Angle Properties of Circles Questions

1. In the figure below PQR and S are points on the circumference of a circle centre O. The point TSO and Q lie on a straight line MPT is a tangent to the circle at P.

Image From EcoleBooks.com

 Find the values of the following angles stating reasons in each case.

 (a) SRP (2mks)

 (b) ORP (2mks)

 (c) RPT (2mks)

 (d) STP (2mks)

 (e) QPM (2mks)

2. In the figure below, TA is a tangent to the circle ABCD with centre O. Image From EcoleBooks.comTAD = 480 and Image From EcoleBooks.comBOD = 1160

ecolebooks.com

Giving reasons calculate:

  1. Image From EcoleBooks.comACD (2mks)
  2. Image From EcoleBooks.comABO (2mks)
  3. Image From EcoleBooks.comADO (2mks)
  4. Image From EcoleBooks.comACB (2mks)
  5. Image From EcoleBooks.comATB (2mks)

3. In the figure below AB = 8cm and O is the centre of the circle. Determine the area of the circle if OAB = 15o (3mks)

Image From EcoleBooks.com

4.

Image From EcoleBooks.com

 The figure above is a cyclic quadrilateral PQRS. Given that TPX is a tangent at P and O is the centre of the circle and that RQX is a straight line with RPQ = 50o and PRS = 25o, giving reason in each case find:

 (a) angle PRQ (2mks)

 (b) angle PSR (2mks)

 (c) angle PXQ (2mks)

 (d) angle TPS (2mks)

 (e) angle POS (2mks)

5. In the figure below ABCD is a circle with centre O. AB and DC meet at a point E outside the circle. DC = BC and Image From EcoleBooks.com

Find the angles

  1. BAD (1mk)
  2. BDC (1mk)
  3. BEC (1mk)

6. In the figure O and P are centres of intersecting circles ABD and DBC respectively. Line ABE is a tangent to circle BCD at B and angle BCD = 420.

Image From EcoleBooks.com

 Giving reasons determine the size of:

 (a) Angle CBD. (2mks)

 (b) Angle ODB. (2mks)

 (c) Angle BAD. (2mks)

 (d) Angle ABD (2mks)

 (e) Angle ODA. (2mks)

7. In the figure below, o is the centre of the circle. Express the angle w in terms of angles p and q. (2mks)

Image From EcoleBooks.com

8. Two circles of radii 4cm and 6cm intersect as shown below. If angle XBY = 30o and

angle XAY = 97.2o.

Find the area of the shaded part.

(Take =
22 )

7

9. In the diagram, O is the centre of the circle and AD is parallel to BC. If angle ACB =50o

and angle ACD = 20o.

Calculate; (i) OAB

(ii) ADC

10. Two intersecting circles have centres S and R. Given that their two radii are 28cm and 35cm,

their common chord AB = 38cm and angles ASB = 85.46° and ARB = 65.76°,

 Calculate the shaded area

11. In the figure below ABCD is a cyclic quadrilateral in which AD = DC and AB is parallel

to CD. Given that angle ABC = 80°, Find the size of:

a) DAC

b) BAC

c) BCD

12. Line QR = 6.5cm is given below:-(Do not use a protractor for this question)

 (a) Draw triangle PQR such that p lies above line QR,  PQR = 30o and PQ = 7cm

(b) By accurate construction on the diagram above, show the locus of a point which lies

within the triangle such that:-

(i) T is more than 2.5cm from line PQ

and

(ii) T is not more than 4.5cm from Q

Shade the region in which T lies


(c) Lines QP and QR are produced to K and M respectively

(i) Show by construction on the diagram above, the locus of a point C which is

equidistant from each of the lines PK, PR and RM

(ii) With centre C and an appropriate radius, draw a circle to touch each of the lines

PK, PR and RM only once

Measure the radius

What name is given to the circle drawn in (c) (ii) with respect to triangle QPR

13. The figure below shows a circle centre O and a cyclic quadrilateral ABCD. AC = CD, angle

Image From EcoleBooks.com ACD is 80o and BOD is a straight line. Giving reasons for your answer, find the size of :-

Image From EcoleBooks.com

Image From EcoleBooks.com

Image From EcoleBooks.com

Image From EcoleBooks.com

Image From EcoleBooks.comImage From EcoleBooks.com

Image From EcoleBooks.com (i) Angle ACB

(ii) Angle AOD

 (iii) Angle CAB

 (iv) Angle ABC

 (v) Angle AXB

14. The figure below shows two circles of equal radius of 9 cm with centres A and B.

Angle CAD = 80o

a) Calculate the area of:-

i) The sector CAD. ii) The triangle CAD. iii) The shaded region.

15. In the diagram below, QOT is a diameter. QTP = 48o, TQR = 46o and SRT = 37o

 Calculate, giving reasons in each case:-

 (a) RST

 (b) SUT

 (c) ROT

 (d) PST

 (e) Reflex SOP

16. The diagram below shows a circle with a chord PQ= 3.4cm and angle PRQ=40o.

Calculate the area of the shaded segment.

