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      Class 10 Maths

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      • Class 10
      • Class 10 Maths
      CoursesClass 10MathsClass 10 Maths
      • 01. Real Numbers
        9
        • Lecture1.1
          Real Numbers and Fundamental Theorem of Arithmetic 59 min
        • Lecture1.2
          Divisibility and Euclid’s Division Lemma 49 min
        • Lecture1.3
          Finding HCF and LCM Using Fundamental Theorem of Arithmetic 02 hour
        • Lecture1.4
          Miscellaneous Questions 31 min
        • Lecture1.5
          Chapter Notes – Real Numbers
        • Lecture1.6
          NCERT Solutions – Real Numbers Exercise 1.1 – 1.4
        • Lecture1.7
          Revision Notes Real Numbers
        • Lecture1.8
          R S Aggarwal Real Numbers
        • Lecture1.9
          R D Sharma Real Numbers
      • 02. Polynomials
        11
        • Lecture2.1
          Introduction to Polynomials and Its Zeroes 44 min
        • Lecture2.2
          Finding Zeroes of Quadratic Polynomial 01 hour
        • Lecture2.3
          Relationship b/w the Zeroes and coefficients of a Polynomial 01 hour
        • Lecture2.4
          Cubic Polynomial 35 min
        • Lecture2.5
          Division and Division Algorithm for Polynomials 01 hour
        • Lecture2.6
          Geometrical Meaning of Zeroes of Polynomial 15 min
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.4
        • Lecture2.9
          Revision Notes Polynomials
        • Lecture2.10
          R S Aggarwal Polynomials
        • Lecture2.11
          R D Sharma Polynomials
      • 03. Linear Equation
        9
        • Lecture3.1
          Solution of linear Equation in Two Variable by Graphical and Algebraic Methods 01 hour
        • Lecture3.2
          Some Basic Questions on Solving Pair of Linear Equations 01 hour
        • Lecture3.3
          Some Basics Problems on Numbers and Ages 40 min
        • Lecture3.4
          Problems on Money Matters, Time, Distance, Speed, Time and Work 01 hour
        • Lecture3.5
          Chapter Notes – Linear Equation
        • Lecture3.6
          NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
        • Lecture3.7
          Revision Notes Linear Equation
        • Lecture3.8
          R S Aggarwal Linear Equation
        • Lecture3.9
          R D Sharma Linear Equation
      • 04. Quadratic Equation
        8
        • Lecture4.1
          Introduction and Finding The Roots of a Quadratic Equation 37 min
        • Lecture4.2
          Miscellaneous Questions 1 02 hour
        • Lecture4.3
          Miscellaneous Questions 2 02 hour
        • Lecture4.4
          Chapter Notes – Quadratic Equation
        • Lecture4.5
          NCERT Solutions – Quadratic Equation Exercise 4.1 – 4.4
        • Lecture4.6
          Revision Notes Quadratic Equation
        • Lecture4.7
          R S Aggarwal Quadratic Equation
        • Lecture4.8
          R D Sharma Quadratic Equation
      • 05. Arithmetic Progressions
        11
        • Lecture5.1
          Introduction to Arithmetic Progression 59 min
        • Lecture5.2
          Miscellaneous Questions Based on nth Term Formula of A.P. 55 min
        • Lecture5.3
          Middle term (s) of Finite A.P. and Arithmetic Mean 01 hour
        • Lecture5.4
          Selection of the Terms and Sum of nth Terms of A.P. 01 hour
        • Lecture5.5
          Miscellaneous Questions Based on Sum of nth Term of A.P. 01 hour
        • Lecture5.6
          Word Problems related to nth term and sum of nth terms of A.P. 01 hour
        • Lecture5.7
          Chapter Notes – Arithmetic Progressions
        • Lecture5.8
          NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
        • Lecture5.9
          Revision Notes Arithmetic Progressions
        • Lecture5.10
          R S Aggarwal Arithmetic Progressions
        • Lecture5.11
          R D Sharma Arithmetic Progressions
      • 06. Some Applications of Trigonometry
        7
        • Lecture6.1
          Angle of Elevation and Depression; Problems involving Elevation and Double Elevation 01 hour
        • Lecture6.2
          Miscellaneous Problems Involving Double Elevation, Depression and Elevation 01 hour
