NCERT Solutions Class 10 Maths Chapter 10

NCERT Solutions for Class 10 Mathematics Chapter 10 Circles

Practicing questions from NCERT Class 10 Mathematics Chapter 10- Circles is important to prepare for your board exams. To help students with solving NCERT book questions and cross-check their answers, Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 10. These solutions are prepared by subject experts to help students solve NCERT textbook questions with accuracy and confidence. They can rely on these reference materials for quick revision and last-minute preparation also.

NCERT Solutions for Class 10 Mathematics Chapter 10 Circles 

NCERT Solutions for Class 10 Mathematics Chapter 10 Circles are prepared as per the CBSE guidelines. The subject matter experts have kept the language simple and explained the answers with relevant examples, wherever possible. Students can access these solutions for online and offline studies.

Get Complete Details about Chapter 10 Mathematics Class 10 NCERT Solutions

Chapter 10 Circles of NCERT Class 10 Mathematics is a branch of Unit 4 Geometry. Unit 4 includes 4 multiple choice questions of 1 mark each, 2 short answer questions of 3 marks each, and 2 long type questions carrying 6 marks each. To ensure that you score 22/22 in questions from this chapter refer to NCERT solutions for Class 10 Mathematics Chapter 10. Only checking out NCERT solutions will not help. It requires practicing the concepts and tips as conveyed in the guide.

Other details about Class 10 Chapter 10 Circles

Chapter 10 Circles of NCERT Class 10 Mathematics consists of the following sub-topics:

  • Introduction 
  • Tangent to a circle
  • Number of Tangents from a point on a circle
  • Summary 

Furthermore, Chapter 10 Circles of NCERT Class 10 Mathematics comes with a set of extensive exercises. These include:

Exercise 10.1 – A total of 4 Questions out of which 1 short format question, 1 Fill in the blanks and 2 long format questions.

Exercise 10.2 – A total of 13 Questions out of which 10 long format questions, 4 descriptive types and 2 short format questions.

NCERT Solutions Class 10 Mathematics Chapter 10 will guide the students appropriately not only to master the concepts but also to solve the problems with utmost precision. This in turn will give students the much-needed confidence to solve questions in their final examination with the same accuracy.

NCERT Solutions for Class 10 Mathematics

The NCERT Solutions for Class 10 Mathematics are available for the following chapters:

[Include Chapter wise Pages]

Chapter 1 – Real Numbers

Chapter 2 – Polynomials

Chapter 3 – Pair of Linear Equations in Two Variables

Chapter 4 – Quadratic Equations

Chapter 5 – Arithmetic Progressions

Chapter 6 – Triangles

Chapter 7 – Coordinate Geometry

Chapter 8 – Introduction to Trigonometry

Chapter 9 – Some Applications of Trigonometry

Chapter 10 – Circles 

Chapter 11 – Constructions

Chapter 12 – Areas Related to Circles

Chapter 13 – Surface Areas and Volumes

Chapter 14 – Statistics

Chapter 15 – Probability

Why Should Students Choose Extramarks?

 Here are some of the reasons why students should pick  Extramarks’ study material:

  • Extramarks’ NCERT Solutions Class 10 Mathematics Chapter 10 are prepared by subject matter experts.
  • The answers are written in simple and easy-to-comprehend language.
  • The solutions are prepared as per CBSE guidelines.
  • The solutions are self-explanatory and do not require any external assistance for studying.

Related Questions

Q1. Fill in the blanks:

 (i) A tangent to a circle intersects it in _______ point(s).

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

(iii) A circle can have ______ parallel tangents at the most.

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

Ans. 

(i) A tangent to a circle intersects it in one point.

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

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

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

Q2. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Ans. From this figure, we can conclude a few points which are:

(i) DR = DS

(ii) BP = BQ

(iii) AP = AS

(iv) CR = CQ

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

Now, adding the LHS and RHS of the above equations we get,

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

By rearranging them we get,

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

By simplifying,

AD+BC= CD+AB

Q3. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Ans. From the figure given in the textbook, join OC. Now the triangles △OPA and △OCA are similar using SSS congruence as:

(i) OP = OC They are the radii of the same circle

(ii) AO = AO It is the common side

(iii) AP = AC These are the tangents from point A

So, △OPA ≅ △OCA

Similarly,

△OQB ≅ △OCB

So,

∠POA = ∠COA … (Equation i)

And, ∠QOB = ∠COB … (Equation ii)

Since the line POQ is a straight line, it can be considered as the diameter of the circle.

