Questions
 In the figure below angle BAC =52^{o}, angle ACB = 40^{o }and AD = DC. The radius of the circle is 7cm. EF is a tangent to the circle
 Find; giving reasons
 angle DCF
 angle AOB (obtuse)
 Calculate the area of the shaded segment AGB
 Find; giving reasons
 In the figure below, O is the centre of the circle. Angle CBA = 50^{o }and 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^{o }BC cuts OE at X
Calculate; angle BOE
 angle BEC
 angle CEF
 angle OXC
 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 Length PQ
 ∠PAS reflex
 Length of arc PYS and QXR
 The total length of the string PQXRSY
 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.
 How long would it take pipes A and B to fill the tank if pipe C is closed?
 Starting with an empty tank, how long would it take to fill the tank with all pipes running?
 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.
 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.
 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: ∠ABD.
 ∠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: ∠BDG
 ∠DGE
 ∠EFG
 ∠CBD
 ∠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: Calculate the size of angles:
 ∠APB
 ∠AQB
 Calculate the area of the shaded region
 Calculate the size of angles:
 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}.
 Giving reasons for your answer, find:
 CBD
 DOB
 DAB
 CDA
 Show that ∠ADB is isosceles
 Giving reasons for your answer, find:

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^{o }and angle MKN = 40^{o}.
Find, giving reasons, the values of angles. ∠MLN
 ∠OLN
 ∠LNP
 ∠MPN
 ∠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 Angle ABR
 Area of sectors ABCR and OAPC
 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: TDC
 TCB
 TCD
 BTC
 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: GEB
 BED
 OBE
 BGE
 GFE
 XYZ is a triangle in which x = 13.4cm, Z= 5cm and ∠XYZ =57.7^{o }. Find:
 Length of XZ
 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 = 750^{0 }∠ NRM = 50^{0} and ∠RPQ = 35^{0}. MRS and PQS are straight lines.
Giving reasons for every statement you write, find the following angles ∠ PQR
 ∠QSR
 Reflex ∠POR
 ∠ MQR
 ∠ PON
 ∠ PQR
 In the diagram below, ATX is a tangent to the circle at point T, ABC is a straight line, ∠ABT= 100^{o}, ∠XTD = 58^{o }and the line AB = BT
Find giving reasons the value of : ∠TDC
 ∠TCB
 ∠TCD
 ∠BTC
 ∠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)

 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 ; ∠PRQ
 ∠RSQ
 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
 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
 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}.
 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 the length of PQ
 the size of;
 angle APB
 angle AQB
 the area of the shaded region
 Triangle ABC is inscribed in the circle. AB= 7.8cm, AC 6.6cm and BC= 5.9cm.
Find: The radius of the circle correct to one decimal place
 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^{o }and angle ABQ = 30^{o}. Find the size of ∠PAQ and PBQ.
 Calculate, to one decimal place the area of:
 Sector APQ and PBQ.
 Triangle APQ and PBQ.
 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
 ∠DCF = 180 – 92 = 44° = < CAD
2  ∠ BAO = 50°
Acute angle AOB = 80°
∴ obtuse angle = 360 – 80 = 280°
 ∠DCF = 180 – 92 = 44° = < CAD
 Area of the sector = (^{80}/_{360 }x ^{22}/_{7 }x 7 x 7)= 34.22cm^{2}
Area of the Δ= ½ x 7 x 7 x sin80= 24.13cm^{2}
Area of the shaded segment = 34.22 – 24.13 = 10.09cm^{2}
 ∠ COB = 2 x 50 = 100°
∠OCA =∠ OAC = 180 – 100 = 40
2
∴∠ BAC = 180 – (50 + 70)
= 60  PB. PA (PT)^{2}PB = PT
PT PA
4 = 12
12 4 + 2r
4(4 +2r) = 12^{2}
4 4
4 + 2r = 36
2r = 32
r = 16 cm 
 ∠BOE = 2 ∠BCE = 2 x 20^{0} = 40^{0}
 ∠ BOE =40^{0}
∠BEC = ½ (360^{0 }– 60^{0 }) = 150^{0}
Angels subtended at the centre is twice at the Circumference.  ∠ CEF = 90^{0 }– 80^{0 }= 10^{0}
 ∠BCO =∠CBO = 60^{0}
Base angles isosceles triangle.
∠ OXC = 180^{0 }– (60^{0 }+ 20^{0})
= 1000  ∠ BCE = 20^{0}
∠ CXE = 180^{0} 100^{0 }= 80^{0}
∠CEX = 80^{0}
∠ OEF = 180^{0 }– (80^{0 }+ 50^{0 }+ 10^{0 })
= 400
 PQ = √(8^{2 } 2^{2})
= 60
= 7.746cm  ∠PAS = 2cos^{1}
= 151^{o}
∴Reflex ∠PAS = 209^{o }OR 360^{o }– 151^{o }= 209^{o}  Length PYS = ^{209}/_{360} x 2 x 6 = 21.89cm
Length QXR = ^{151}/_{360} x 2 x 4 = 10.54cm  Length of belt = 7.74 x 2 + 21.89 + 10.54
= 47.92cm
 PQ = √(8^{2 } 2^{2})


