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Geometry Exercise 2

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  1. Question 1 of 19
    1. Question

    Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the outer circle which is outside the inner circle is of length

    Hint

    geometry-38681.png

    AB = geometry-38675.png

    ∴ AC = geometry-38671.pngcm

  2. Question 2 of 19
    2. Question

    Two circles of radii 10 cm. 8 cm. intersect and length of the common chord is 12 cm. Find the distance between their centres.

    Hint

    Here, OP = 10 cm; O’P = 8 cm

    geometry-38666.png

    PQ = 12 cm

    ∴ PL = 1/2 PQ

    ⇒ PL = ½ × 12

    ⇒ PL = 6 cm

    In rt. ∆ OLP, OP² = OL² + LP²

    (using Pythagoras theorem)

    ⇒ 10² = OL² + 6²

    ⇒ OL² + 64; OL = 8

    In ∆ O'LP, (O'L)² = O'P² – LP²

    = 64 – 36 = 28

    O’L² = 28 geometry-38638.png

    O’L = 5.29 cm

    ∴ OO'= OL + O'L

    OO'= 13.29 cm

  3. Question 3 of 19
    3. Question

    Arc ADC is a semicircle and DB ⊥ AC. If AB = 9 and BC = 4, find DB.

    Hint

    m ∠ ADC = 90⁰ (Angle subtended by the diameter on a circle is 90°)

    geometry-38626.png

    ∴ ∆ ADC is a right angled triangle.

    ∴ (DB)² = BA × BC

    [since, DB is the perpendicular to the hypotenuse]

    = 9 × 4 = 36

    ∴ DB = 6

  4. Question 4 of 19
    4. Question

    A point P is 13 cm. from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the radius of the circle.
    38791.png

    Hint

    Tangent at any point of a circle is ⊥ to the radius

    In ∆OPT, OP² = PT² + OT²

    (13)² = (12)² + OT²

    ⇒ 169 – 144 =OT²

    ⇒ 25 = OT²

    ⇒ 5 = OT

  5. Question 5 of 19
    5. Question

    In the figure given below, O is the centre of the circle. If ∠OBC = 37°, then ∠BAC is equal to :
    38785.png

    Hint

    We have, ∠OBC = ∠OCB = 37°

    (equal angles of an isosceles triangle)

    ⇒ ∠COB = 180° – (37° + 37°) = 106°

    Therefore, ∠BAC geometry-38921.png

  6. Question 6 of 19
    6. Question

    PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR=OS. Then, the ratio of area of the circle to the area of the square is
    39011.png

    Hint

    geometry-39115.png

    In geometry-39108.png

    geometry-39102.png geometry-39096.png

    ∴ geometry-39090.png

  7. Question 7 of 19
    7. Question

    In the given figure, AB is chord of the circle with centre O, BT is tangent to the circle. The values of x and y are
    38981.png

    Hint

    Given AB is a circle and BT is a tangent,

    ∠BAO = 32°

    Here, ∠ OBT = 90°

    [Since tangent is ⊥ to the radius at the point of contact]

    OA = OB [ Radii of the same circle ]

    ∴ ∠ OBA = ∠OAB = 32° [ Angles opposite to equal side are equal]

    ∴ ∠ OBT = ∠ OBA + ∠ ABT = 90°

    ⇒ 32° + x = 90°.

    ⇒ ∠ x = 90° – 32° = 58° .

    Also, ∠ AOB = 180° – ∠OAB – ∠OBA

    = 180° – 32° – 32° = 116°

    Now Y = ½ AOB [ Angle formed at the center of a circle is double the angle formed in the remaining part of the circle] = ½ × 116° = 58° .

  8. Question 8 of 19
    8. Question

    In the figure given below, AB is a diameter of the semicircle APQB, centre O, ∠ POQ = 48° cuts BP at X, calculate ∠AXP.
    38975.png

    Hint

    b = ½(48°) (∠at centre = 2 at circumference on same PQ) = 24°

    ∠AQB = 90° (∠In semi- circle)

    ∠QXB = 180° – 90° – 24° = 66°

  9. Question 9 of 19
    9. Question

    OA is perpendicular to the chord PQ of a circle with centre O. If QR is a diameter, AQ = 4 cm, OQ = 5 cm, then PR is equal to
    38969.png

    Hint

    AO =geometry-39083.png

    = geometry-39077.png = geometry-39071.png= 3

    Now, from similar ∆s QAO and QPR

    OR = 2OA = 2× 3 = 6 cm.

  10. Question 10 of 19
    10. Question

    In the given figure, m ∠ EDC = 54°. m ∠ DCA = 40°.Find x, y and z.
    38963.png

    Hint

    m ∠ ACD = ½ M(are CXD) = m ∠ DEC

    ∴ m ∠ DEC = x = 40°

    ∴ m ∠ ECB = ½ m (are EYC) = m ∠ EDC

    ∴ m ∠ ECB = y = 54°

    54 + x + z = 180° [Sum of all the angles of a triangle]

    54 + 40 + z = 180°

    ∴ z = 86°.

