogram circumscribing the circle is a rhombus.

Answer 28:

Consider a parallelogram ABCD that is circumscribing a circle with a centre O. Now, since ABCD is a parallelogram, AB = CD and BC = AD.

From the above given figure, we see that,

(i) DR = DS

(ii) BP = BQ

(iii) CR = CQ

(iv) AP = AS

These are the tangents to the circle at D, B, C, and A, respectively.

Adding all these, we observe,

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

By the rearranging them, we observe,

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

Again by rearranging them, we get,

AB+CD = BC+AD

then, since AB = CD and BC = AD, the above equation becomes

2AB = 2BC

∴ AB = BC

hence AB = BC = CD = DA, it can be said that ABCD is a rhombus.

 

Question 29. Prove that the opposite sides of the quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer 29:

First, draw a quadrilateral ABCD that will circumscribe a circle with its centre O in a way that it touches the circle at points P, Q, R, and S. Now, after joining the vertices of ABCD. We get the following figure:

Now, consider the triangles OAP and OAS

AP = AS (They are the tangent lines from the same point A)

OA = OA (these are the common side)

OP = OS (it is the radii of the circle)

thus, by SSS congruency △OAP ≅ △OAS

thus, ∠POA = ∠AOS.

Which implies that∠1 = ∠8

Similarly, other angles will be,

∠4 = ∠5

∠2 = ∠3

∠6 = ∠7

then by adding these angles, we observe,

∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360°

then by rearranging,

(∠1+∠8)+(∠2+∠3)+(∠4+∠5)+(∠6+∠7) = 360°

2∠1+2∠2+2∠5+2∠6 = 360°

we are taking 2 as common and solving we observe,

(∠1+∠2)+(∠5+∠6) = 180°

hence, ∠AOB+∠COD = 180°

same as it can be proved that ∠BOC+∠DOA = 180°

thus, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the centre of the circle.

 

Question 30. The length of the tangent from the external point P on a circle with centre O is always less than OP.

Answer 30:

True

clarify:

Consider the following figure of a circle with the centre O.

Let assume PT be a tangent drawn from external point P.

then, Joint OT.

IT ⏊ PT

We know that,

The tangent can be at any point on the circle is perpendicular to the radius through the point of contact.

Hence, O.P.T. is a right-angled triangle formed.

We also know that,

In the right-angled triangle, the hypotenuse is always greater than any of the two sides of the triangle.

Hence,

OP > PT or PT < OP

Hence, the length of the tangent from an external point P on a circle with centre O is always less than OP.

 

Question 31. The angle formed between the two tangents to the circle may be 0°.

Answer 31:

True

clarify:

The angle formed between the two tangents to the circle may be 0°only when both of the tangent lines coincide or are parallel to each other.

 

Question 32. when the angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°, then OP = a√2.

Answer 32:

True.

Tangent is always perpendicular to the radius at the point of contact.

therefore, ∠RPT = 90

when 2 tangents are drawn from the external point, they are equally inclined to the line segment joining the centre to that point.

Consider the following figure,

From point P, two tangents are drawn.

It is given that OT = a

And line OP bisects ∠RPT.

So,

∠TPO = ∠RPO = 45o

We know that OT ⊥ PT.

In right-angled triangle OTP,

Sin 45o = OT/OP

= 1/√2 = a/OP

Hence, OP = a√2

 

Question 33. Outside from the two concentric circles, the radius for the outer circle is 5 cm, and the chord AC of the length of 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Answer 33:

From the given figure,

Chord AB = 8 cm

OC is perpendicular to the chord AB

AC = CB = 4 cm

In the right triangle OCA

OC2 + CA2 = OA2

OC2 = 52 – 42 = 25 – 16 = 9

OC = 3 cm

 

Question 34. Two tangents, PQ and PR, are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

Answer 34:

We know that,

Radius ⊥ Tangent = OR ⊥ PR

i.e., ∠ORP = 90°

Likewise,

Radius ⊥ Tangent = OQ ⊥PQ

∠OQP = 90°

In quadrilateral ORPQ,

Sum of all interior angles = 360º

∠ORP + ∠RPQ+ ∠PQO + ∠QOR = 360º

90º + ∠RPQ + 90º + ∠QOR = 360º

Hence, ∠O + ∠P = 180o

PROQ is a cyclic quadrilateral.

