Circles –chords and tangents Questions and Answers

Circles –chords and tangents Questions

1.  In the figure below not drawn to scale. DC is a tangent to the circle. DC = 6cm, AB = 5cm. calculate BC. (3mks)

2.

The figure above shows a circle in which chords AD and BC intersect at G. chords AB and CD produced meet at K.

(a) If and, determine the size of (2 mks)

(b) Given that KB = 5 cm, KC = 15 cm and KD = 7 cm, determine the length of KA (3 mks)

(c) Giving reasons for your answer, show that triangle KDA and KBC are similar  (5 mks)

3.  The figure below shows a circle with secants ABE and CDE, If AB = 4cm and BE = 6 cm and DE = 4 cm. Find the length of CD.  (2mks)

4.  In the figure below, CF is a tangent to the circle. BC = 3cm, ED = 4cm and DC = 2cm.

Find:-

1. AB (2 mks)
2. FC (2 mks)

5.  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

 (a) Find; giving reasons

  (i) angle DCF  

  (ii) angle AOB (obtuse)

1. Calculate the area of the shaded segment AGB  

6.  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

7.  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

8.  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;

 (a) angle BOE

 (b) angle BEC

 (c) angle CEF

 (d) angle OXC

(e) angle OFE

9.  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:

 (a) Length PQ

 (b) PAS reflex  

 (c) Length of arc PYS and QXR

 (d) The total length of the string PQXRSY  

10.  a) 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.

i) How long would it take pipes A and B to fill the tank if pipe C is closed?

ii) Starting with an empty tank, how long would it take to fill the tank with all pipes running?

 b) 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.

11.  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:

 a)  ABD.

 b)  CBD

12.  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

13.  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

14.  In the figure below ABC is a tangent to the circle at B. given that

Find the sizes of the following angles giving reasons in each case:

 a) BDG

  b)DGE

  c)EFG

 d)CBD

 e)BCD

15.  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:

 (a) Calculate the size of angles:-

  (i) APB

(ii) AQB

 (b) Calculate the area of the shaded region  

16.  The figure 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:-

(i) CBD

(ii) DOB

(iii) DAB

(iv) CDA

b) Show that ADB is isosceles

17.  

 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.

1. MLN
2. OLN
3. LNP
4. MPN
5. LMO

18.  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

 (a) Angle ABR  

 (b) Area of sectors ABCR and OAPC  

 (c) Area of the shaded part

19.  In the figure below, ATX is a tangent to the circle at point T, ABC is a straight line, angle

ABT = 100^o, angle XTD = 58^o 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

20.  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:

 a) GEB  

 b) BED  

 c) OBE  

 d) BGE  

 e) GFE

21.  XYZ is a triangle in which x = 13.4cm, Z= 5cm and XYZ =57.7^o . Find:

 (i) Length of XZ  

 (ii) The circum radius of the triangle

22.  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  as 3.142)

23.  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

(a) ∠ PQR

(b) ∠QSR  

(c) Reflex ∠POR  

(d) ∠ MQR

(e) ∠ PON

24.  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 :

 (a) TDC

 (b) TCB

  (c) TCD

 (d) BTC

 (e) DTC

25.  

 In the figure above AB = 6 cm, BC = 4 cm DC = 5 cm. Find the length DE.  

26.  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  

27.  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  

28.  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)  

29.

(a) 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

 (i) PRQ

(ii) RSQ

(b) 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  

(c) 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  

30.  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;

:

 (a) the length of PQ  

 (b) the size of;

  (i) angle APB  

  (ii) angle AQB  

 (c) the area of the shaded region

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

(a) The radius of the circle correct to one decimal place

(b) The area of the shaded region

32.  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 = 42⁰ and angle ABQ = 30⁰.

(a) Find the size of ∠PAQ and PBQ.  

(b) Calculate, to one decimal place the area of:

(i) Sector APQ and PBQ.

(ii) Triangle APQ and PBQ.

 (iii) The shaded area (take π 22)  

33.  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 dpl.  