        • Lecture6.3
          Miscellaneous Questions (Level-1, 2, 3, 4) 49 min
        • Lecture6.4
          Chapter Notes – Some Applications of Trigonometry
        • Lecture6.5
          NCERT Solutions – Some Applications of Trigonometry
        • Lecture6.6
          Revision Notes Some Applications of Trigonometry
        • Lecture6.7
          R D Sharma Some Applications of Trigonometry
      • 07. Coordinate Geometry
        17
        • Lecture7.1
          Introduction to Terms related to Coordinate Geometry 49 min
        • Lecture7.2
          Locating Coordinates of Point on the Axes 37 min
        • Lecture7.3
          Finding vertices of a Geometrical Fig. and Distance b/w Two Points 58 min
        • Lecture7.4
          Distance b/w Two Points Formula Based Examples 44 min
        • Lecture7.5
          Finding the type of Triangle and Quadrilateral 01 hour
        • Lecture7.6
          Find the Collinear or Non-collinear points and Missing Vertex of a Geometrical Figure 42 min
        • Lecture7.7
          Section Formula and Corollary 01 hour
        • Lecture7.8
          Finding the Section Ratio and Involving Equation of Line 53 min
        • Lecture7.9
          Proof Related to Mid-points and Centroid of a Triangle 59 min
        • Lecture7.10
          Finding Area of Triangle and Quadrilateral 53 min
        • Lecture7.11
          Collinearity 34 min
        • Lecture7.12
          Examples Based on If Areas are Given 34 min
        • Lecture7.13
          Chapter Notes – Coordinate Geometry
        • Lecture7.14
          NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
        • Lecture7.15
          Revision Notes Coordinate Geometry
        • Lecture7.16
          R S Aggarwal Coordinate Geometry
        • Lecture7.17
          R D Sharma Coordinate Geometry
      • 08. Triangles
        15
        • Lecture8.1
          Introduction, Similarity of Triangles, Properties of Triangles, Thales’s Theorem 52 min
        • Lecture8.2
          Questions Based on Thales’s Theorem, Converse Thales’s Theorem 01 hour
        • Lecture8.3
          Questions Based on Converse Thales’s Theorem, Questions Based On Mid-points and its Converse 59 min
        • Lecture8.4
          Questions Based on Internal Bisectors of an Angle, Similar Triangles, AAA criterion for Similarity 52 min
        • Lecture8.5
          SSS & SAS Criterion of Similarity, Question Based on Criteria for Similarity 02 hour
        • Lecture8.6
          Questions Based on Criteria for Similarity Conti. 01 hour
        • Lecture8.7
          Questions Based On Area of Similar Triangles 02 hour
        • Lecture8.8
          Questions Based On Area of Similar Triangles cont., Pythagoras Theorem Level-1 01 hour
        • Lecture8.9
          Pythagoras Theorem Level-1 cont. 01 hour
        • Lecture8.10
          Miscellaneous Questions 01 hour
        • Lecture8.11
          Chapter Notes – Triangles
        • Lecture8.12
          NCERT Solutions – Triangles Exercise 8.1 – 8.6
        • Lecture8.13
          Revision Notes Triangles
        • Lecture8.14
          R S Aggarwal Triangles
        • Lecture8.15
          R D Sharma Triangles
      • 09. Circles
        8
        • Lecture9.1
          Introduction to Circle and Tangent and Theorems-1, 2 01 hour
        • Lecture9.2
          Property of Chord of Circle and Theorem-3 52 min
        • Lecture9.3
          Theorem 4 and Angle in Semicircle 01 hour
        • Lecture9.4
          Chapter Notes – Circles
        • Lecture9.5
          NCERT Solutions – Circles Exercise
        • Lecture9.6
          Revision Notes Circles
        • Lecture9.7
          R S Aggarwal Circles
        • Lecture9.8
          R D Sharma Circles
      • 10. Areas Related to Circles
        10
        • Lecture10.1
          Area and Perimeter of Circle, Semicircle and Quarter circle 56 min
        • Lecture10.2
          Touching Circles and Example on Bending Wire, Wheel Rotation 58 min
        • Lecture10.3
          Sector of A Circle 51 min
        • Lecture10.4
          Segment of Circle and Area of Combination of Plane Figures 01 hour
        • Lecture10.5
