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.

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. TAD = 480 and BOD = 1160

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Giving reasons calculate:

1. ACD  (2mks)
2. ABO  (2mks)
4. ACB  (2mks)
5. ATB  (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)

4.

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

Find the angles

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.

Giving reasons determine the size of:

(a) Angle CBD.  (2mks)

(b) Angle ODB.  (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)

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

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°,

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

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

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

(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.

a) Calculate the area of:-

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.

FAD = 65o and CDE = 35o

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

(a) ABC

(b) BCD

(c) DCE

(d) ACD

18.  In the figure below BD is the diameter of the circle and O is the centre.

Find the size of

(b) ∠ AEB

 1. 0 Base angles of isosc. Triangle0 00 Base angles of Isoc triangle0 diameter subtended right angle at the circumference(a) 0 – 30 – 300  = 3000Diam. Subt 900 at circumference(b) ORP = 600 Base angle of isosceles triangle(c) OP to MPT 0Radius meets tangent at 9000 –  = 900 – 300  = 600(d) 0 – 0 – 0  Angle sum of triangle  = 300(e) 0Angles in alternate segment B1B1 B1 B1B1  B1 B1B1 B1B1 10 2. 0 Base angles of isosc. Triangle0 00 Base angles of Isoc triangle0 diameter subtended right angle at the circumference(a) 0 – 30 – 300  = 3000Diam. Subt 900 at circumference(b) ORP = 600 Base angle of isosceles triangle(c) OP to MPT 0Radius meets tangent at 9000 –  = 900 – 300  = 600(d) 0 – 0 – 0  Angle sum of triangle  = 300(e) 0Angles in alternate segment B1B1 B1 B1B1  B1 B1B1 B1B1 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

= (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

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

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.

= 48o + 57o = 105o

b) Consider ∆ BCE

∡ AEB is an exterior opposite angle

∡ AEB = 33o + 48o = 81o

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