Circles –chords and tangents Answers

1.  a) i) 180 – 92 = 44° = < CAD

2

ii) < BAO = 50°

   Acute angle AOB = 80°

 obtuse angle = 360 – 80 = 280°

b) Area of the sector = (⁸⁰/₃₆₀ x ²²/₇ x 7 x 7)= 34.22cm²

Area of the  = ½ x 7 x 7 x sin80= 24.13cm²

Area of the shaded segment = 34.22 –

24.13

  10.09cm²

2.  < COB = 2 x 50 = 100 °

< OCA =< OAC = 180 – 100 = 40

  2

 < BAC = 180 – (50 + 70)

= 60

3.  PB. PA (PT)²

PB = PT

PT PA

4 = 12

12 4 + 2r

4(4 +2r) = 12²

  4 4

 4 + 2r = 36

  2r = 32

  r = 16 cm

4.  (a) ∠BOE = 2 ∠BCE = 2 x 20⁰ = 40⁰

(b)∠ BOE =40⁰

∠BEC = ½ (360⁰ – 60⁰) = 150⁰

   Angels subtended at the centre is twice at the Circumference.

c) ∠ CEF = 90⁰ – 80⁰ = 10⁰

d) ∠BCO =∠CBO = 60⁰

Base angles isosceles triangle.

∠ OXC = 180⁰ – (60⁰ + 20⁰)

= 100⁰

 e) ∠ BCE = 20⁰

  ∠ CXE = 180⁰ -100⁰ = 80⁰

 ∠CEX = 80⁰

 ∠ OEF = 180⁰ – (80⁰ + 50⁰ + 10⁰)

= 40⁰

5.  (a) PQ = 8² – 2²

= 60

   = 7.746cm

(b) PAS = 2cos^-1

  = 151^o

Reflex  PAS = 209^o OR 360^o – 151^o = 209^o

(c) Length PYS = 209 x 2 x 6 = 21.89cm

  360

Length QXR = 151 x 2 x 4 = 10.54cm

 360

(d) Length of belt = 7.74 x 2 + 21.89 + 10.54

= 47.92cm

6.  a) i)  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.

 ii) ¹/₃ + ¼ – ¹/₆ = ⁵/₁₂
⇒ in one hr

⁵/₁₂ = 1hr

    1 = 1 x 1 x ¹²/₅

 = 2 ²/₅ hrs

7.  ∠ABD = 31⁰

∠ CBD = 37⁰

8.  x (x+9) = 4×9

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

9.  PO. OQ = BO .OA

8 x 6 = 4.5 x y

y = 8 x 6

  4.5

  = 10.67

10.   < DGB = < ABG = 40° (alt.seg <,s)

1. < DGE = < DBE = 25° (

< GEB = 40°, =

 In  GED,



d) 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)

e) Angle BCD in  BCD, Angle BDC = 70 º Angles in a straight line

  Angle BCD = 180 – (70 + 45) Angles of a triangle = 65º

11.  (a)Sin = 4.5 = 0.5025

8

 = Sin^-1 0.5625

   = 34.23^o

Apb = 68.46^o

Sin = 4-5 = 0.75

6

 = Sin-10.75

= 48.59^o

Aqb = 97.18^o

(b) Area Of Segment PAB = 68.46 X 22 X 8 X 8 – ½ X 8 Sin 68.46

  360 7

 = 38.25 – 29.77

  = 8.48cm²

Area Of Segment AQB = 97.18 X 22 X 36 – ½ 36 Sin 97.18

  360 7

  = 30.65 – 17.86 = 12.68cm²

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  

12.  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 180^o

DAB = 138^o = 69^o

  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=69^o

CDA = 90 + 69^o = 159^o

Show ADB is asoccesters

DAB = 69^o

DAB = 69^o

ADB = 69^o

ABD = 42^o

So two angles are equal hence it is asoccesters

13.  

 a) MLN = 40^o angles subtended by same chord in the same segment are equal.

 b) OLN = 90 – 65 = 25^o

  Angle sum of ∆ is 180^o or angle subtended by > diameter is 90^o.

 c) 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.

d) MPN = 180 – 170 = 10^o angle sum of a ∆ is 180^o

e) LMO = 65^o angles subtended by same chord.