So, ∠POA +∠COA +∠COB +∠QOB = 180°

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

2∠COA+2∠COB = 180°

∠COA+∠COB = 90°

∴∠AOB = 90°

Q.1 How many tangents can a circle have?

Ans.

A circle can have infinite tangents.

Q.2 Fill in the blanks:
(i) A tangent to a circle intersects it in ­______ point (s).
(ii) A line intersecting a circle in two points is called a _________.
(iii) A circle can have _______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _______.

Ans.

(i) A tangent to a circle intersects it in one point.
(ii) A line intersecting a circle in two points is called a secant.
(iii) A circle can have two parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called point of contact.

Q.3

A tangent PQ at a point P of a circle of radius 5 cm meets a line throughthe centre O at a point Q so that OQ = 12 cm. Length PQ isA 12 cm            B 13 cmC 8.5 cm           D 119 cm

Ans.

We know that the line drawn from the center of the circle to the tangent is perpendicular to the tangent. OPPQ

By applying Pythagoras theorem in ΔOPQ, we get OP2+PQ2 = OQ2or 52+PQ2 = 122or PQ2 = 14425 = 119or PQ = 119 cm.Hence, the correct option is (D).

Q.4 Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Ans.

Here, we draw a circle centred at O and having AB as diameter. We draw two lines CD and XY parallel to diameter AB. CD is a secant and XY is a tangent to the circle at point P.

Q.5 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm (B) 12 cm
(C) 15 cm (D) 24.5 cm

Ans.

We know that the line drawn from the center of the circle to the tangent is perpendicular to the tangent. OPPQBy applying Pythagoras theorem in Δ OPQ, we get OP2+PQ2 = OQ2or OP2+242 = 252or OP2 = 625576 = 49or OP = 7 cm.Therefore, radius of the circle is 7 cm.Hence, the correct option is (A).

Q.6

In the following figure, if TP and TQ are the two tangents to a circle with centre O so that POQ=110°, then PTQ is equal to (A) 60° (B) 70° (C) 80° (D) 90°

Ans.