 

Question 35. when from an external point B of a circle with centre O, two tangents BC and BD are drawn such that angle DBC = 120°, proving that BC + BD. = B.O., i.e., B.O. = 2BC.

Answer 35:

According to the question,

By RHS rule,

ΔOBC and ΔOBD are congruent

By CPCT.

∠OBC and ∠ OBD are equal

Therefore,

∠OBC = ∠OBD =60°

In triangle OBC,

cos 60°=BC/OB

½ =BC/OB

OB=2BC

Hence proved

 

Question 36. Prove that the centre of the circle touching two intersecting lines lies on the angle bisector of the lines.

Answer 36:

Let assume the lines be l1 and l2.

we know that O touches l₁ and l₂ at M and N,

We 

observe,

OM = ON (Radius of the circle)

hence,

From the centre “O” of a circle, it has an same distance from l₁ & l₂.

In Δ OPM & OPN,

OM = ON (Radius of the circle)

∠OMP = ∠ONP (Radius is perpendicular to its tangent)

OP = OP (Common sides)

hence,

Δ OPM = ΔOPN (SSS congruence rule)

By CPCT,

∠MPO = ∠NPO

So, l bisects ∠MPN.

Hence, O lies on the bisector of the angle between l₁ & l₂.

Therefore, we prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

 

Question 37. In Fig., AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.

Answer 37:

as per the question,

AB = CD

Construction: Draw the side AB and CD to intersect at the point P.

Proof:

Consider the circle with a greater radius.

Tangents drawn from the external point to a circle are equal

AP = CP …(1)

Also,

Consider the circle with a smaller radius.

Tangents drawn from the external point to a circle are equal

BP = BD …(2)

Subtract Equation (2) from (1). We Get

AP – BP = CP – BD

AB = CD

Hence Proved.

Exercise 9.4 Page No: 110

 

Question 38. when a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.

Answer 38:

According to the question,

A Hexagon ABCDEF circumscribes a circle.

To prove:

AB + CD + EF = BC + DE + FA

Proof:

Tangents drawn from the external point to a circle are equal.

therefore, we get

AM = RA … eq 1 [tangents having point A]

BM = BN … eq 2 [tangents having point B]

CO = NC … eq 3 [tangents having point C]

OD = DP … eq 4 [tangents having point D]

EQ = PE … eq 5 [tangents having point E]

QF = FR … eq 6 [tangents having point F] [eq 1]+[eq 2]+[eq 3]+[eq 4]+[eq 5]+[eq 6]

AM + BM + CO + OD + EQ + QF = RA + BN + NC + DP + PE + FR

On rearranging, we observe,

(AM + BM) + (CO + OD) + (EQ + QF) = (BN + NC) + (DP + PE) + (FR + RA)

AB + CD + EF = BC + DE + FA

thus Proved!

 

Question 39. Let’s denote the semi-perimeter of the triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, and AB at D, E, and F, respectively, prove that BD. = s – b.

Answer 39:

As per the given question,

A triangle ABC with the side BC = a , CA = b and AB = c . And a circle is inside that touches the sides BC, CA and AB at the points D, E and F, respectively and s is the semi- perimeter for the triangle

To Prove that BD. = s – b

Proof:

as per question,

We have,

Semi Perimeter = s

Perimeter = 2s

2s = AB + BC + AC [1]

As we know that,

Tangents drawn from the external point to a circle are equal

So we have

AF = AE [2] [Tangents having point A]

BF = BD [3] [Tangents having point B]

CD = CE [4] [Tangents having point C]

sum of [2] [3] and [4]

AF + BF + CD = AE + BD + CE

AB + CD = AC + BD

sum of BD both the side

AB + CD + BD = AC + BD + BD

AB + BC – AC = 2BD

AB + BC + AC – AC – AC = 2BD

2s – 2AC = 2BD [From 1]

2BD = 2s – 2b [as AC = b]

BD = s – b

thus Proved.

