Circles, Chords and Tangents Questions and Answers - Form 3 Topical Mathematics

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Questions

  1. In the figure below angle BAC =52o, angle ACB = 40o and AD = DC. The radius of the circle is 7cm. EF is a tangent to the circle
    circles q1
    1.  Find; giving reasons
      1. angle DCF
      2. angle AOB (obtuse)
    2. Calculate the area of the shaded segment AGB
  2. In the figure below, O is the centre of the circle. Angle CBA = 50o and angle BCO = 30o.
    circle q2
    Find the size of the angle BAC
  3. In the given figure, O is the centre of the circle and AOBP is a straight line. PT is a tangent to the circle. If PT = 12cm and BP = 4cm. find the radius of the circle
    circle q3
  4. In the figure below AOD is a diameter of the circle cetre O. BC is a chord parallel to AD. FE is a tangent to the circle. OF bisects angle COD. Angle BCE = angle COE = 20o BC cuts OE at X
    circle q4
    Calculate;
    1. angle BOE
    2. angle BEC
    3. angle CEF
    4. angle OXC
    5. angle OFE
  5. The figure below shows two pulleys of radii 6cm and 4cm with centres A and B respectively. AB = 8cm. The pulleys are connected by a string PQXRSY
    circle q5
    Calculate
    1. Length PQ
    2. ∠PAS reflex
    3. Length of arc PYS and QXR
    4. The total length of the string PQXRSY
    1. Two pipes A and B can fill a tank in 3hrs and 4 hrs respectively. Pipe C can empty the full tank in 6 hrs.
      1. How long would it take pipes A and B to fill the tank if pipe C is closed?
      2. Starting with an empty tank, how long would it take to fill the tank with all pipes running?
    2. The high quality Kencoffee is a mixture of pure Arabica coffee and pure Robusta coffee in the ratio 1 : 3 by mass. Pure Arabica coffee costs shs. 180 per kg and pure Robusta coffee costs sh 120 per kg. Calculate the percentage profit when the coffee is sold at sh 162 per kg.
  6. In the figure below, ABCD is a cyclic quadrilateral and BD is a diagonal. EADF is a straight line, ∠CDF = 68o, ∠BDC = 45o and ∠BAE = 98o.
    circle q7
    Calculate the size of:
    1. ∠ABD.
    2. ∠CBD
  7. The figure below shows a circle centre O. AB and PQ are chords intersecting externally at a point C. AB = 9cm, PQ = 5cm and QC = 4cm.
    circle q8
    Find the value of
    x
  8. The chords AB and PQ intersects internally at O. Given that the length of OP=8cm, OA= 4.5cm and OQ=6cm.
    circle q9
    Calculate the length of OB
  9. In the figure below ABC is a tangent to the circle at B. given that ∠ABG=40º, ∠BGD=45º, and ∠DBE=25º as shown below. 
    circle q10
    Find the sizes of the following angles giving reasons in each case:
    1. BDG
    2. DGE
    3. EFG
    4. CBD
    5. BCD
  10. The figure below shows two intersecting circles radii 8 cm and 6 cm respectively.
    circles q11
    The common chord AB = 9cm ad P and Q are the centres as shown:
    1. Calculate the size of angles:-
      1. ∠APB
      2. ∠AQB
    2. Calculate the area of the shaded region
  11. In the figure below, O and P are centres of two intersecting circles. ABE is tangent to circle BCD at B angle BCD is 42o.
    circles q12
    1. Giving reasons for your answer, find:-
      1. CBD
      2. DOB
      3. DAB
      4. CDA
    2. Show that ∠ADB is isosceles
  12.  
    circles q13
    In the figure above K, M & P are points on a straight line. PN is a tangent of the circle centre O. Angle KOL = 130o and angle MKN = 40o.
    Find, giving reasons, the values of angles.
    1. ∠MLN
    2. ∠OLN
    3. ∠LNP
    4. ∠MPN
    5. ∠LMO
  13.  In the diagram below, O is the centre of the circle of radius 8cm. BA and BC are tangents to the circle at A and C respectively. PD is the diameter and AC is a chord of length 8cm. Angle ADC = 120o. ARC is an arc of the circle, Centre B and radius 4.6cm.
    circles q14
    Calculate correct to 2 decimal places
    1. Angle ABR
    2. Area of sectors ABCR and OAPC
    3. Area of the shaded part
  14. In the figure below, ATX is a tangent to the circle at point T, ABC is a straight line, angle ABT = 100o, angle XTD = 58o and line AB = line BT. C and D lie on the circle
    circles q15
    Find by giving reasons, the value of angle:
    1. TDC
    2. TCB
    3. TCD
    4. BTC
    5. DTC
  15. In the figure below, B, D, E, F and G are on the circumference of the circle centre O. A, B and C form a tangent to the circle at point B. GD is the diameter of the circle.