          Questions Based on Area of Combination of Plane Figures 01 hour
        • Lecture10.6
          Chapter Notes – Areas Related to Circles
        • Lecture10.7
          NCERT Solutions – Areas Related to Circles
        • Lecture10.8
          Revision Notes Areas Related to Circles
        • Lecture10.9
          R S Aggarwal Areas Related to Circles
        • Lecture10.10
          R D Sharma Areas Related to Circles
      • 11. Introduction to Trigonometry
        7
        • Lecture11.1
          Introduction, Trigonometric Ratios, Sums when Sides of Triangle are given, Miscellaneous Questions 01 hour
        • Lecture11.2
          Trigonometric Ratios for Some Specific Angles, Trigonometric Ratio for 30 & 60, Value Table, Word Problems 48 min
        • Lecture11.3
          Complimentary angles in Inverse Multiplication, Miscellaneous Problems 01 hour
        • Lecture11.4
          Miscellaneous Questions 01 hour
        • Lecture11.5
          Chapter Notes – Introduction to Trigonometry
        • Lecture11.6
          NCERT Solutions – Introduction to Trigonometry
        • Lecture11.7
          Revision Notes Introduction to Trigonometry
      • 12. Surface Areas and Volumes
        9
        • Lecture12.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere, Type-1 : Surface Area and Volume of A Solid 01 hour
        • Lecture12.2
          Volume Related Questions, Based On Embarkment 48 min
        • Lecture12.3
          Based on the Rate of Flowing, Frustum of Cone (A) Surface Area (B) Volume 52 min
        • Lecture12.4
          Quantity, Percentage, Very Short Answer Types Questions 44 min
        • Lecture12.5
          Chapter Notes – Surface Areas and Volumes
        • Lecture12.6
          NCERT Solutions – Surface Areas and Volumes
        • Lecture12.7
          Revision Notes Surface Areas and Volumes
        • Lecture12.8
          R S Aggarwal Surface Areas and Volumes
        • Lecture12.9
          R D Sharma Surface Areas and Volumes
      • 13. Statistics
        12
        • Lecture13.1
          Introduction and Mean of Ungrouped Data 46 min
        • Lecture13.2
          Mean of Grouped Data 58 min
        • Lecture13.3
          Median of Ungrouped and grouped Data 01 hour
        • Lecture13.4
          Mode of Ungrouped and Grouped Data 49 min
        • Lecture13.5
          Mean, Median and Mode & Cumulative Frequency Curve-Less Than Type 53 min
        • Lecture13.6
          Cumulative Frequency Curve (Ogive)-More Than Type 52 min
        • Lecture13.7
          Miscellaneous Questions 38 min
        • Lecture13.8
          Chapter Notes – Statistics
        • Lecture13.9
          NCERT Solutions – Statistics
        • Lecture13.10
          Revision Notes Statistics
        • Lecture13.11
          R S Aggarwal Statistics
        • Lecture13.12
          R D Sharma Statistics
      • 14. Probability
        9
        • Lecture14.1
          Activities, Understanding term Probability, Experiments 54 min
        • Lecture14.2
          Calculations 43 min
        • Lecture14.3
          Event, Favorable Outcomes, Probability, Some important experiments- Tossing a coin, Rolling dice, Drawing a card 01 hour
        • Lecture14.4
          Complement of an event, Elementary Events, Equally Likely events, Some problems of Probability 58 min
        • Lecture14.5
          Chapter Notes – Probability
        • Lecture14.6
          NCERT Solutions – Probability
        • Lecture14.7
          Revision Notes Probability
        • Lecture14.8
          R S Aggarwal Probability
        • Lecture14.9
          R D Sharma Probability
      • 15. Construction
        7
        • Lecture15.1
          Introduction, Construction to Divide a Line Segment in Certain Ratio, Construction of Similar Triangles 44 min
        • Lecture15.2
          Construction of Different Angles, Construction of Angle Bisector 41 min
        • Lecture15.3
          Construction of Tangent of Circle at a Given Point 35 min
        • Lecture15.4
          Chapter Notes – Construction
        • Lecture15.5
          NCERT Solutions – Construction
        • Lecture15.6
          R S Aggarwal Construction
        • Lecture15.7
          R D Sharma Construction

