Circles

  • Question 1
    CBSEENMA9002726

    The angles of a quadrilateral are in the ratio 3:5:9: 13. Find all the angles of the quadrilateral.  

    Solution

    Let ABCD be a quadrilateral in which
    ∠A : ∠B : ∠C : ∠D = 3 : 5 : 9 : 13
    Sum of the ratios = 3 + 5 + 9+ 13 = 30
    Also, ∠A + ∠B + ∠C + ∠D = 360°
    Sum of all the angles of a quadrilateral is 360°
    therefore space space space space space space space space space space space space space space space space angle straight A equals 3 over 30 cross times 360 degree equals 36 degree
space space space space space space space space space space space space space space space space space space space angle straight A equals 5 over 30 cross times 360 degree equals 60 degree
space space space space space space space space space space space space space space space space space space space angle straight C equals 9 over 30 cross times 360 degree equals 108 degree
and space space space space space space space space space space space space space space space angle straight D equals 13 over 30 cross times 360 degree equals 156 degree
space space space space space space space space space space space space space space space space space space space

    Question 2
    CBSEENMA9002727

    If the diagonals of a parallelogram are equal, then show that it is a rectangle.

    Solution

    Given: In parallelogram ABCD, AC = BD.
    To Prove: ||gm ABCD is a rectangle.

    Proof: In ∆ACB and ∆BDA,
    AC = BD    | Given
    AB = BA    | Common
    BC = AD
    | Opposite sides of || gm ABCD
    ∴ ∆ACB ≅ ∆BDA
    | SSS Congruence Rule
    ∴ ∠ABC = ∠BAD ...(1) C.P.C.T.
    Again, ∵ AD || BC
    | Opp. sides of || gm ABCD and transversal AB intersects them.
    ∴ ∠BAD + ∠ABC = 180°    ...(2)
    | Sum of consecutive interior angles on the same side of a transversal is 180°
    From (1) and (2),
    ∠BAD = ∠ABC = 90°
    ∴ ∠A = 90°
    ∴ || gm ABCD is a rectangle.

    Question 3
    CBSEENMA9002728

    Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

    Solution

    Given: ABCD is a quadrilateral whose diagonals AC and BD intersect each other at right angles at O.
    To Prove: Quadrilateral ABCD is a rhombus.


    Proof: In ∆AOB and ∆AOD,
    AO = AO    | Common
    OB = OD    | Given
    ∠AOB = ∠AOD    | Each = 90°
    ∴ ∆AOB ≅ ∆AOD
    | SSS Congruence Rule
    ∴ AB = AD    ...(1) | C.P.C.T.
    Similarly, we can prove that
    AB = BC    ...(2)
    BC = CD    ...(3)
    CD = AD    ...(4)
    In view of (1), (2), (3) and (4), we obtain
    AB = BC = CD = DA
    ∴ Quadrilateral ABCD is a rhombus.

     

    Question 4
    CBSEENMA9002729

     Show that the diagonals of a square are equal and bisect each other at right angles.

    Solution

    Given: ABCD is a square.
    To Prove: (i) AC = BD
    (ii) AC and BD bisect each other at right angles.
    Proof: (i) In ∆ABC and ∆BAD,

    AB = BA    | Common
    BC = AD    Opp. sides of square ABCD
    ∠ABC = ∠BAD | Each = 90°
    (∵ ABCD is a square)
    ∴ ∆ABC ≅ ∆BAD
    | SAS Congruence Rule
    ∴ AC = BD    | C.P.C.T
    (ii) In ∆OAD and ∆OCB,
    AD = CB
    | Opp. sides of square ABCD
    ∠OAD = ∠OCB
    | ∵    AD || BC and transversal AC intersects them
    ∠ODA = ∠OBC
    | ∵    AD || BC and transversal BD intersects them
    ∴ ∆OAD ≅ ∆OCB
    | ASA Congruence Rule
    ∴ OA = OC    ...(1)
    Similarly, we can prove that
    OB = OD    ...(2)
    In view of (1) and (2),
    AC and BD bisect each other.
    Again, in ∆OBA and ∆ODA,
    OB = OD | From (2) above
    BA = DA
    | Opp. sides of square ABCD
    OA = OA    | Common
    ∴ ∆OBA ≅ ∆ODA
    | SSS Congruence Rule
    ∴ ∠AOB = ∠AOD    | C.P.C.T.
    But ∠AOB + ∠AOD = 180°
    | Linear Pair Axiom
    ∴ ∠AOB = ∠AOD = 90°
    ∴ AC and BD bisect each other at right angles.  

    Sponsor Area

    Mock Test Series

    Sponsor Area

    NCERT Book Store

    NCERT Sample Papers

    Entrance Exams Preparation