17. The figure below shows circle ABCD. The line EDF is a tangent to the circle at D.

Image From EcoleBooks.comImage From EcoleBooks.com ADF = 70o
FAD = 65o and CDE = 35o

Image From EcoleBooks.com

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com

Image From EcoleBooks.com

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com

Find the values of the following angles, stating your reasons in each case

(a) ABC

(b) BCD

(c) DCE

(d) ACD

Image From EcoleBooks.com18. In the figure below BD is the diameter of the circle and O is the centre.

Image From EcoleBooks.comImage From EcoleBooks.com

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com

 Find the size of

Image From EcoleBooks.com (a) ∠ADC

 (b) ∠ AEB

Image From EcoleBooks.com

Angle properties of circles Answers

1.

0 Base angles of isosc. Triangle

0

0

0 Base angles of Isoc triangle

0 diameter subtended right angle at the circumference

(a) 0 – 30 – 300

 = 300

0

Diam. Subt 900 at circumference

(b) ORP = 600 Base angle of isosceles triangle

(c) OP to MPT 0

Radius meets tangent at 900

0 –

 = 900 – 300

 = 600

(d) 0 – 0 – 0

 Angle sum of triangle

 = 300

(e) 0

Angles in alternate segment

B1

B1

B1 B1

B1

B1

B1

B1

B1

B1

10

2.

0 Base angles of isosc. Triangle

0

0

0 Base angles of Isoc triangle

0 diameter subtended right angle at the circumference

(a) 0 – 30 – 300

 = 300

0

Diam. Subt 900 at circumference

(b) ORP = 600 Base angle of isosceles triangle

(c) OP to MPT 0

Radius meets tangent at 900

0 –

 = 900 – 300

 = 600

(d) 0 – 0 – 0

 Angle sum of triangle

 = 300

(e) 0

Angles in alternate segment

B1

B1

B1 B1

B1

B1

B1

B1

B1

B1

10

1. Area of ∆AXY = ½ x 42 x sin 97.20

= 7.94 cm2

 Area of sector AXY = 97.2 x  x 42

360

= 13.57 cm2

Area of shaded part = 13.57 – 7.94 = 5.63 cm2

 Area of ∆ BXY = ½ x 62 sin 30

= 9 cm2

Area of sector BXY = 30 x  x 62

360

= 9.42 cm2

 Area of shaded part

= (9.42 – 9) cm2= 0.42 cm2

 Area of shaded region = (5.63 + 42) cm2 = 6.05 cm2

2. (i) ∠ AOB = 2 ∠ ACB

= 100o

∠ OAB = 180 – 100 Base angles of Isosceles ∆

2

 = 400

 (ii) ∠ ADC = 1800 – 700

= 1100

3. 2/5 ÷ ½ 0f 4/9 – 11/10

= 2/5 ÷ ½ X 4/9 11/10

= 2/5 x 9/211/10

 = 9/511/10 = 18 -11/ 10 = 7/10

1/8 1/6 X 3/8 = 1/81/16

= 2-1/16 = 1/16

2/5 ÷ ½ 0f 4/9 – 11/10 = 7/10

1/8 1/6 of 3/8 1/16

 = 7/10 X 16/1

= 56/5 = 111/5

4. a) DAC =DCA = ½ (180 – 100) (base sios = 40o

(b) BAC = DCA alt ,s AB//AD)

= 40o

(b) DAB = DAC + BAC = 40 + 40 = 80o

BCD = 180o – 80o

= 100o

Image From EcoleBooks.com

5. c) (ii) Radius = 2.3  0.1cm

Name of QPR : Escribed circle


6. (i) ACB = 10o (s subtended by chord AB)

(ii) AOD = 160o ( at centre line at circumference)

(iii) CAB = 40o (s subtended by chord AB)

(iv) ABC = 130o ( Opposite s of cyclic quadrilateral)

(v) AXB = 60o(sum angle of triangle


7. i) 80 x 22 x 9 x 9

360 7

 = 63.6429 cm2


 ii) ½ ab Sin C

= ½ x 9 x 9 Sin 800

 = 39.8847 cm2

iii) 180 x 22 x 9 x 9

360 7

= 127.2857 cm2

Segment: 63.6429 – 39.8847

 = 23.7582 x 2 = 47.5164 cm2

∴ 127.2857 – 47.5164

= 79.7693cm2 = 79.77 cm2

8. (a) RST = 180o – 46o Opposite angel in cyclic quadrilateral

= 134o

(b) SUT = 180o – 46o – 27o (Sum of angles in a traingle QRU)

= 180o – 173o = 7o

(c) ROT = 2 x 46o (angle substended by chord RT at the centre

= 92o

(d) PST = 180o – 37o – 48o – 53o

Sum of angles in a triangle PST

(e) Reflex SOP = (2x 37o) + 2x 42o) = 158o

Angle subtended chord at centres is twice angle at circle

9. POQ = 80o

 Radius = 1.7

 Sin 40 = 2.645cm

Area of the triangle = ½ x 2.6452 sin 80 = 3.445cm2

Area of the sector = ( 80 x  x 2.6452)

 360 = 4.884cm2

Area of the shaded segment = (4.884 – 3.445) = 1.439cm2

10. a) ∢BDC = 90o -33o, 3rd angle of

= 57o∆BCD, ∡BCD = 90.

∡ADC = ∡ADB + ∡BDC

= 48o + 57o = 105o

b) Consider ∆ BCE

∡ AEB is an exterior opposite angle


∡ AEB = 33o + 48o = 81o





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