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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
Giving reasons calculate:
- ACD (2mks)
- ABO (2mks)
- ADO (2mks)
- ACB (2mks)
- 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
- BAD (1mk)
- BDC (1mk)
- 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)
(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)
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
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.
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.
ADF = 70o
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
(a) ∠ADC
(b) ∠ AEB
Angle properties of circles Answers
1. | (a) = 300 Diam. Subt 900 at circumference (b) ORP = 600 Base angle of isosceles triangle (c) OP to MPT Radius meets tangent at 900 = 900 – 300 = 600 (d) Angle sum of triangle = 300 (e) Angles in alternate segment |
B1 B1
B1 B1 B1
B1
B1 B1
B1 B1 | |
10 | |||
2. | (a) = 300 Diam. Subt 900 at circumference (b) ORP = 600 Base angle of isosceles triangle (c) OP to MPT Radius meets tangent at 900 = 900 – 300 = 600 (d) Angle sum of triangle = 300 (e) 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 9/2 – 11/10
= 9/5 – 11/10 = 18 -11/ 10 = 7/10
1/8 – 1/6 X 3/8 = 1/8 – 1/16
= 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
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 √