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

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Questions

In the figure below angle BAC =52^o, angle ACB = 40^oand AD = DC. The radius of the circle is 7cm. EF is a tangent to the circle
1. Find; giving reasons
  1. angle DCF
  2. angle AOB (obtuse)
2. Calculate the area of the shaded segment AGB
In the figure below, O is the centre of the circle. Angle CBA = 50^oand angle BCO = 30^o.

Find the size of the angle BAC
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
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 = 20^oBC cuts OE at X

Calculate;
1. angle BOE
2. angle BEC
3. angle CEF
4. angle OXC
5. angle OFE
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

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.
In the figure below, ABCD is a cyclic quadrilateral and BD is a diagonal. EADF is a straight line, ∠CDF = 68^o, ∠BDC = 45^o and ∠BAE = 98^o.

Calculate the size of:
1. ∠ABD.
2. ∠CBD
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.

Find the value of x
The chords AB and PQ intersects internally at O. Given that the length of OP=8cm, OA= 4.5cm and OQ=6cm.

Calculate the length of OB
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.

Find the sizes of the following angles giving reasons in each case:
1. ∠BDG
2. ∠DGE
3. ∠EFG
4. ∠CBD
5. ∠BCD
The figure below shows two intersecting circles radii 8 cm and 6 cm respectively.

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
In the figure below, O and P are centres of two intersecting circles. ABE is tangent to circle BCD at B angle BCD is 42^o.
1. Giving reasons for your answer, find:-
  1. CBD
  2. DOB
  3. DAB
  4. CDA
2. Show that ∠ADB is isosceles
In the figure above K, M & P are points on a straight line. PN is a tangent of the circle centre O. Angle KOL = 130^oand angle MKN = 40^o.
Find, giving reasons, the values of angles.
1. ∠MLN
2. ∠OLN
3. ∠LNP
4. ∠MPN
5. ∠LMO
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 = 120^o. ARC is an arc of the circle, Centre B and radius 4.6cm.

Calculate correct to 2 decimal places
1. Angle ABR
2. Area of sectors ABCR and OAPC
3. Area of the shaded part
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

Find by giving reasons, the value of angle:
1. TDC
2. TCB
3. TCD
4. BTC
5. DTC
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.

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
XYZ is a triangle in which x = 13.4cm, Z= 5cm and ∠XYZ =57.7^o. Find:
1. Length of XZ
2. The circum radius of the triangle
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

Calculate the area of the shaded region (take πas 3.142)
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.

Giving reasons for every statement you write, find the following angles
1. ∠ PQR
2. ∠QSR
3. Reflex ∠POR
4. ∠ MQR
5. ∠ PON
In the diagram below, ATX is a tangent to the circle at point T, ABC is a straight line, ∠ABT= 100^o, ∠XTD = 58^oand the line AB = BT

Find giving reasons the value of :
1. ∠TDC
2. ∠TCB
3. ∠TCD
4. ∠BTC
5. ∠DTC
In the figure above AB = 6 cm, BC = 4 cm DC = 5 cm.

Find the length DE.
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
In the diagram below ABE is a tangent to a circle at B and DCE is a straight line.

If ABD = 60^o, BOC = 80^o and O is the centre of the circle, find with reasons ∠BEC
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)
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 = 38^o and angle QTR = 56^o.
  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
The figure below shows two intersecting circles with centres P and Q and radius 5cm and 6cm respectively.

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
Triangle ABC is inscribed in the circle. AB= 7.8cm, AC 6.6cm and BC= 5.9cm.

Find:
1. The radius of the circle correct to one decimal place
2. The area of the shaded region
The figure below shows two circles centres A and B and radii 6 cm and 8 cm respectively.