We know that the line drawn from the center of the circle to the tangent is perpendicular to the tangent. OPPT and OQTQ In quadrilateral OPTQ, P = Q = 90°. Now, we have P+Q+T+O = 360° or 90°+90°+110°+T = 360° or 290°+T = 360° or T = 360°-290° = 70° Therefore, PTQ = 70° Hence, the correct option is (B). MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafeaakq aabeqaaiaabEfacaqGLbGaaeiiaiaabUgacaqGUbGaae4BaiaabEha caqGGaGaaeiDaiaabIgacaqGHbGaaeiDaiaabccacaqG0bGaaeiAai aabwgacaqGGaGaaeiBaiaabMgacaqGUbGaaeyzaiaabccacaqGKbGa aeOCaiaabggacaqG3bGaaeOBaiaabccacaqGMbGaaeOCaiaab+gaca qGTbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaaeyzaiaa b6gacaqG0bGaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaGaae iDaiaabIgacaqGLbGaaeiiaiaabogacaqGPbGaaeOCaiaabogacaqG SbGaaeyzaiaabccaaeaacaqG0bGaae4BaiaabccacaqG0bGaaeiAai aabwgacaqGGaGaaeiDaiaabggacaqGUbGaae4zaiaabwgacaqGUbGa aeiDaiaabccacaqGPbGaae4CaiaabccacaqGWbGaaeyzaiaabkhaca qGWbGaaeyzaiaab6gacaqGKbGaaeyAaiaabogacaqG1bGaaeiBaiaa bggacaqGYbGaaeiiaiaabshacaqGVbGaaeiiaiaabshacaqGObGaae yzaiaabccacaqG0bGaaeyyaiaab6gacaqGNbGaaeyzaiaab6gacaqG 0bGaaeOlaaqaaiabgsJiCjaabccacaqGpbGaaeiuaiabgwQiEjaabc facaqGubGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGpbGaaeyu aiabgwQiEjaabsfacaqGrbaabaGaaeysaiaab6gacaqGGaGaaeyCai aabwhacaqGHbGaaeizaiaabkhacaqGPbGaaeiBaiaabggacaqG0bGa aeyzaiaabkhacaqGHbGaaeiBaiaabccacaqGpbGaaeiuaiaabsfaca qGrbGaaeilaiaabccacqGHGic0caqGqbGaaeiiaiaab2dacaqGGaGa eyiiIaTaaeyuaiaabccacaqG9aGaaeiiaiaabMdacaqGWaGaaeiSai aab6caaeaacaqGobGaae4BaiaabEhacaqGSaGaaeiiaiaabEhacaqG LbGaaeiiaiaabIgacaqGHbGaaeODaiaabwgaaeaacaqGGaGaaeiiai aabccacaqGGaGaaeiiaiaabccacaqGGaGaeyiiIaTaaeiuaiaabUca cqGHGic0caqGrbGaae4kaiabgcIiqlaabsfacaqGRaGaeyiiIaTaae 4taiaabccacaqG9aGaaeiiaiaabodacaqG2aGaaeimaiaabclaaeaa caqGVbGaaeOCaiaabccacaqGGaGaaeiiaiaabccacaqG5aGaaeimai aabclacaqGRaGaaeyoaiaabcdacaqGWcGaae4kaiaabgdacaqGXaGa aeimaiaabclacaqGRaGaeyiiIaTaaeivaiaabccacaqG9aGaaeiiai aabodacaqG2aGaaeimaiaabclaaeaacaqGVbGaaeOCaiaabccacaqG GaGaaeiiaiaabccacaqGYaGaaeyoaiaabcdacaqGWcGaae4kaiabgc IiqlaabsfacaqGGaGaaeypaiaabccacaqGZaGaaeOnaiaabcdacaqG WcaabaGaae4BaiaabkhacaqGGaGaaeiiaiaabccacaqGGaGaeyiiIa TaaeivaiaabccacaqG9aGaaeiiaiaabodacaqG2aGaaeimaiaabcla caqGTaGaaeOmaiaabMdacaqGWaGaaeiSaiaabccacaqG9aGaaeiiai aabEdacaqGWaGaaeiSaaqaaiaabsfacaqGObGaaeyzaiaabkhacaqG LbGaaeOzaiaab+gacaqGYbGaaeyzaiaabYcacaqGGaGaeyiiIaTaae iuaiaabsfacaqGrbGaaeiiaiaab2dacaqGGaGaae4naiaabcdacaqG WcaabaGaaeisaiaabwgacaqGUbGaae4yaiaabwgacaqGSaGaaeiiai aabshacaqGObGaaeyzaiaabccacaqGJbGaae4BaiaabkhacaqGYbGa aeyzaiaabogacaqG0bGaaeiiaiaab+gacaqGWbGaaeiDaiaabMgaca qGVbGaaeOBaiaabccacaqGPbGaae4CaiaabccacaqGOaGaaeOqaiaa bMcacaqGUaaaaaa@4F32@

Q.7

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then POA is equal to(A) 50° (B) 60°(C) 70° (D) 80°

Ans.

We know that the line drawn from the center of the circle to the tangent is perpendicular to the tangent. OAAP and OBPBIn quadrilateral OAPB, A = B = 90°.Now, we have P+A+B+O = 360°or 80°+90°+90°+O = 360°or 260°+O = 360°or O = 360°260° = 100°Therefore, AOB = 100°Now, in Δ AOP and Δ BOP, we haveAO = BO [Radius]AP = BP [Tangents drawn from P]OP = OPΔ AOPΔ BOP [By SSS congruence criterion]So, AOP = BOP     [By CPCT]        AOB = AOP+BOP = 100°or AOP+AOP = 100° [AOP = BOP]or   2AOP = 100°or   AOP = 50°or   POA = 50°Hence, the correct option is (A).

Q.8 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Ans.

Let AB is a diameter and PQ and RS be two tangents atthe ends of diameter AB.We know that the line drawn from the center of the circle to the tangent is perpendicular to the tangent. OARS and OBPQThus,we have OAR = OAS = 90° = OBP = OBQor OAR = OBQ and OAS = OBP i.e., alternate interior angles are equal.Therefore, by converse of parallel lines axiom, RS||PQ.

Q.9 Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Ans.

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

Let the perpendicular to AB at P does not pass through the centre O. Let it pass through another point R. We join OP and PR.