 

Question 40. From an external point P, two tangents, PA and PB, are drawn to a circle with the centre O. At one point, E on the circle tangent is drawn, that intersects PA and PB at point C and D, respectively. When PA = 10 cm, find out the perimeter of the triangle PCD.

as per question,

From the external point P, two tangents, PA as well as PB, are drawn to a circle with centre O. At a point E on the circle, the tangent is drawn, that intersects PA and PB at point C and D, respectively. And PA = 10 cm

To Find: Perimeter of △PCD

As we know that, Tangents drawn from the external point to a circle are equal.

So we have

AC = CE [1] [Tangents having point C]

ED = DB [2] [Tangents having point D]

then Perimeter of the Triangle PCD

= PC + CD + DP

= PC + CE + ED + DP

= PC + AC + DB + DP [From 1 and 2]

= PA + PB

then,

PA = PB = 10 cm as tangents drawn from an external point to a circle are equal.

So we have

Perimeter = PA + PB = 10 + 10 = 20 cm

 

Question 41. when AB is a chord of a circle with centre O, AOC is diameter as well as AT is the tangent at A as shown in the given Fig.  . Prove that ∠BAT = ∠ACB.

Answer 41:

According to the question,

A circle with centre O and AC as diameter and AB and BC as two chords. also, AT is a tangent at the point A

To Prove : ∠BAT = ∠ACB

Proof :

∠ABC = 90° [Angle in a semicircle is the right angle]

In △ABC By angle addition property of triangle

∠ABC + ∠ BAC + ∠ACB = 180 °

∠ACB + 90° = 180° – ∠BAC

∠ACB = 90 – ∠BAC [1]

then,

OA ⏊ AT [Tangent at the point on the circle is perpendicular to the radius through point of contact ]

∠OAT = ∠CAT = 90°

∠BAC + ∠BAT = 90°

∠BAT = 90° – ∠BAC [2]

From [1] and [2]

∠BAT = ∠ACB [Proved]

 

Question 42. Two circles with the centres O and O‘ of the radii of 3 cm and 4 cm, respectively, intersect at the two points, P and Q, such that OP and O‘P are tangents to the two circles. Find the length of the common chord PQ.

According to the question,

Two circles with the centres O and O’ of radii 3 cm and 4 cm, respectively, intersect at the two points, P and Q, such that OP and O’P are tangents to the two circles, and PQ is a common chord.

To Find out: Length of common chord PQ

∠OPO’ = 90° [Tangent at the point on the circle is the perpendicular to the radius through point of contact]

thus, OPO is the right-angled triangle at P

Using the Pythagoras in △ OPO’, we know

(OO’)2= (O’P)2+ (OP)2

(OO’)2 = (4)2 + (3)2

(OO’)2 = 25

OO’ = 5 cm

Let assume ON = x cm and NO’ = 5 – x cm

In the right angled triangle ONP

(ON)2+ (PN)2= (OP)2

x2 + (PN)2 = (3)2

(PN)2= 9 – x2[1]

In the right angled triangle O’NP

(O’N)2 + (PN)2= (O’P)2

(5 – x)2 + (PN)2 = (4)2

25 – 10x + x2 + (PN)2 = 16

(PN)2 = -x2+ 10x – 9[2]

From [1] and [2]

9 – x2 = -x2 + 10x – 9

10x = 18

x = 1.8

From (1) we have

(PN)2 = 9 – (1.8)2

=9 – 3.24 = 5.76

PN = 2.4 cm

PQ = 2PN = 2(2.4) = 4.8 cm

 

Question 43. In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.

Answer 43:

According to the question,

At a right angle, ΔABC is which ∠B = 90°, and a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Also, PQ is a tangent at P

To Prove: PQ bisects BC, i.e. BQ. = QC.