    circles q16
    Given that 
    FG = DE, reflex angle GOB = 252°, angles DBC = 36° and FEG = 20°
    Giving reasons in each case find the angles:
    1. GEB
    2. BED
    3. OBE
    4. BGE
    5. GFE
  16. XYZ is a triangle in which x = 13.4cm, Z= 5cm and ∠XYZ =57.7o . Find:
    1. Length of XZ
    2. The circum radius of the triangle
  17. In the figure shown below, the centers of the two circles are A and B. PQ is a common chord to the two circles. AP = 6cm, BP=4cm and PQ =5cm
    circles q18
    Calculate the area of the shaded region (take π
    as 3.142)
  18. In the figure below NR is a diameter of the circle centre O. Angle PNR = 7500 NRM = 500 and RPQ = 350. MRS and PQS are straight lines.
    circles q19
    Giving reasons for every statement you write, find the following angles
    1. PQR
    2. QSR
    3. Reflex POR
    4. MQR
    5. PON
  19. In the diagram below, ATX is a tangent to the circle at point T, ABC is a straight line, ∠ABT= 100o, ∠XTD = 58o and the line AB = BT
    circles q20
    Find giving reasons the value of :
    1. ∠TDC
    2. ∠TCB
    3. ∠TCD
    4. ∠BTC
    5. ∠DTC
  20. In the figure above AB = 6 cm, BC = 4 cm DC = 5 cm.
    circles q21
    Find the length DE.
  21. The eleventh term of an AP is four times the second term. If the sum of the first seven terms of the AP is 175, find the first term and the common difference
  22. In the diagram below ABE is a tangent to a circle at B and DCE is a straight line. 
    circles q23
    If ABD = 60o, BOC = 80o and O is the centre of the circle, find with reasons ∠BEC
  23. The circle below circumscribes a triangle ABC where AB = 6.3cm, BC = 5.7cm and AC = 4.8cm. Find the area of the shaded part (use π = 3.142)
    circles q24
  24.  
    circles q25
    1. O is the centre of the circle and QOTS is a diameter. P, Q, R and S are points on the circumference of the circle. Angle PQS = 38o and angle QTR = 56o.
      Calculate the size of ;
      1. ∠PRQ
      2. ∠RSQ
    2. Given that A varies directly as B and inversely as the cube of C and that; A = 12 when B = 3 and C = 2. Find B when A = 10 and C = 1.5
    3. A quantity y is partly constant and partly varies inversely as the square of x. The quantity y=7 when x=10 and y=5½ when x=20. Find the value of y when x=18
  25. The figure below shows two intersecting circles with centres P and Q and radius 5cm and 6cm respectively.
    circles q26
    AB is a common chord of length 8cm. Calculate
    1. the length of PQ
    2. the size of;
      1. angle APB
      2. angle AQB
    3. the area of the shaded region
  26. Triangle ABC is inscribed in the circle. AB= 7.8cm, AC 6.6cm and BC= 5.9cm.
    circles q27
    Find:
    1. The radius of the circle correct to one decimal place
    2. The area of the shaded region
  27. The figure below shows two circles centres A and B and radii 6 cm and 8 cm respectively.
    circles q28
    The 
    circles intersect at P and Q. Angle PAB = 42o and angle ABQ = 30o.
    1. Find the size of PAQ and PBQ.
    2. Calculate, to one decimal place the area of:
      1. Sector APQ and PBQ.
      2. Triangle APQ and PBQ.
      3. The shaded area (take π = 22/7)
  28. The minute hand of a clock is 6.5 cm long. Calculate the distance in cm moved by its tip between 10.30 am. and 10. 45 a.m. to 2 dp

Answers

      1. ∠DCF = 180 – 92 = 44° = < CAD
                          2
      2. ∠ BAO = 50°
        Acute angle AOB = 80°
        ∴ obtuse angle = 360 – 80 = 280°
    1. Area of the sector = (80/360 x 22/7 x 7 x 7)= 34.22cm2
      Area of the Δ= ½ x 7 x 7 x sin80= 24.13cm2
      Area of the shaded segment = 34.22 – 24.13 = 10.09cm2
  1. ∠ COB = 2 x 50 = 100°
    ∠OCA =∠ OAC = 180 – 100 = 40
                                   2
     BAC = 180 – (50 + 70)
    = 60
  2. PB. PA (PT)2
    PBPT
    PT      PA
    4    =    12   
    12      4 + 2r
    4(4 +2r) = 122
         4           4
    4 + 2r = 36
    2r = 32
    r = 16 cm
  3.  