Perpendicular to AB at P passes through R. RPB = 90° (1) We know that the line joining the centre and point of contact to the tangent to a circle are perpendicular to each other. OPB = 90° (2) From (1) and (2), we get OPB = RPB From figure, we find that RPB < OPB OPB = RPB is not possible. It is only possible when the line RP coincides with OP. Therefore, the perpendicular at the point of contact to the tangent to a circle passes through the centre. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafeaakq aabeqaaiaabcfacaqGLbGaaeOCaiaabchacaqGLbGaaeOBaiaabsga caqGPbGaae4yaiaabwhacaqGSbGaaeyyaiaabkhacaqGGaGaaeiDai aab+gacaqGGaGaaeyqaiaabkeacaqGGaGaaeyyaiaabshacaqGGaGa aeiuaiaabccacaqGWbGaaeyyaiaabohacaqGZbGaaeyzaiaabohaca qGGaGaaeiDaiaabIgacaqGYbGaae4BaiaabwhacaqGNbGaaeiAaiaa bccacaqGsbGaaeOlaiaabccaaeaacqGH0icxcqGHGic0caqGsbGaae iuaiaabkeacaqGGaGaaeypaiaabccacaqG5aGaaeimaiaabclacaqG GaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabc cacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeii aiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeOlaiaab6cacaqGUa GaaeikaiaabgdacaqGPaaabaGaae4vaiaabwgacaqGGaGaae4Aaiaa b6gacaqGVbGaae4DaiaabccacaqG0bGaaeiAaiaabggacaqG0bGaae iiaiaabshacaqGObGaaeyzaiaabccacaqGSbGaaeyAaiaab6gacaqG LbGaaeiiaiaabQgacaqGVbGaaeyAaiaab6gacaqGPbGaaeOBaiaabE gacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogacaqGLbGaaeOB aiaabshacaqGYbGaaeyzaiaabccacaqGHbGaaeOBaiaabsgacaqGGa GaaeiCaiaab+gacaqGPbGaaeOBaiaabshacaqGGaGaae4BaiaabAga aeaacaqGJbGaae4Baiaab6gacaqG0bGaaeyyaiaabogacaqG0bGaae iiaiaabshacaqGVbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG 0bGaaeyyaiaab6gacaqGNbGaaeyzaiaab6gacaqG0bGaaeiiaiaabs hacaqGVbGaaeiiaiaabggacaqGGaGaae4yaiaabMgacaqGYbGaae4y aiaabYgacaqGLbGaaeiiaiaabggacaqGYbGaaeyzaiaabccacaqGWb GaaeyzaiaabkhacaqGWbGaaeyzaiaab6gacaqGKbGaaeyAaiaaboga caqG1bGaaeiBaiaabggacaqGYbGaaeiiaiaabshacaqGVbGaaeiiaa qaaiaabwgacaqGHbGaae4yaiaabIgacaqGGaGaae4BaiaabshacaqG ObGaaeyzaiaabkhacaqGUaaabaGaeyinIWLaeyiiIaTaae4taiaabc facaqGcbGaaeiiaiaab2dacaqGGaGaaeyoaiaabcdacaqGWcGaaeii aiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGa GaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabcca caqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaab6cacaqGUaGaaeOlai aabIcacaqGYaGaaeykaaqaaiaabAeacaqGYbGaae4Baiaab2gacaqG GaGaaeiiaiaabIcacaqGXaGaaeykaiaabccacaqGHbGaaeOBaiaabs gacaqGGaGaaeikaiaabkdacaqGPaGaaeilaiaabccacaqG3bGaaeyz aiaabccacaqGNbGaaeyzaiaabshaaeaacqGHGic0caqGpbGaaeiuai aabkeacaqGGaGaaeypaiaabccacqGHGic0caqGsbGaaeiuaiaabkea aeaacaqGgbGaaeOCaiaab+gacaqGTbGaaeiiaiaabAgacaqGPbGaae 4zaiaabwhacaqGYbGaaeyzaiaabYcacaqGGaGaae4DaiaabwgacaqG GaGaaeOzaiaabMgacaqGUbGaaeizaiaabccacaqG0bGaaeiAaiaabg gacaqG0baabaGaeyiiIaTaaeOuaiaabcfacaqGcbGaaeiiaiaabYda caqGGaGaeyiiIaTaae4taiaabcfacaqGcbaabaGaeyinIWLaaeiiai abgcIiqlaab+eacaqGqbGaaeOqaiaabccacaqG9aGaaeiiaiabgcIi qlaabkfacaqGqbGaaeOqaiaabccacaqGPbGaae4CaiaabccacaqGUb Gaae4BaiaabshacaqGGaGaaeiCaiaab+gacaqGZbGaae4CaiaabMga caqGIbGaaeiBaiaabwgacaqGUaGaaeiiaiaabMeacaqG0bGaaeiiai aabMgacaqGZbGaaeiiaiaab+gacaqGUbGaaeiBaiaabMhacaqGGaGa aeiCaiaab+gacaqGZbGaae4CaiaabMgacaqGIbGaaeiBaiaabwgaca qGGaGaae4DaiaabIgacaqGLbGaaeOBaiaabccacaqG0bGaaeiAaiaa bwgacaqGGaaabaGaaeiBaiaabMgacaqGUbGaaeyzaiaabccacaqGsb GaaeiuaiaabccacaqGJbGaae4BaiaabMgacaqGUbGaae4yaiaabMga caqGKbGaaeyzaiaabohacaqGGaGaae4DaiaabMgacaqG0bGaaeiAai aabccacaqGpbGaaeiuaiaab6caaeaacaqGubGaaeiAaiaabwgacaqG YbGaaeyzaiaabAgacaqGVbGaaeOCaiaabwgacaqGSaGaaeiiaiaabs hacaqGObGaaeyzaiaabccacaqGWbGaaeyzaiaabkhacaqGWbGaaeyz aiaab6gacaqGKbGaaeyAaiaabogacaqG1bGaaeiBaiaabggacaqGYb GaaeiiaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzaiaabcca caqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabccacaqGVbGaaeOzai aabccacaqGJbGaae4Baiaab6gacaqG0bGaaeyyaiaabogacaqG0bGa aeiiaiaabshacaqGVbaabaGaaeiiaiaabshacaqGObGaaeyzaiaabc cacaqG0bGaaeyyaiaab6gacaqGNbGaaeyzaiaab6gacaqG0bGaaeii aiaabshacaqGVbGaaeiiaiaabggacaqGGaGaae4yaiaabMgacaqGYb Gaae4yaiaabYgacaqGLbGaaeiiaiaabchacaqGHbGaae4Caiaaboha caqGLbGaae4CaiaabccacaqG0bGaaeiAaiaabkhacaqGVbGaaeyDai aabEgacaqGObGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGa aeyzaiaab6gacaqG0bGaaeOCaiaabwgacaqGUaaaaaa@E502@