Proof:

∠APB = 90° [Angle in the semicircle is a right-angle]

∠BPC = 90° [Linear Pair]

∠3 + ∠4 = 90 [1]

then, ∠ABC = 90°

So in △ABC

∠ABC + ∠BAC + ∠ACB = 180°

90 + ∠1 + ∠5 = 180

∠1 + ∠5 = 90 [2]

then,

∠ 1 = ∠ 3[angle between the tangent and the chord equals angle made by the chord in the alternate segment]

Using this in [2] we have

∠3 + ∠5 = 90 [3]

From [1] and [3] we have

∠3 + ∠4 = ∠3 + ∠5

∠4 = ∠5

QC = PQ [Sides opposite to equal angles are equal]

But Also PQ = BQ. [Tangents drawn from the external point to a circle are equal]

So, B

.Q. = QC.

i.e. PQ bisects BC.

 

Question 44. In Fig., tangents PQ as well as PR are drawn to a circle such that ∠RPQ. = 30°. A chord RS has drawn parallel to the tangent PQ. Find out the ∠RQS.

[Hint: Draw a line passing through Q and perpendicular to the QP.]

Answer 44:

According to the question,

Tangents PQ and PR are drawn to a circle such that ∠RPQ. = 30°. A chord RS is drawn parallel to the tangent PQ.

To Find ∠RQS.

PQ = PR [Tangents drawn from the external point to a circle are equal]

∠PRQ = ∠PQR [Angles opposite to the equal sides are equal] [1]

In the triangle △PQR

∠PRQ + ∠PQR + ∠QPR = 180°

∠PQR + ∠PQR + ∠QPR = 180° [Using 1]

2∠PQR + ∠RPQ = 180°

2∠PQR + 30 = 180

2∠PQR = 150

∠PQR = 75°

∠QRS = ∠PQR = 75° [Alternate interior angles]

∠QSR = ∠PQR = 75° [angle between tangent and the chord equals angle made up by the chord in the alternate segment]

thus In △RQS

∠RQS + ∠QRS + ∠QSR = 180

∠RQS + 75 + 75 = 180

∠RQS = 30°

 

Question 45: In the figure, two tangents, TP and TQ, are drawn to a circle with centre O from the external point T, proving that ∠PTQ. = 2OPQ.

Answer 45:

Given that two tangents, TP and TQ, are drawn to a circle with centre O from the external point T

Let ∠PTQ = θ.

Then, by using the theorem “the lengths of tangents drawn from an external point to a circle are equal”, we can say TP = TQ. So, TPQ is an isosceles triangle.

hence, 

The angle ∠TPQ = ∠TQP = ½ (180°− θ ) = 90° – (½) θ

Applying the theorem for “the tangent at any point of the circle is perpendicular towards the radius passing through the point of contact”, we have that∠OPT = 90°

hence,

∠OPQ = ∠OPT – ∠TPQ = 90° – [90° – (½) θ]

∠OPQ = (½)θ

∠OPQ = (½) ∠PTQ

⇒ ∠PTQ = 2 ∠OPQ.

thus proved.

 

Question 46: Prove that the lengths of tangents drawn from an external point to a circle are equal.

Answer 46:

Consider the circle with the centre “O”, and P is the point that lies outside the circle. Therefore, the two tangents formed are PQ and PR.

We need to prove: PQ = PR.

To prove the tangent PQ is equal to PR, join OP, OQ and OR. Hence, ∠QOP and ∠ORP are the right angles.

Therefore, OQ = OR (Radii)

OP = OP (Common side)

By using the RHS rule, we can say ∆ QOP ≅ ∆ ORP.

Thus, by using the CPCT rule, the tangent PQ = PR.

Hence proved.

 

Question 47: In the figure, from an external point P, two tangents, PT and PS, are drawn to the circle with centre O and radius r. If OP = 2r, shows that ∠OTS = ∠OST = 30°.

Answer 47:

Given that from the external point P, Two tangents, PT and PS, are drawn to the circle with centre O and radius r and OP = 2r

OS = OT {radii of same circle}

∠OTS = ∠OST {angles opposite to the equal sides are equal} ….(i)

A tangent drawn at a point on the circle is perpendicular to the radius through the point of contact.