14. (a) Sin = ⁴/_4.6 = 0.869565

= sin-10.89565 = 60.408^o

ABR = 2 x 60.408^o = 120.8163^oC

  120.82^o (2d.p)

(b) Area of sector ABCR

= 120.8163 ^o x  x 4.62)cm²

360^o

= 22.30994cm²

Area of sector OAPC

= 60^o x  x 8²)cm²

360^o

= 33.51032cm²

 = 33.51cm²(2d.p)

Area of ABC = ( ½ x 4.6²sin 120.8163)cm² = 9.08625cm²

Area of AOC = ( ½ x 82 sin 60) cm² = 27.7128cm²

Sum of area of s = 36.799cm² 36.80cm²

Area of shaded part = area of sectors – area of s

= (22.31 + 33.51 – 36.80)cm² = 19.02cm²(2dp)

15.  (a) TDC = ABT (exterior opp. angle of a cyclic quadrilateral)

= 100^o

(b) BAT = ATB (base s of isosceles ATB)

  = 180 – 100 = 40^o

(c) TCD = X TD (angles in alternate segments)

  = 60^o

Or BTC + 40^o = 100^o(exterior angle of a )

  BTC = 100^o – 40^o= 60^o

(d) DTC = 180^o– (58^o + 100^o) (angles in TDC = 12^o

16.  a) GBD = 90°

ABG = 180 – (90 + 36)

= 180 – 126 = 54°

GEB = ABG = 54°

b) BED = CBD = 36°

c) DGE = FEG = 20°

OEB = 90 – (36 +20)

= 90 – 56 = 34°

 OBE = OEB = 34°

d) BGE = 36 + 20 = 56°

e) GFE = 180 – EDG

= 180 – 70 = 110°

17.  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

(ii) 2R = 11.5304

Sin 57.7^o

2R = 11.5304

0.8453

 2R = 13.60866

 R = 6.08043cm

18.  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 triangle 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

19.  a) ∡ PQR = 180^o – 75^o

= 105^o. NPQR is cyclic quadrilateral.

(b) ∡ NRP = 90^o -75^o

= 15^o, Third angle of ∆ NRP.

 ∡PRS = 180^o -65^o, Angles on a

  = 115^o, straight line.

∴∡QSR = 180^o – (115^o – 35^o)

  = 30^o, 3^rd angle of triangle PRS.

(c) Reflex ∡POR = 2 ∡PQR

= 2 x 105^o = 210^o

(d) ∡MQR = ∡MNR = 40^o

Subtended by same chord MR

20.

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)
   

21.  4 x 10 = 5(5 + x)

 40 = 25 + 5x

 3 = x

22.  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

23.   CBE = 40⁰ ( alt.segiment theoren)

  BCE = 120⁰ (Suppl. To BCD = 60⁰alt. seg.)

  (40 + 120 + E) = 180⁰ (Angle sum of  )

 BEC = 20⁰

24.  Taxable income p.a = 36,000+53142.86

=sh.412142.86

 Monthly salary = 413142.86 + 12,000

12

= 34428.57 + 1200 = Sh 35628.57

25.  a) (i)  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 = 52^o

(ii)  RSQ =  RPQ = 18^o

b) A α B. 1

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

c) 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

26.  a) PN² = 5² – 4²

   PN= 3cm

QN² = 6² – 4²

QN = 4.47cm

 PQ = 3 + 4.47= 7.47

b)i) < APB

  Sin ½ 
⁴/₅ = 0.8

  ½ sin  = 53.13

 < APB

ii)  Sin ½  = 4/6 = 0.6667

½  = 41.81

  83.62

< AQB = 83.62°

c) 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

27.  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

  = 4cm

Area of circle = 3.142 x 4² = 50.27

Area of PQR = ½ (6.6) (5.9) sin 77

 = 18.97

Area of shaded region = 50.27 – 18.97 = 31.30cm²

28.  a)  PAQ = 2 PAB = 42^o x 2 = 84^o

   PBQ = 2  ABQ = 30⁰ x 2 = 60^o

(b) (i) 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

 (ii) Area of ∆APQ = ½ x 6 x 65:- 84^o = 18 x 0.9945

= 17.9 cm²

Area of ∆ PBQ = ½ x 8 x 85: = 60^o = 32 x 0.8660

= 27.7 cm²

  (iii) 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^nd 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

29.  90 x 3.142 x 2 x 6.5

 360

  10.2115 cm

  = 10.21 cm