Q.10 The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Ans.

Let centre of the circle be at O and point of contact of the tangent from A be P.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
null.

Q.11 Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Ans.

Let centre of the circles be at O and point of contact of the tangent PQ be A.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

We know that the line drawn from the center of the circle to the tangent is perpendicular to the tangent. OAPQ It is given that OP = OQ = 5cm and OA = 3 cm By applying Pythagoras theorem in ΔOAP, we get OA 2 +PA 2 = OP 2 or 3 2 +PA 2 = 5 2 or PA 2 = 2516 = 9 or PA = 3 cm Similarly, we have AQ = 3 cm PQ = PA+AQ = 3+3 = 6 cm Therefore, length of the chord PQ is 6 cm. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafeaakq aabeqaaiaabEfacaqGLbGaaeiiaiaabUgacaqGUbGaae4BaiaabEha caqGGaGaaeiDaiaabIgacaqGHbGaaeiDaiaabccacaqG0bGaaeiAai aabwgacaqGGaGaaeiBaiaabMgacaqGUbGaaeyzaiaabccacaqGKbGa aeOCaiaabggacaqG3bGaaeOBaiaabccacaqGMbGaaeOCaiaab+gaca qGTbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaaeyzaiaa b6gacaqG0bGaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaGaae iDaiaabIgacaqGLbGaaeiiaiaabogacaqGPbGaaeOCaiaabogacaqG SbGaaeyzaiaabccaaeaacaqG0bGaae4BaiaabccacaqG0bGaaeiAai aabwgacaqGGaGaaeiDaiaabggacaqGUbGaae4zaiaabwgacaqGUbGa aeiDaiaabccacaqGPbGaae4CaiaabccacaqGWbGaaeyzaiaabkhaca qGWbGaaeyzaiaab6gacaqGKbGaaeyAaiaabogacaqG1bGaaeiBaiaa bggacaqGYbGaaeiiaiaabshacaqGVbGaaeiiaiaabshacaqGObGaae yzaiaabccacaqG0bGaaeyyaiaab6gacaqGNbGaaeyzaiaab6gacaqG 0bGaaeOlaaqaaiabgsJiCjaabccacaqGpbGaaeyqaiabgwQiEjaabc facaqGrbaabaGaaeysaiaabshacaqGGaGaaeyAaiaabohacaqGGaGa ae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeiDaiaabIgaca qGHbGaaeiDaiaabccacaqGpbGaaeiuaiaabccacaqG9aGaaeiiaiaa b+eacaqGsbGaaeiiaiaab2dacaqGGaGaaeynaiaaykW7caaMc8Uaae 4yaiaab2gacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaab+eacaqG bbGaaeiiaiaab2dacaqGGaGaae4maiaabccacaqGJbGaaeyBaaqaai aabkeacaqG5bGaaeiiaiaabggacaqGWbGaaeiCaiaabYgacaqG5bGa aeyAaiaab6gacaqGNbGaaeiiaiaabcfacaqG5bGaaeiDaiaabIgaca qGHbGaae4zaiaab+gacaqGYbGaaeyyaiaabohacaqGGaGaaeiDaiaa bIgacaqGLbGaae4BaiaabkhacaqGLbGaaeyBaiaabccacaqGPbGaae OBaiaabccacaqGuoGaaGPaVlaab+eacaqGbbGaaeiuaiaabYcacaqG GaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGaaeiiai aabccacaqGGaGaaeiiaiaabccacaqGGaGaae4taiaabgeadaahaaWc beqaaiaabkdaaaGccaqGRaGaaeiuaiaabgeadaahaaWcbeqaaiaabk daaaGccaqGGaGaaeypaiaabccacaqGpbGaaeiuamaaCaaaleqabaGa aeOmaaaaaOqaaiaab+gacaqGYbGaaeiiaiaabccacaqGGaGaae4mam aaCaaaleqabaGaaeOmaaaakiaabUcacaqGqbGaaeyqamaaCaaaleqa baGaaeOmaaaakiaabccacaqG9aGaaeiiaiaabwdadaahaaWcbeqaai aabkdaaaaakeaacaqGVbGaaeOCaiaabccacaqGGaGaaeiiaiaabcfa caqGbbWaaWbaaSqabeaacaqGYaaaaOGaaeiiaiaab2dacaqGGaGaae OmaiaabwdacqGHsislcaqGXaGaaeOnaiaabccacaqG9aGaaeiiaiaa bMdaaeaacaqGVbGaaeOCaiaabccacaqGGaGaaeiiaiaabcfacaqGbb Gaaeiiaiaab2dacaqGGaGaae4maiaabccacaqGJbGaaeyBaaqaaiaa bofacaqGPbGaaeyBaiaabMgacaqGSbGaaeyyaiaabkhacaqGSbGaae yEaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaaeiAaiaabggacaqG 2bGaaeyzaiaabccacaqGbbGaaeyuaiaabccacaqG9aGaaeiiaiaabo dacaqGGaGaae4yaiaab2gaaeaacaqGqbGaaeyuaiaabccacaqG9aGa aeiiaiaabcfacaqGbbGaae4kaiaabgeacaqGrbGaaeiiaiaab2daca qGGaGaae4maiaabUcacaqGZaGaaeiiaiaab2dacaqGGaGaaeOnaiaa bccacaqGJbGaaeyBaaqaaiaabsfacaqGObGaaeyzaiaabkhacaqGLb GaaeOzaiaab+gacaqGYbGaaeyzaiaabYcacaqGGaGaaeiBaiaabwga caqGUbGaae4zaiaabshacaqGObGaaeiiaiaab+gacaqGMbGaaeiiai aabshacaqGObGaaeyzaiaabccacaqGJbGaaeiAaiaab+gacaqGYbGa aeizaiaabccacaqGqbGaaeyuaiaabccacaqGPbGaae4Caiaabccaca qG2aGaaeiiaiaabogacaqGTbGaaeOlaaaaaa@6BB0@

Q.12

A quadrilateral ABCD is drawn to circumscribe a circle(see the following figure). Prove thatAB+CD=AD+BC.

Ans.