    1. BOE = 2 BCE = 2 x 200 = 400
    2. BOE =400
      BEC = ½ (3600 – 600 ) = 1500
      Angels subtended at the centre is twice at the Circumference.
    3. CEF = 900 – 800 = 100
    4. BCO =CBO = 600
      Base angles isosceles triangle.
      OXC = 1800 – (600 + 200)
      = 1000
    5. BCE = 200
      CXE = 1800 -1000 = 800
      CEX = 800
      OEF = 1800 – (800 + 500 + 100 )
      = 400

    1. PQ = √(82 - 22)
      = 60
      = 7.746cm
    2. ∠PAS = 2cos-1
      = 151o
      ∴Reflex ∠PAS = 209o OR 360o – 151o = 209o
    3. Length PYS = 209/360 x 2 x 6 = 21.89cm
      Length QXR = 151/360 x 2 x 4 = 10.54cm
    4. Length of belt = 7.74 x 2 + 21.89 + 10.54
      = 47.92cm
  4.  
    1.  
      1. In 1 hr; Tap A fills 1/3
        B - ¼
        Capacity filled in 1 hr = 1/3 + ¼
        =7/12
        7/12 = 1 hr
        1 = 1 x 1 x 12/7
        = 1 5/7 hrs.
      2. 1/3 + ¼ - 1/6 = 5/12 in one hr
        5/12 = 1hr
        1 = 1 x 1 x 12/5
        = 2 2/5 hrs
  5. ABD = 310
    CBD = 370
  6. x (x+9) = 4x9
    x2+ 9x – 36 = 0
    (x2 – 3x) + (12x -36=0)
    x(x-3) + 12(x-3) =0
    (x+12) (x-3) = 0
    x - 3 = 0
    x = 3 only
  7. PO. OQ = BO.OA
    8 x 6 = 4.5 x y
    y = 8 x 6
           4.5
    = 10.67
  8.  
    1. ∠DGB = ∠ ABG = 40° (alt.seg ∠,s)
    2. ∠ DGE = ∠ DBE = 25° (∠s in same segment)
    3. ∠EFG
      ∠ GEB = 40°, = ∠BDG and ∠ BED = 45° = ∠BGD
      ∴ 
      In ∠GED, ∠GDE = 180 – (25 + 40 + 45) = 70°
      ∴∠GFE = 180 – 70 = 110º (Sup angles)
    4. Angle CBD in ∠BGE, Angle GBE = 180 – (110) = 70º
      ∠Angle CBD = 180 – (40 + 70 + 25) = 45º
      Or Angle CBD = Angle BGD = 45º (Angles in Alt segment)
    5. Angle BCD in BCD, Angle BDC = 70 º Angles in a straight line
      ∠Angle BCD = 180 – (70 + 45) Angles of a triangle = 65º
  9.  
    1. Sinθ= 4.5 = 0.5025
        8
      θ = Sin-1 0.5625
      = 34.23o
      ∠Apb = 68.46o
      Sin α= 4.5 = 0.75
                 6
      α = Sin-10.75
      = ∠ 48.59o
      ∠ Aqb = 97.18o
    2. Area Of Segment PAB = 68.46/360 X 22/7 X 8 X 8 – ½ X 8 Sin 68.46
      = 38.25 – 29.77
      = 8.48cm2
      Area Of Segment AQB = 97.18/360 X 22/7 X 36 – ½ 36 Sin 97.18
      = 30.65 - 17.86 
      = 12.68 cm2
      Area of quadrilateral APBQ = ½ 64 sin 68.46 + ½ x 36sin 92.18
      = 29.77 + 17.86 = 47.63
      Shaded area = 47.63 – (8.48 + 12.68) = 26.47cm2
  10. CBD = 90 - 42 = 48o
    Angle of triangle add to 180o
    DOB = 180o – 42 = 138o
    Opposite angles of cyclic quadrilateral add to 180o
    DAB = 138o = 69o
                2
    Angle at circumference is half the nagle substended at centre by same chord
    CDA
    ABD= 90 – 48 = 42o
    ADB = 180 –(69+42)
    180 - 111=69o
    CDA = 90 + 69
    Show ∠ADB is asoccesters= 159o
    ∠DAB = 69o 
    ∠DAB = 69o
    ∠ADB = 69o
    ABD = 42o
    So two angles are equal hence it is asoccesters
  11.  
    circles ans 13
    1. MLN = 40o angles subtended by same chord in the same segment are equal.
    2. OLN = 90 – 65 = 25o
      Angle sum of is 180o or angle subtended by > diameter is 90o.