OT ⏊ TP and OS ⏊ SP

∠OSP = 90°

∠OST + ∠PST = 90°

∠PST = 90° – ∠OST….(ii)

In triangle PTS.

PT = PS {tangents drawn from the external point to a circle are equal}

∠PST = ∠PTS = 90° – ∠OST {from (ii)}

In △PTS

∠PTS + ∠PST + ∠SPT = 180° {angle sum property of the triangle}

90° – ∠OST + 90° – ∠OST + ∠SPT = 180°

∠SPT = 2∠OST….(iii)

here, △OTP, OT⏊TP

sin(∠OPT) = OT/OP = r/2r = 1/2

sin(∠OPT) = sin 30°

∠OPT = 30°….(iv)

same as, 

In △OSP,

∠OPS = 30°….(v)

Adding (iv) and (v),

∠OPT + ∠OPS = 30° + 30°

∠SPT = 60°

then substituting this value in (iii),

∠SPT = 2∠OST

60° = 2∠OST

∠OST = 30°….(vi)

From (i) and (vi),

∠OST = ∠OTS = 30°

 

Question 48: In the figure, two tangents, RQ and RP, are drawn from the external point R to the circle with the centre O. If ∠PRQ. = 120°, then prove that OR = P.R. + R.Q.

Answer 48:

Given, two tangents, RQ and RP, are drawn from an external point R to the circle with centre O. ∠PRQ. = 120°

Join OP, OQ and OR.

∠PRQ = ∠QRO = 120°/2 = 60°

RQ and RP are the tangents to the circle.

OQ and OP are radii

OQ ⊥ QR and OP ⊥ PR

Form right ∆OPR,

∠POR = 180° – (90° + 60°) = 30°

here, ∠QOR = 30°

cos a = PR/OR (suppose ‘a’ be an angle)

cos 60° =PR/OR

1/2 = PR/OR

OR = 2 PR

Again from the right ∆OQR,

OR = 2 QR

From both the results, we have

2 PR + 2 QR = 2OR

or OR = PR + RQ

Hence Proved.

 

Question 49: Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the endpoints of the arc.

Answer 49:

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

Now, join AB, AM and MB.

hence, arc AM = arc MB

⇒ Chord AM = Chord MB

In ∆AMB, AM = MB

⇒ ∠MAB = ∠MBA….(i) {angles corresponding to the equal sides are equal}

Hence, TMT’ is a tangent line.

∠AMT = ∠MBA {angles in alternate segment are equal}

so, ∠AMT = ∠MAB {from (i)}

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

Thus, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the endpoints of the arc.

Thus proved.

 

Question 50. A triangle ABC is drawn to circumscribe the circle of radius 4 cm such that the segments BD. As well as DC, into which BC is divided by the point of contact D is of lengths 8 cm as well as 6 cm respectively (see Fig.). Find the sides AB and AC.

Answer 50:

The figure given is as follows:

Consider the triangle ABC,

We get that the length of any two tangents which are drawn from the same point to the circle is equal.

thus,

(i) CF = CD = 6 cm

(ii) BE = BD = 8 cm

(iii) AE = AF = x

then, it can be observe that,

(i) AB = EB+AE = 8+x

(ii) CA = CF+FA = 6+x

(iii) BC = DC+BD = 6+8 = 14

then the semi perimeter “s” will be calculated as follows

2s = AB+CA+BC

By putting the respective values we get,

2s = 28+2x

s = 14+x

Area of △ABC = √[s(s-a)(s-b)(s-c)]

By solving this we observe,

= √(14+x)48x ……… (i)

Again, the area of the △ABC = 2 × area of (△AOF + △COD + △DOB)

= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]

= 2×½(4x+24+32) = 56+4x …………..(ii)

then from (i) and (ii) we get,

√(14+x)48x = 56+4x

Now, square both the sides,

48x(14+x) = (56+4x)2

48x = [4(14+x)]2/(14+x)

48x = 16(14+x)

48x = 224+16x

32x = 224

x = 7 cm

So, AB = 8+x

i.e. AB = 15 cm

And, CA = x+6 =13 cm.

 

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