We have to prove that AB+CD = AD+BC We knowthat lengths of tangents drawn from a point to a circle are equal. Therefore, from figure, we have DR = DS, CR = CQ,AS = AP,BP = BQ Now, LHS = AB+CD = (AP+BP)+(CR+DR) = (AS+BQ)+(CQ+DS) = AS+DS+BQ+CQ = AD+BC = RHS MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeaabaWaaqaafeaakq aabeqaaiaabEfacaqGLbGaaeiiaiaabIgacaqGHbGaaeODaiaabwga caqGGaGaaeiDaiaab+gacaqGGaGaaeiCaiaabkhacaqGVbGaaeODai aabwgacaqGGaGaaeiDaiaabIgacaqGHbGaaeiDaaqaaiaabgeacaqG cbGaae4kaiaaboeacaqGebGaaeiiaiaab2dacaqGGaGaaeyqaiaabs eacaqGRaGaaeOqaiaaboeaaeaacaqGxbGaaeyzaiaabccacaqGRbGa aeOBaiaab+gacaqG3bGaaGjbVlaabshacaqGObGaaeyyaiaabshaca qGGaGaaeiBaiaabwgacaqGUbGaae4zaiaabshacaqGObGaae4Caiaa bccacaqGVbGaaeOzaiaabccacaqG0bGaaeyyaiaab6gacaqGNbGaae yzaiaab6gacaqG0bGaae4CaiaabccacaqGKbGaaeOCaiaabggacaqG 3bGaaeOBaiaabccacaqGMbGaaeOCaiaab+gacaqGTbGaaeiiaiaabg gacaqGGaGaaeiCaiaab+gacaqGPbGaaeOBaiaabshacaqGGaGaaeiD aiaab+gacaqGGaGaaeyyaaqaaiaabogacaqGPbGaaeOCaiaabogaca qGSbGaaeyzaiaabccacaqGHbGaaeOCaiaabwgacaqGGaGaaeyzaiaa bghacaqG1bGaaeyyaiaabYgacaqGUaaabaGaaeivaiaabIgacaqGLb GaaeOCaiaabwgacaqGMbGaae4BaiaabkhacaqGLbGaaeilaiaabcca caqGMbGaaeOCaiaab+gacaqGTbGaaeiiaiaabAgacaqGPbGaae4zai aabwhacaqGYbGaaeyzaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGa aeiAaiaabggacaqG2bGaaeyzaaqaaiaabseacaqGsbGaaeiiaiaab2 dacaqGGaGaaeiraiaabofacaqGSaGaaeiiaiaaboeacaqGsbGaaeii aiaab2dacaqGGaGaae4qaiaabgfacaqGSaGaaGPaVlaaykW7caqGbb Gaae4uaiaabccacaqG9aGaaeiiaiaabgeacaqGqbGaaeilaiaaykW7 caaMc8UaaeOqaiaabcfacaqGGaGaaeypaiaabccacaqGcbGaaeyuaa qaaiaab6eacaqGVbGaae4DaiaabYcaaeaacaqGmbGaaeisaiaabofa caqGGaGaaeypaiaabccacaqGbbGaaeOqaiaabUcacaqGdbGaaeirai aabccacaqG9aGaaeiiaiaabIcacaqGbbGaaeiuaiaabUcacaqGcbGa aeiuaiaabMcacaqGRaGaaeikaiaaboeacaqGsbGaae4kaiaabseaca qGsbGaaeykaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGjbVlaab2dacaqGGaGaaeikaiaabgeacaqGtbGa ae4kaiaabkeacaqGrbGaaeykaiaabUcacaqGOaGaae4qaiaabgfaca qGRaGaaeiraiaabofacaqGPaaabaGaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMe8UaaeypaiaabccacaqGbbGa ae4uaiaabUcacaqGebGaae4uaiaabUcacaqGcbGaaeyuaiaabUcaca qGdbGaaeyuaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGjbVlaab2dacaqGGaGaaeyqaiaabseacaqGRaGa aeOqaiaaboeaaeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGjbVlaaykW7caqG9aGaaeiiaiaabkfacaqGibGaae4uaa aaaa@D285@

Q.13

In the following figure. XY and X’Y’ are two parallelTangents to a circle with center O and another tangent AB with point of contact C intersecting XY atA and X’Y’ at B. Prove that AOB = 90.

Ans.

Let us join points O and C.In Δ OPA and Δ OCA,OP = OC (Radii of the given circle)AP = AC (Tangents from point A)AO = AO (Common side) Δ OPAΔ OCA (SSS congruence criterion)So, POA = COA ...(1)Similarly, we have Δ OQBΔ OCB and QOB = COB ...(2)Now, POQ is a straight line. So, we have         POA+COA+QOB+COB = 180°or    COA+COA+COB+COB = 180° [From (1) and (2)]or    2COA+COB = 180°or    COA+COB = 180°2 = 90°or    AOB = 90°

Q.14 Prove that the angle between the 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 centre.