 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.  ^{1}/_{3 }+ ¼  ^{1}/_{6 }= ^{5}/_{12 }⇒ in one hr
^{5}/_{12 }= 1hr
1 = 1 x 1 x ^{12}/_{5}= 2 ^{2}/_{5 }hrs
 In 1 hr; Tap A fills ^{1}/_{3}

 ∠ABD = 31^{0}∠ CBD = 37^{0}
 x (x+9) = 4x9
x^{2}+ 9x – 36 = 0
(x^{2 }– 3x) + (12x 36=0)
x(x3) + 12(x3) =0
(x+12) (x3) = 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 
 ∠DGB = ∠ ABG = 40° (alt.seg ∠,s)
 ∠ DGE = ∠ DBE = 25° (∠s in same segment)
 ∠EFG
∠ GEB = 40°, = ∠BDG and ∠ BED = 45° = ∠BGD
∴ In ∠GED, ∠GDE = 180 – (25 + 40 + 45) = 70°
∴∠GFE = 180 – 70 = 110º (Sup angles)  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)  Angle BCD in BCD, Angle BDC = 70 º Angles in a straight line
∠Angle BCD = 180 – (70 + 45) Angles of a triangle = 65º

 Sinθ= 4.5 = 0.5025
8
θ = Sin1 0.5625
= 34.23^{o}∠Apb = 68.46^{o}Sin α= 4.5 = 0.75
6
α = Sin^{1}0.75
= ∠ 48.59^{o}∠ Aqb = 97.18^{o}  ^{}Area Of Segment PAB = ^{68.46}/_{360} X ^{22}/_{7} X 8 X 8 – ½ X 8 Sin 68.46
= 38.25 – 29.77
= 8.48cm^{2}Area Of Segment AQB = ^{97.18}/_{360} X ^{22}/_{7} X 36 – ½ 36 Sin 97.18
= 30.65  17.86
= 12.68 cm^{2}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) = 26.47cm^{2}
 Sinθ= 4.5 = 0.5025
 ^{}CBD = 90  42 = 48^{o}Angle of triangle add to 180^{o}
DOB = 180^{o }– 42 = 138^{o}Opposite 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^{o}ADB = 180 –(69+42)
180  111=69^{o}CDA = 90 + 69^{o }Show ∠ADB is asoccesters= 159^{o}∠DAB = 69^{o}
∠DAB = 69^{o}∠ADB = 69^{o}∠ABD = 42^{o}So two angles are equal hence it is asoccesters 
 MLN = 40^{o }angles subtended by same chord in the same segment are equal.
 OLN = 90 – 65 = 25^{o}Angle sum of ∆ is 180^{o }or angle subtended by > diameter is 90^{o}.
 LNP = 65^{o }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.
 MPN = 180 – 170 = 10^{o }angle sum of a ∆ is 180o
 LMO = 65^{o }angles subtended by same chord.


Sin = ^{4}/_{4.6 }= 0.869565
= sin10.89565 = 60.408^{o}
ABR = 2 x 60.408^{o }= 120.8163^{o}
∠120.82^{o }(2d.p)  Area of sector ABCR
= 120.8163^{o }x πx 4.62)cm^{2}
360^{o}
= 22.30994cm^{2}
Area of sector OAPC
=(60^{o }x πx 8^{2})cm^{2}
360^{o}= 33.51032cm^{2}
= 33.51cm^{2}(2d.p)
Area of ΔABC = ( ½ x 4.62sin 120.8163)cm^{2 }= 9.08625cm^{2}
Area of AOC = ( ½ x 82 sin 60) cm^{2 }= 27.7128cm^{2}Sum of area ofΔs = 36.799cm2 36.80cm2
∴Area of shaded part = area of sectors – area of Δs
= (22.31 + 33.51 – 36.80)cm^{2 }= 19.02cm^{2}(2dp)


 ∠TDC = ABT (exterior opp. angle of a cyclic quadrilateral)
= 100^{o}  ^{} BAT = ATB (base s of isosceles ATB)
= 180 – 100 = 40^{o}  ∠TCD = ∠XTD (angles in alternate segments)
= 60^{o}Or ∠BTC + 40^{o }= 100o(exterior angle of a Δ)
∠BTC = 100^{o} – 40^{o }= 60^{o}  DTC = 180o (58^{o }+ 100^{o}) (angles in ∠TDC = 12^{o}
 ∠TDC = ABT (exterior opp. angle of a cyclic quadrilateral)

 GBD = 90°
ABG = 180 – (90 + 36)
= 180 – 126 = 54°
GEB = ABG = 54°  BED = CBD = 36°
 DGE = FEG = 20°
OEB = 90 – (36 +20)
= 90 – 56 = 34°
OBE = OEB = 34°  BGE = 36 + 20 = 56°
 GFE = 180 – EDG
= 180 – 70 = 110°
 GBD = 90°