        NCERT Solutions – Circles Exercise

        Q.1      How many tangents can a circle have ?
        Sol.         A circle can have an infinite number of 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 ______.
        Sol.         (i) exactly one
                       (ii) secant
                       (iii) two
                       (iv) point of contact


         

        Q.3      A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length of PQ is :
                     (A) 12 cm                  (B) 13 cm                   (C) 8.5 cm                    (D) 119−−−√cm
        Sol.         (D). Because,
        PQ=OQ2−OP2−−−−−−−−−−√=122−52−−−−−−−√=144−25−−−−−−−√
        =119−−−√cm


        Q.4      Draw a circle and two lines parallel to a given line such that one is tangent and the other, a secant to the circle.
        Sol.        The required figure is as under :
        Here ℓ is the given line and a circle with centre O is drawn.
        Line PT is drawn || to line ℓ.
        PT is the tangent to the circle.
        AB is drawn || to line ℓ and is the secant.

        1

        Q.1      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
        Sol.          Since QT is a tangent to the circle at T and OT is radius,
        Therefore, OT⊥QT
        It is given that OQ = 25 cm and QT = 24 cm,
        By Pythagoras theorem, we have 

        2
        OQ2=QT2+OT2
        ⇒OT2=OQ2−QT2
        ⇒OT2=252−242
        =(25+24)(25-24)
        =49×1=49
        ⇒OT=49−−√=7
        Hence, radius of the circle is 7 cm , i.e., (A).


        Q.2      In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ=1100, then ∠PTQ is equal to
                    (A) 600                 (B) 700                  (C) 800                      (D) 900

        4
        Sol.       Since TP and TQ are tangents to a circle with centre O so that ∠POQ=1100,
        Therefore, OP⊥PT and OQ⊥QT
        ⇒∠OPT=900 and ∠OQT=900
        In the quadrilateral TPOQ, we have
        ∠PTQ+∠TPO+∠POQ+∠OQT=3600
        ⇒∠PTQ+900+1100+900=3600
        ⇒∠PTQ+2900=3600
        ⇒∠PTQ=3600−2900
        = 700, i.e. , (B).

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        Q.3      If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 800, then ∠POA is equal to
                  (A) 500               (B) 600                 (C) 700                    (D) 800
        Sol.         Since PA and PB are tangents to a circle with centre O,

        3
        Therefore, OA⊥AP, and OB⊥BP
        ⇒∠OAP=900and∠OBP=900
        In the quadrilateral PAOB, we have
        ∠ABP+∠PAO+∠AOB+∠OBP=3600
        ⇒800+900+∠AOB+900=3600
        ⇒2600+∠AOB=3600
        ⇒AOB=3600−2600
        =1000
        In the right Δs OAP and OBP, we have
        OP = OP                           [Common]
        OA = OB                          [Radii]
        ∠OAP=∠OBP   [Each = 90°]
        Therefore, ΔOAP≅ΔOBP   ( By SAS Criterion)
        ⇒∠POA=∠POB      [C.P.C.T.]
        Therefore, ∠POA=12∠AOB=12×1000
        =500,i.e.,(A).


        Q.4       Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
        Sol.         Let PQ be a diameter of the given circle with centre O.
        6
        Let AB and CD be the tangents drawn to the circle at the end points of the diameter PQ respectively.
        Since, tangent at a point to a circle is perpendicular to the radius through the point,
        Therefore, PQ⊥ABandPQ⊥CD
        ⇒∠APQ=∠PQD
        ⇒AB||CD                     [Because ∠APQand∠PQD are alternate angles]


        Q.5      Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
        Sol.         Let AB be the tangent drawn at the point P on the circle with O.