    3. LNP = 65o exterior is equal to opposite interior angle or angle btwn a chord and a tangent is equal to angle subtended by the same chord in the alternate segment.
    4. MPN = 180 – 170 = 10o angle sum of a is 180o
    5. LMO = 65o angles subtended by same chord.
  12.  
    1.  
      circles ans 14
      Sin =
      4/4.6 = 0.869565
      = sin-10.89565 = 60.408o
      ABR = 2 x 60.408o = 120.8163o
      ∠120.82o (2d.p)
    2. Area of sector ABCR
      = 120.8163o x πx 4.62)cm2
           360o
      = 22.30994cm2
      Area of sector OAPC
      =(60o x πx 82)cm2
         360o
      = 33.51032cm2
      = 33.51cm2(2d.p)
      Area of ΔABC = ( ½ x 4.62sin 120.8163)cm2 = 9.08625cm2
      Area of AOC = ( ½ x 82 sin 60) cm2 = 27.7128cm2
      Sum of area ofΔs = 36.799cm2 36.80cm2
      Area of shaded part = area of sectors – area of Δs
      = (22.31 + 33.51 – 36.80)cm2 = 19.02cm2(2dp)
  13.  
    1. TDC = ABT (exterior opp. angle of a cyclic quadrilateral)
      = 100o
    2.  BAT = ATB (base s of isosceles ATB)
      = 180 – 100 = 40o
    3. ∠TCD = ∠XTD (angles in alternate segments)
      = 60o
      Or ∠BTC + 40o = 100o(exterior angle of a Δ)
      ∠BTC = 100o – 40o = 60o
    4. DTC = 180o- (58o + 100o) (angles in ∠TDC = 12o
  14.  
    1. GBD = 90°
      ABG = 180 – (90 + 36)
      = 180 – 126 = 54°
      GEB = ABG = 54°
    2. BED = CBD = 36°
    3. DGE = FEG = 20°
      OEB = 90 – (36 +20)
      = 90 – 56 = 34°
      OBE = OEB = 34°
    4. BGE = 36 + 20 = 56°
    5. GFE = 180 – EDG
      = 180 – 70 = 110°
  15.  
    1. XZ2 = 13.42 + 52 – 2 x 13.4 x 5 cos 57.7o
      = 170.56 + 25 – 134 x 0.5344
      = 204.56 – 71.6096
      XZ2 = 132.9504
      XZ = 11.5304cm
    2. 2R = 11.5304
      Sin 57.7o
      2R = 11.5304
               0.8453
      2R = 13.60866
      R = 6.08043cm
      circles ans 17
  16. 52 = 62 + 62 – 2 x 6 x 6 cos A
    72 cos A = 72 – 25 = 46
    Cos A = 46/72 = 0.6389
    A = Cos-1 0.6389 = 50.29º
    Area of the minor sector APQ
    = 50.29 x 3.142 x 62
        360
    = 15.801cm2
    Area of the triangle APQ = ½ x 6 x 6 sin 50.29 = 13.847cm2
    Area of the minor segment = = (15.801 – 13.847)cm2 = 1.954cm2
    Area of triange PBQ 
    √6.5 (6.5 – 4) (6.5 – 4) (6.5 – 5)
    √6.5 x 2.5 x 2.5 x 1.5 = 7.806cm2
    Area of shaded region = (7.806 – 1.954)cm2 = 5.852cm2
  17.  
    1. PQR = 180o – 75o
      = 105o. NPQR is cyclic quadrilateral.
    2. NRP = 90o -75o
      = 15o, Third angle of NRP.
      PRS = 180o -65o , Angles on a
      = 115o, straight line.
      QSR = 180o – (115o – 35o )
      = 30o, 3rd angle of triangle PRS.
    3. Reflex POR = 2 PQR
      = 2 x 105o = 210o
    4. MQR = MNR = 40o
      Subtended by same chord MR
  18.  