 XZ^{2 }= 13.4^{2 }+ 5^{2 }– 2 x 13.4 x 5 cos 57.7^{o}= 170.56 + 25 – 134 x 0.5344
= 204.56 – 71.6096
XZ^{2 }= 132.9504
XZ = 11.5304cm  2R = 11.5304
Sin 57.7^{o}
2R = 11.5304
0.845^{3}2R = 13.60866
R = 6.08043cm
 XZ^{2 }= 13.4^{2 }+ 5^{2 }– 2 x 13.4 x 5 cos 57.7^{o}= 170.56 + 25 – 134 x 0.5344
 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
= 15.801cm^{2}Area of the triangle APQ = ½ x 6 x 6 sin 50.29 = 13.847cm^{2}
Area of the minor segment = = (15.801 – 13.847)cm^{2} = 1.954cm^{2}
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 = 7.806cm^{2}
Area of shaded region = (7.806 – 1.954)cm^{2 }= 5.852cm^{2} 
 ∡ PQR = 180^{o} – 75^{o}
= 105^{o}. NPQR is cyclic quadrilateral.  ∡ 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.  Reflex ∡POR = 2 ∡PQR
= 2 x 105^{o }= 210^{o}  ∡MQR = ∡MNR = 40^{o}Subtended by same chord MR
 ∡ PQR = 180^{o} – 75^{o}

 ∠ TDC = 100^{o }(Cyclic quadrilateral)
 ∠ TCB = 40^{o} (Cyclic quadrilateral)
 ∠ TCD = 58^{o }(Cyclic quadrilateral)
 ∠ BTC = 60^{o} (Sum angle of a ∆ add upto 180^{o})
 ∠DTC = 22^{o }( angle sum of a straight line add upto 180^{o})
 4 x 10 = 5(5 + x)
40 = 25 + 5x
3 = x  T_{11 }= a + 10d
T_{2 }= 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 = 40^{0 }( alt.segiment theoren)
∠BCE = 120^{0 }(Suppl. To BCD = 60^{0} alt. seg.)
∴ (40 + 120 + E) = 1800 (Angle sum of Δ)
∠ BEC = 20^{0}  Taxable income p.a = 36,000+53142.86
=sh.412142.86
Monthly salary = 413142.86 + 12,000
12
= 34428.57 + 1200 = Sh 35628.57 

 ∠PTQ = 180^{o } 56^{o }= 124^{o}124 + 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  ∠RSQ = ∠RPQ = 18^{o}
 ∠PTQ = 180^{o } 56^{o }= 124^{o}124 + 38 = 162^{o}
 ^{}A α B. 1
C3
A = K.B
C^{3}12 = 3K
2^{3}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  y = K + Mx^{2 }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 18^{2}
y = 7.5 – 1.62
y = 5.88


 PN^{2 }= 5^{2}  4^{2}
PN= 3cm
QN^{2 }= 6^{2 }– 4^{2}
QN = 4.47cm
∴ PQ = 3 + 4.47= 7.47  ∠ APB
Sin ½ θ^{4}/_{5 }= 0.8
½ sin θ= 53.13
∠APB  Sin ½ α= 4/6 = 0.6667
½ α= 41.81
α = 83.62
∴∠ AQB = 83.62°
 ∠ APB
 Area of the shaded region – Area of the segments
= 106.3 x 22 x 5^{2 }– 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
 PN^{2 }= 5^{2}  4^{2}
 Using cosine rule
7.8^{2 }= 6.6^{2 }+ 5.9^{2 }– 2 x 6.6 x 5.9 cos R
Cos C = 6.6^{2 }+ 5.9^{2 }– 7.8^{2} 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 4^{2}Area of ΔPQR = ½ (6.6) (5.9) sin 77
= 18.97
∴Area of shaded region = 50.27 – 18.97 = 31.30cm^{2} 
 ∠ PAQ = 2 ∠PAB = 42^{o }x 2 = 84o
∠PBQ = 2 ∠ABQ = 30^{o }x 2 = 60^{o}  ^{}
 Area of sector APQ = 84 x 22 x 6 x 6 = 26.4 cm^{2}
360 7
Area of sector PBQ 60 x 22 x 8 x = 33.5 cm^{2}
360 7  Area of ∆APQ = ½ x 6 x 65 = 84^{o} = 18 x 0.9945
= 17.9 cm^{2}Area of ∆ PBQ = ½ x 8 x 85: = 60^{o }= 32 x 0.8660
= 27.7 cm2  For each circle, shaded area = sector area – triangle Area.
= area of sector APQ – area of triangle APQ
= 26.4 – 17.9 = 8.5 cm^{2}
2^{nd }circle, shaded area
= area of sector PBQ – area of ∆PBQ
= 33.5 – 27.7 = 5.8 cm^{2}
Total shaded area = 8.5 + 5.8 = 14.3 cm^{2}
 Area of sector APQ = 84 x 22 x 6 x 6 = 26.4 cm^{2}
 ∠ PAQ = 2 ∠PAB = 42^{o }x 2 = 84o
 90 x 3.142 x 2 x 6.5
360
10.2115 cm
10.21 cm
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