The circles intersect at P and Q. Angle PAB = 42^oand angle ABQ = 30^o.
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)
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. 1. ∠DCF = 180 – 92 = 44° = < CAD
    2
  2. ∠ BAO = 50°
    Acute angle AOB = 80°
    ∴ obtuse angle = 360 – 80 = 280°
2. Area of the sector = (⁸⁰/₃₆₀x ²²/₇x 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
∠ COB = 2 x 50 = 100°
∠OCA =∠ OAC = 180 – 100 = 40
2
∴∠ BAC = 180 – (50 + 70)
= 60
PB. PA (PT)²PB = PT
PT PA
4 = 12
12 4 + 2r
4(4 +2r) = 12²
4 4
4 + 2r = 36
2r = 32
r = 16 cm
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
  = 151^o
  ∴Reflex ∠PAS = 209^oOR 360^o– 151^o= 209^o
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
1. 1. In 1 hr; Tap A fills ¹/₃
    B - ¼
    Capacity filled in 1 hr = ¹/₃+ ¼
    =⁷/₁₂7/12 = 1 hr
    1 = 1 x 1 x ¹²/₇= 1 ⁵/₇hrs.
  2. ¹/₃+ ¼ - ¹/₆= ⁵/₁₂⇒ in one hr
    ⁵/₁₂= 1hr
    1 = 1 x 1 x ¹²/₅= 2 ²/₅hrs
∠ABD = 31⁰∠ CBD = 37⁰
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
PO. OQ = BO.OA
8 x 6 = 4.5 x y
y = 8 x 6
4.5
= 10.67
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
1. Sinθ= 4.5 = 0.5025
  8
  θ = Sin-1 0.5625
  = 34.23^o∠Apb = 68.46^oSin α= 4.5 = 0.75
  6
  α = Sin^-10.75
  = ∠ 48.59^o∠ Aqb = 97.18^o
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
  = 8.48cm²Area Of Segment AQB = ^97.18/₃₆₀ X ²²/₇ X 36 – ½ 36 Sin 97.18
  = 30.65 - 17.86
  = 12.68 cm²Area 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
CBD = 90 - 42 = 48^oAngle of triangle add to 180^o
DOB = 180^o– 42 = 138^oOpposite angles of cyclic quadrilateral add to 180o
DAB = 138^o= 69^o
2
Angle at circumference is half the nagle substended at centre by same chord
CDA
ABD= 90 – 48 = 42^oADB = 180 –(69+42)
180 - 111=69^oCDA = 90 + 69^oShow ∠ADB is asoccesters= 159^o∠DAB = 69^o
∠DAB = 69^o∠ADB = 69^o∠ABD = 42^oSo two angles are equal hence it is asoccesters
1. MLN = 40^oangles subtended by same chord in the same segment are equal.
2. OLN = 90 – 65 = 25^oAngle sum of ∆ is 180^oor angle subtended by > diameter is 90^o.
3. LNP = 65^oexterior ∆ 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 = 10^oangle sum of a ∆ is 180o
5. LMO = 65^oangles subtended by same chord.
1. Sin = ⁴/_4.6= 0.869565
  = sin-10.89565 = 60.408^o
  ABR = 2 x 60.408^o= 120.8163^o
  ∠120.82^o(2d.p)
2. Area of sector ABCR
  = 120.8163^ox πx 4.62)cm²
  360^o
  = 22.30994cm²
  Area of sector OAPC
  =(60^ox πx 8²)cm²
  360^o= 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= 27.7128cm²Sum 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)
1. ∠TDC = ABT (exterior opp. angle of a cyclic quadrilateral)
  = 100^o
2. BAT = ATB (base s of isosceles ATB)
  = 180 – 100 = 40^o
3. ∠TCD = ∠XTD (angles in alternate segments)
  = 60^oOr ∠BTC + 40^o= 100o(exterior angle of a Δ)
  ∠BTC = 100^o – 40^o= 60^o
4. DTC = 180o- (58^o+ 100^o) (angles in ∠TDC = 12^o
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°
1. XZ²= 13.4²+ 5²– 2 x 13.4 x 5 cos 57.7^o= 170.56 + 25 – 134 x 0.5344
  = 204.56 – 71.6096
  XZ²= 132.9504
  XZ = 11.5304cm
2. 2R = 11.5304
  Sin 57.7^o
  2R = 11.5304
  0.845³2R = 13.60866
  R = 6.08043cm
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
= 15.801cm²Area 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
1. ∡ PQR = 180^o – 75^o
  = 105^o. NPQR is cyclic quadrilateral.
2. ∡ NRP = 90^o -75^o
  = 15o, Third angle of ∆ NRP.
  ∡PRS = 180^o -65^o , Angles on a
  = 115o, straight line.
  ∴∡QSR = 180^o– (115^o– 35^o)
  = 30^o, 3rd angle of triangle PRS.
3. Reflex ∡POR = 2 ∡PQR
  = 2 x 105^o= 210^o
4. ∡MQR = ∡MNR = 40^oSubtended by same chord MR
1. ∠ TDC = 100^o(Cyclic quadrilateral)
2. ∠ TCB = 40^o (Cyclic quadrilateral)
3. ∠ TCD = 58^o(Cyclic quadrilateral)
4. ∠ BTC = 60^o (Sum angle of a ∆ add upto 180^o)
5. ∠DTC = 22^o( angle sum of a straight line add upto 180^o)
4 x 10 = 5(5 + x)
40 = 25 + 5x
3 = x
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
∠CBE = 40⁰( alt.segiment theoren)
∠BCE = 120⁰(Suppl. To BCD = 60⁰ alt. seg.)
∴ (40 + 120 + E) = 1800 (Angle sum of Δ)
∠ BEC = 20⁰
Taxable income p.a = 36,000+53142.86
=sh.412142.86
Monthly salary = 413142.86 + 12,000
12
= 34428.57 + 1200 = Sh 35628.57
1. 1. ∠PTQ = 180^o- 56^o= 124^o124 + 38 = 162^o
    180^o– 162^o= 18^o
    90^o + 18^o= 108^o
    180^o– 108^o= 72^o
    180^o- (72^o+ 56^o) = 52^o∠PRS = 52o
  2. ∠RSQ = ∠RPQ = 18^o
2. A α B. 1
  C3
  A = K.B
  C³12 = 3K
  2³K = 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 + Mx²where 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
1. PN²= 5² - 4²
  PN= 3cm
  QN²= 6²– 4²
  QN = 4.47cm
  ∴ PQ = 3 + 4.47= 7.47
2. 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°
3. 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
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 = 77^o 7.8 = 2r ⇒ r = 7.8
Sin 77 2 x sin 77
= 4 cm
Area of circle = 3.142 x 4²Area of ΔPQR = ½ (6.6) (5.9) sin 77
= 18.97
∴Area of shaded region = 50.27 – 18.97 = 31.30cm²
1. ∠ PAQ = 2 ∠PAB = 42^ox 2 = 84o
  ∠PBQ = 2 ∠ABQ = 30^ox 2 = 60^o
2. 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 = 84^o = 18 x 0.9945
    = 17.9 cm²Area of ∆ PBQ = ½ x 8 x 85: = 60^o= 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
    2^ndcircle, 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
90 x 3.142 x 2 x 6.5
360
10.2115 cm
10.21 cm