        5
        If possible, let PQ be perpendicular to AB, not passing through O.
        Join OP.
        Since, tangent at a point to a circle is perpendicular to the radius through the point,
        Therefore, AB⊥OP i.e., ∠OPB=90∘
        Also, ∠QPB=90∘    (Construction)
        Therefore, ∠QPB=∠OPB, which is not possible as a part cannot be equal to whole.
        Thus, it contradicts our supposition.
        Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

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        Q.6      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.
        Sol.         Since tangent to a circle is perpendicular to the radius through the point of contact,

        8                Therefore, ∠OTA=90∘
        In right ΔOTA, we have
        OA2=OT2+AT2
        ⇒52=OT2+42
        ⇒OT2=52−42=25−16=9
        ⇒OT=9–√=3
        Hence, radius of the circle is 3 cm.


        Q.7       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.
        Sol.          Let O be the common centre of two concentric circles, and let AB be a chord of the larger circle touching the smaller circle at P.
        10
        Join OP.
        Since, OP is the radius of the smaller circle and AB is tangent to this circle at P,
        Therefore, OP⊥AB.
        We know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.
        So, OP⊥AB and AP = BP.
        In right ΔAPO, we have
        OA2=AP2+OP2
        ⇒52=AP2+32
        ⇒AP2=52−32=25−9=16
        ⇒AP=16−−√=4
        Now, AB = 2AP                           [Because AP = BP]
        =2×4=8
        Hence, the length of the chord of the larger circle which touches the smaller circle is 8 cm.


        Q.8       A quadrilateral ABCD is drawn to circumscribe a circle (see figure).

        7
        Prove that  AB + CD = AD + BC.
        Sol.           Let the quadrilateral ABCD be drawn to circumscribe a circle as shown in the figure.
        i.e., the circle touches the sides AB, BC, CD and DA at P, Q, R and S respectively.
        Since, lengths of two tangents drawn from an external point of circle are equal,
        AP = AS
        BP = BQ
        DR = DS
        CR = CQ
        Adding these all, we get
        (AP + BP) + (CR +RD) = (BQ+QC) + (DS+SA)
        ⇒AB+CD=BC+DA
        which proves the result.

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        Q.9       In figure, 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.

        9
        Prove that ∠AOB=90∘.
        Sol.      Since tangents drawn from an external point to a circle are equal,
        Therefore, AP = AC. Thus in Δs PAO and AOC, we have
        AP = AC
        AO = AO [Common]
        and, PO = OC [Radii of the same circle]
        By SSS-criterion of congruence, we have
        ΔPAO≅ΔAOC
        ⇒∠PAO=∠CAO
        ⇒∠PAC=2∠CAO
        Similarly, we can prove that
        ∠QBO=∠CBO
        ⇒∠CBQ=2∠CBO
        Now, ∠PAC+∠CBQ=180∘ [Sum of the interior angles on the same side of transversal is 180°]
        ⇒2∠CAO+2∠CBO=180∘
        ⇒∠CAO+∠CBO=90∘
        ⇒   180o−∠AOB=90o
                     [Since ∠CAO,∠CBO  and ∠AOBare∠s  of a triangle  Therefore ∠CAO+∠CBO+∠AOB=180o]
                     ⇒∠AOB=90∘


        Q.10      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.
        Sol.           Let PA and PB be two tangents drawn from an external point P to a circle with centre O. We have to prove that-

        11
        ∠AOB+∠APB=180∘
        In right Δs OAP and OBP, we have
        PA = PB                              [Tangents drawn from an external point are equal]
        OA = OB                             [Radii of the same circle]
        and, OP = OP                    [Common]
        Therefore, by SSS – criterion of congruence
        ΔAOP≅ΔOBP
        ⇒∠OPA=∠OPB
        and, ∠AOP=∠BOP⇒∠APB=2∠OPA
        and, ∠AOB=2∠AOP
        But ∠AOP=90∘−∠OPA              [Because ΔOAP is right triangle]
        Therefore, 2∠AOP=180∘−2∠OPA
        ⇒∠AOB=180∘−∠APB
        ⇒∠AOB+∠APB=180∘

        Q.11.      Prove that the paralelogram circumscribing a circle is a rhombus.
        Sol.            Let ABCD be a parallelogram such that its sides touch a circle with centre O.
        We know that the tangents to a circle from an exterior point are equal in length.

