## 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 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. 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 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 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 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. 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. 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. 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. 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. 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. 1. Giving reasons for your answer, find:-
1. CBD
2. DOB
3. DAB
4. CDA
2. Show that ∠ADB is isosceles
12. 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. 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 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. 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 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. 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 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. 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. 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) 24. 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. 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. 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. 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

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
ABD = 42o
So two angles are equal hence it is asoccesters
11. 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. 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 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 #### Download Circles, Chords and Tangents Questions and Answers - Form 3 Topical Mathematics.

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