  11. Question 11 of 19
    11. Question

    ABCD is a cyclic quadrilateral in which BC || AD, ∠ADC = 110° and ∠BAC = 50° find ∠DAC

    Hint

    ∠ABC + ∠ADC = 180° (sum of opposites angles of cyclic quadrilateral is 180°)

    geometry-39065.png

    ⇒ ∠ABC + 110° = 180° [ABCD is a cyclic quadrilateral]

    ⇒ ∠ABC = 180 – 110

    ⇒ ∠ABC = 70°

    Since AD || BC

    ∴ ∠ABC + ∠BAD = 180° [Sum of the interior angles on the same side of transversal is 180°]

    70° + ∠ BAD = 180°

    ⇒ ∠BAD = 180° – 70° = 110°

    ⇒ ∠BAC + ∠DAC = 110°

    ⇒ 50° + ∠DAC = 110°

    ⇒ ∠DAC = 110° – 50 ° = 60°

  12. Question 12 of 19
    12. Question

    In a circle of radius 17 cm, two parallel chords are drawn on opposite sides of a diameter. The distance between the chords is 23 cm. If length of one chord is 16 cm, then the length of the other one is :

    Hint

    Let PQ and RS be two parallel chords of the circle on the opposite sides of the diameter AB and PQ = 16 cm

    geometry-39561.png

    Now, PN = 8 (Since ON is the perpendicular bisector)

    In ∆ PON, ON² = OP² – PN²

    = (17)² – (8)² = 289 – 64 = 225

    ⇒ ON = 15

    ∴ OM = 23 – 15 = 8

    In ∆ ORM, RM² = OR² – OM²

    = (17)² – (8)² = 289 – 64 = 225

    ⇒ RM = 15

    ⇒ RS = 15 × 2 = 30 cm

  13. Question 13 of 19
    13. Question

    In given fig, if ∠BAC = 60° and ∠BCA = 20° find ∠ADC
    39261.png

    Hint

    In ∆ABC, ∠B = 180° – ( 60° + 20°) [By ASP]

    geometry-39523.png

    ⇒ ∠B = 100°

    But ∠B + ∠D = 180°

    [Since ABCD is a cyclic quadrilateral; Sum of opposite is 180°]

    100° + ∠D = 180°

    ⇒ ∠ADC = 80°

  14. Question 14 of 19
    14. Question

    A point P is 26 cm. away form the centre O of a circle and the length PT of the tangent draw from P to the circle is 10cm. Find radius of the circle

    Hint

    24 cm

  15. Question 15 of 19
    15. Question

    In a circle of radius 10 cm, a chord is drawn 6 cm from its centre. If an another chord, half the length of the original chord were drawn, its distance in centimeters from the centre would be:

    Hint

    geometry-39516.png

    In a triangle ∆AMO,

    geometry-39510.png= 8

    Therefore, the length of the another chord A’B′ = 8 cm.

    Now, A’N = 4

    In ∆ OA’N, ON² = (OA’)² – A’N²

    = 10² – 4² = 100 – 16 = 84

    ⇒ ON = geometry-39492.png

  16. Question 16 of 19
    16. Question

    AB and CD two chords of a circle such that AB = 6 cm, CD = 12 cm. And AB || CD. The distance between AB and CD is 3 cm. Find the radius of the circle.

    Hint

    Draw OE ⊥ CD and OF ⊥ AB

    geometry-39473.png

    AB||CD [Given]

    Let ‘r’ be the radius of the circle

    Now in rt. ∆ OED, (OD)² = (OE)² + (ED)²

    [using Pythagoras theorem]

    geometry-39448.png

    ⇒ r² = x² + 36 …(1)

    In rt. ∆ OFB, (OB)² = (OF)² + (EB)²

    ⇒ r² = (x + 3)² + (3)²

    ⇒ r² = x² + 6x + 9 + 9

    ⇒ r² = x² + 6x + 18 …(2)

    From (1) and (2), we get

    x² + 36 = x² + 6x + 18

    ⇒ 36 = 6x + 18

    ⇒ 36 – 18 = 6x

    ⇒ 6x = 18

    ⇒ x = 3

    For (1), r² = (3)² + (6)²

    r² = 9 + 36

    ⇒ r² = 45

    geometry-39374.pngcm

  17. Question 17 of 19
    17. Question

    A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?
    39225.png

    Hint

    geometry-39368.png

    OABC is square with side = 2

    geometry-39362.png

    OB = 2geometry-39356.png = OD + r + O’B = 2 + r + rgeometry-39350.png

    ⇒ r(geometry-39344.png+1) = 2(geometry-39338.png–1)

    geometry-39332.png

    geometry-39326.png

  18. Question 18 of 19
    18. Question

    In the figure below, which of the following is the relationship between ‘x’ and ‘y’ if the equal circles shown are tangents to each other and to the sides of the rectangle
    39194.png

    Hint

    Diameter of circle = x

    ∴ y = 4x

    ∴ x = ¼ y

  19. Question 19 of 19
    19. Question

    In the cyclic quadrilateral ABCD BCD =120° , m(arc DZC) = 7°, find DAB and m (arc CXB).
    39188.png

    Hint

    m ∠ DAB + 180° – 120° =60°

    [Opposite angles of a cyclic quadrilateral]

    m (arc BCD) = 2m ∠ DAB = 120°.

    geometry-39320.png

    ∴ m (arc CXB) = m (BCD) – m (arc DZC) = 120° – 70° = 50° .

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