        13
        Therefore, AP = AS
        BP = BQ
        CR = CQ
        and, DR = DS
        Adding these, we get
        (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
        ⇒ AB + CD = AD + BC
        ⇒ 2AB = 2BC [Because ABCD is a ||gm Therefore AB = CD and BC = AD]
        ⇒ AB = BC
        Thus, AB = BC = CD = AD
        Hence, ABCD is a rhombus

        Q.12      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 figure). Find the sides AB and AC.
        12
        Sol.      Let the ΔABC be drawn to circumscribe a circle with centre O and radius 4 cm.
        i.e., the circle touches the sides BC, CA and AB at D, E and F respectively.

        14
        It is given that BD = 8cm, CD = 6 cm.
        Since, lengths of two tangents drawn from an external point of circle are equal.
        Therefore, BF = BD = 8 cm,
        CE = CD = 6 cm
        and let AF = AE = x cm.
        Then, the sides of the triangle are 14 cm, (x+6) cm and (x+8) cm.
        Therefore, 2s = 14 + (x + 6) + (x+8)  = 28 + 2x
        ⇒ s = 14 + x
        s−a=14+x−14=x,
        s−b=14+x−x−6=8
        and,s−c=14+x−x−8=6
        Therefore Area(ΔABC)=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−−√
        =(14+x)(x)(8)(6)−−−−−−−−−−−−−−−√
        =48(x2+14x)−−−−−−−−−−−√
        Also, Area (ΔABC) = Area (ΔOBC) + Area (ΔOCA) + Area  (ΔOAB)
                         =12×BC×OD+12×CA×OE+12×AB×OF
                         =12×14×4+12×(x+6)×4+12×(x+8)×4
                         =2(14+x+6+x+8)
                         =2(28+2x)
                         Therefore, 48(x2+14x)−−−−−−−−−−−√=2(28+2x),i.e.,4(14+x)
                         Squaring, we get
                         48(x2+14x)=16(14+x)2
                         ⇒3(x2+14x)=196+28x+x2
                         ⇒2x2+14x−196=0
                         ⇒x2+7x−98=0
                         ⇒(x−7)(x+14)=0
                         ⇒x=7orx=−14
                         But x cannot be negative,
                         Therefore, x = 7
                         Hence, the sides AB and AC are 15 cm and 13 cm respectively.


        Q.13      Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
        Sol.           Let a circle with centre O touch the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and s respectively.
        We have to prove that-
        ∠AOB+∠COD=180∘
        and, ∠AOD+∠BOC=180∘
        Join OP, OQ, OR and OS.
        Since, the two tangents drawn from an external point to a circle subtend equal angles at the centre.

        15
        Therefore, ∠1=∠2,∠3=∠4,∠5=∠6and∠7=∠8
        Now, ∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360∘
        [Because Sum of all the ∠s around a point is 360°]
        ⇒2(∠2+∠3+∠6+∠7)=360∘
        and, 2(∠1+∠8+∠4+∠5)=360∘
        ⇒(∠2+∠3+∠6+∠7)=180∘
        and (∠1+∠8)+(∠4+∠5)=180∘
        ⇒∠AOB+∠COD=180∘
                          ⇒[Since,∠2+∠3=∠AOB,∠6+∠7=∠COD∠1+∠8=∠AODand∠4+∠5=∠BOC]
        and, ∠AOD+∠BOC=180∘

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          6 Comments

        1. Priyanshu Kumar
          May 28, 2021

          How we access live tutorial classes please tell me

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          September 15, 2021

          I am not able to access any test of coordinate geometry, can you pls resolve the issue?

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            September 20, 2021

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          November 12, 2021

          Hi!
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        4. Nucleon IIT JEE KOTA
          December 17, 2022

          Great article, the information provided to us. Thank you

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