    1. TDC = 100o (Cyclic quadrilateral)
    2. TCB = 40o (Cyclic quadrilateral)
    3. TCD = 58o (Cyclic quadrilateral)
    4. BTC = 60o (Sum angle of a add upto 180o)
    5. DTC = 22o ( angle sum of a straight line add upto 180o)
  19. 4 x 10 = 5(5 + x)
    40 = 25 + 5x
    3 = x
  20. T11 = a + 10d
    T2 = a + d
    a + 10d = 4a + 4d …...........(i)
    3a – 6d = 0
    S7 = 7/2{2a + 6d} = 175 …(ii)
    2a + 6d = 50
    3a – 6d = 0
    5a        = 50
    a = 10 d = 5
  21. CBE = 400 ( alt.segiment theoren)
    ∠BCE = 1200 (Suppl. To BCD = 600 alt. seg.)
    ∴ (40 + 120 + E) = 1800 (Angle sum of Δ)
    BEC = 200
  22. Taxable income p.a = 36,000+53142.86
    =sh.412142.86
    Monthly salary = 413142.86 + 12,000
                                    12
    = 34428.57 + 1200 = Sh 35628.57
  23.  
    1.  
      1. ∠PTQ = 180o - 56o = 124o
        124 + 38 = 162o
        180o – 162o = 18o
        90o + 18o = 108o
        180o – 108o = 72o
        180o- (72o + 56o) = 52o
        ∠PRS = 52o
      2. RSQ = ∠RPQ = 18o
    2. A α B. 
                C3
      A = K.B
             C3
      12 = 3K
              23
      K = 12 x 8 = 32
                3
      A = 32B
              C3
      10 x (1.5)3 = B
           32
      B = 1.055
      Value of the constant.
      Substitution Formulation
      Values of constants.
      Substitution
    3. y = K + Mx2 where K and M are constants
      7 = K + 100 M                    100 x 0.005 + K = 7
      5.5 = K + 400M                              -0.5 + K = 7
      1.5 = 300M                                               K = 7.5
      M= 0.005
      y = 7.5 – 0.005 x 182
      y = 7.5 – 1.62
      y = 5.88
  24.  
    1. PN2 = 52 - 42
      PN= 3cm
      QN2 = 62 – 42
      QN = 4.47cm
      PQ = 3 + 4.47= 7.47
      1. ∠ APB
        Sin ½ θ4/5 = 0.8
        ½ sin θ= 53.13
        ∠APB
      2. Sin ½ α= 4/6 = 0.6667
        ½ α= 41.81
        α = 83.62
        ∴∠ AQB = 83.62°
    2. Area of the shaded region – Area of the segments
      = 106.3 x 22 x 52 – ½ x 5 x 5 sin 106.3
          360       7
      = 23.19 – 11.998 = 11.192
      83.6 x 22 x 6x 6 – ½ x 6 x 6 sin 83.6 = 8.38
      360     7
      Total 11.192 + 8.38 = 19.52
  25. Using cosine rule
    7.82 = 6.62 + 5.92 – 2 x 6.6 x 5.9 cos R
    Cos C = 6.62 + 5.92 – 7.82
                   2 x 6.6 x 5.9
    = 43.59 + 34.81 – 60.84 = 78.37 – 60.84
                 77.88                        77.88
    = 17.53 = 0.2251
       77.88
    C = 77o
       7.8  = 2r ⇒ r =     7.8     
    Sin 77                 2 x sin 77
                                = 4 cm
    Area of circle = 3.142 x 42
    Area of ΔPQR = ½ (6.6) (5.9) sin 77
    = 18.97
    ∴Area of shaded region = 50.27 – 18.97 = 31.30cm2
  26.  
    1. PAQ = 2 ∠PAB = 42o x 2 = 84o
      ∠PBQ = 2 ∠ABQ = 30o x 2 = 60o
      1. Area of sector APQ = 84 x 22 x 6 x 6 = 26.4 cm2
                                       360   7
        Area of sector PBQ 60 x 22 x 8 x = 33.5 cm2
                                    360  7
      2. Area of APQ = ½ x 6 x 65 = 84o = 18 x 0.9945
        = 17.9 cm2
        Area of PBQ = ½ x 8 x 85: = 60o = 32 x 0.8660
        = 27.7 cm2
      3. For each circle, shaded area = sector area – triangle Area.
        = area of sector APQ – area of triangle APQ
        = 26.4 – 17.9 = 8.5 cm2
        2nd circle, shaded area
        = area of sector PBQ – area of PBQ
        = 33.5 – 27.7 = 5.8 cm2
        Total shaded area = 8.5 + 5.8 = 14.3 cm2
  27. 90 x 3.142 x 2 x 6.5         
    360 
    10.2115 cm
    10.21 cm
    circles ans 29

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