# Angle Properties of Circles Questions and Answers - Form 2 Topical Mathematics

## Questions

1. 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 = π22/7 )

2. 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;
1. ∠OAB
3. 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:
1. ∠DAC
2. ∠BAC
3. ∠BCD
4. Line QR = 6.5cm is given below:-(Do not use a protractor for this question)
1. Draw triangle PQR such that p lies above line QR, ∠PQR = 30o and PQ = 7cm

2. By accurate construction on the diagram above, show the locus of a point which lies within the triangle such that:-
1. T is more than 2.5cm from line PQ and
2. T is not more than 4.5cm from Q
3. Shade the region in which T lies
3. Lines QP and QR are produced to K and M respectively
1. 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
2. With centre C and an appropriate radius, draw a circle to touch each of the lines PK, PR and RM only once
4. What name is given to the circle drawn in (c) (ii) with respect to triangle QPR
5. 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.

1. Angle ACB
2. Angle AOD
3. Angle CAB
4. Angle ABC
5. Angle AXB
6. The figure below shows two circles of equal radius of 9 cm with centres A and B. Angle CAD = 80o
1. Calculate the area of:-
7. In the diagram below, ∠QOT is a diameter. ∠QTP = 48o, ∠TQR = 46o and ∠SRT = 37o

Calculate, giving reasons in each case:-
1. ∠RST
2. ∠SUT
3. ∠ROT
4. ∠PST
5. Reflex ∠SOP
8. The diagram below shows a circle with a chord PQ= 3.4cm and angle PRQ=40o

Calculate the area of the shaded segment
9. In the figure below BD is the diameter of the circle and O is the centre.

Find the size of
2. AEB
10. The figure below shows circle ABCD. The line EDF is a tangent to the circle at D. ∠ADF = 70o ∠FAD = 65and ∠CDE = 35o

Find the values of the following angles, stating your reasons in each case
1. ∠ABC
2. ∠BCD
3. ∠DCE
4. ∠ACD
11. 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°,

1. Area of AXY = ½ x 42 x sin 97.20
= 7.94 cm2
Area of sector AXY = 97.2/360x 42
= 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 = 30/360x πx 62
= 9.42 cm2
= (9.42 – 9) cm2= 0.42 cm2
Area of shaded region = (5.63 + 42) cm2 = 6.05 cm2
2.
1. AOB = 2ACB
= 100o
OAB = (180 – 100)/2 Base angles of Isosceles
= 400
2. ADC = 1800 - 700
= 1100
1. DAC =DCA = ½ (180 – 100) (base sios = 40o
2. BAC = DCA alt ,∠s AB//AD)
= 40o
3. DAB = DAC + BAC = 40 + 40 = 80o
BCD = 180o – 80o
= 100o
3. c) (ii) Radius = 2.3 ±0.1cm
Name of QPR : Escribed circle

1. ∠ACB = 10o (∠s subtended by chord AB)
2. ∠AOD = 160o (∠at centre line at circumference)
3. ∠CAB = 40o (∠s subtended by chord AB)
4. ∠ABC = 130o ( Opposite ∠s of cyclic quadrilateral)
5. ∠AXB = 60o(sum angle of triangle
1. 80/360 x 22/7 x 9 x 9
= 63.6429 cm2
2. ½ ab Sin C
= ½ x 9 x 9 Sin 800
= 39.8847 cm2
3. 180/360 x 22/7 x 9 x 9
= 127.2857 cm2
Segment: 63.6429 – 39.8847
= 23.7582 x 2 = 47.5164 cm2
127.2857 – 47.5164
= 79.7693cm2 = 79.77 cm2
4.
1. ∠RST = 180o – 46o Opposite angel in cyclic quadrilateral
= 134o
2. SUT = 180o – 46o – 27o (Sum of angles in a traingle QRU)
= 180o – 173o = 7o
3. ∠ROT = 2 x 46o (angle substended by chord RT at the centre
= 92o
4. ∠PST = 180o – 37o – 48o – 53o
Sum of angles in a triangle PST
5. Reflex ∠SOP = (2x 37o) + 2x 42o) = 158o
Angle subtended chord at centres is twice angle at circle
5. POQ = 80o
Radius = 1.7/Sin 40 = 2.645cm
Area of the triangle = ½ x 2.6452sin 80 = 3.445cm2
Area of the sector = (80/360 x 2.6452
= 4.884cm2
Area of the shaded segment = (4.884 – 3.445)
= 1.439cm2
1. BDC = 90o -33o, 3rd angle of BCD
= 57o, BCD = 90.
= 48o + 57o = 105o
2. Consider BCE
AEB is an exterior opposite angle
AEB = 33o + 48o = 81o

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