Ans.

We have the above figure as per given information.We know that a line from the centre of a circle is perpendicularto the point of contact of a tangent.Therefore, in quadrilateral PAOB, we have         OAP+OBP+AOB+APB = 360°or 90°+90°+O+P = 360°or AOB+APB = 360°-180° = 180°or AOB+APB = 180°

Q.15 Prove that the parallelogram circumscribing a circle is a rhombus.

Ans.

It is given that ABCD is a rhombus. Therefore, we have             AB = CD ...(1) and  AD = BC ...(2)We know that lengths of tangents drawn from a point to acircle are equal.Therefore, from figure, we haveDR = DS, CR = CQ,  AS = AP,  BP = BQNow,         AB+CD = (AP+BP)+(CR+DR)or    AB+CD  = (AS+BQ)+(CQ+DS)or    AB+CD  = AS+DS+BQ+CQor    AB+CD  = AD+BC ...(3)From equations (1), (2) and (3), we getor    AB+AB = AD+AD or    2AB = 2ADor    AB = AD ...(4)On comparing equations (1), (2) and (4), we getAB = BC = CD = ADHence,ABCD is a rhombus.

Q.16 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see the following figure). Find the sides AB and AC.


Ans.

Let the given circle touch the sides AB and AC of the triangleat points E and F respectively and length of the line segment AF be x.In Δ ABC, we haveCF = CD = 6 cm [Tangents on the circle from point C]BE = BD = 8 cm  [Tangents on the circle from point B]AE = AF = x         [Tangents on the circle from point A]AB = AE+EB = x+8BC = BD+DC = 8+6 = 14 cmCA = CF+FA = 6+xNow, perimeter of Δ ABC = 2s = AB+BC+CA                                                               = (x+8)+14+(x+6)                                                               = 2x+28 = 2(x+14)So,     s = x+14Area of Δ ABC = s(sa)(sb)(sc)                              = (x+14)(x+14x8)(x+1414)(x+146x)                              = (x+14)48x                              = 43(x2+14x)Area of Δ OBC = 12OD×BC = 12×4×14 = 28 cm2Area of Δ OCA = 12OF×AC = 12×4×(6+x) = 12+2xArea of Δ OAB = 12OE×AB = 12×4×(8+x) = 16+2xNow,         Area of Δ ABC = Area of Δ OBC+Area of Δ OCA+Area of Δ OABor 43(x2+14x) = 28+12+2x+16+2x = 56+4x = 4(x+14)or 3x2+42x = x+142 = x2+28x+196or 3x2+42xx228x = 196or 2x2+14x = 196or 2(x2+7x98) = 0or x2+7x98=0or x2+14x7x-98=0or x(x+14)7(x+14)=0or (x+14)(x7)=0or x=7 [Length can not be negative, so x¹14.]Hence,  AB=x+8=7+8=15 cm ​​​​​​​​​​                CA=6+x=6+7=13 cm

Q.17 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Ans.

Let ABCD be a quadrilateral circumscribing a circle centred at Osuch that it touches the circle at points P, Q, R and S. Let us jointhe vertices of the quadrilateral ABCD to the centre of the circle.In ΔOAP and Δ OAS, we haveAP = AS [Tangents drawn from point A]OP = OS  [Radii of the same circle]OA = OA [Common side]Δ OAPΔ OAS [SSS congruence criterion]Thus, POA = AOS or   1 = 2Similarly, 3 = 4, 5 = 6,   7 = 8Now, from the above figure, we have          1+2+3+4+5+6+ 7+8 = 360°or     (1+2)+(3+4)+(5+6)+(7+8) = 360°or     (2+2)+(3+3)+(6+6)+( 7+7) = 360°or     22+3+6+7 = 360°or     2+3+6+7 = 360°2 = 180°or     2+3+6+7 = 180°or AOB+COD = 180°Similarly, we  have         1+8+4+5 = 180°or AOD+BOC = 180°   AOB+COD = 180° = AOD+BOCi.e., opposite sides of a quadrilateral circumscribing a circlesubtend supplementary angles at the centre of the circle.

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