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 = 30^o and angle XAY = 97.2^o. 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 =50^o and angle ACD = 20^o. 
   
   Calculate;
   1. ∠OAB
   2. ∠ADC
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
      3. Measure the radius
         
      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 80^o and BOD is a straight line.
   
   Giving reasons for your answer, find the size of :-
   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 = 80^o
   
   1. Calculate the area of:-
      1. The sector CAD.
      2. The triangle CAD. 
      3. The shaded region.
7. In the diagram below, ∠QOT is a diameter. ∠QTP = 48^o, ∠TQR = 46^o and ∠SRT = 37^o 
   
   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=40^o. 
   
   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
   1. ∠ADC
   2. ∠ AEB
10. The figure below shows circle ABCD. The line EDF is a tangent to the circle at D. ∠ADF = 70^o ∠FAD = 65^o and ∠CDE = 35^o
    
    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°, 
    
    Calculate the shaded area
    
    

Answers

1. Area of ∆AXY = ½ x 42 x sin 97.20
   = 7.94 cm²
   Area of sector AXY = ^97.2/₃₆₀xπx 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
   Area of shaded part
   = (9.42 – 9) cm²= 0.42 cm²
   Area of shaded region = (5.63 + 42) cm2 = 6.05 cm2
2.  
   1. ∠ AOB = 2∠ACB
      = 100^o
      ∠OAB = ^{(180 – 100)}/₂ Base angles of Isosceles ∆
      = 40⁰
   2. ∠ ADC = 180⁰ - 70⁰
      = 1100
3. 1. DAC =DCA = ½ (180 – 100) (base sios = 40^o
   2. BAC = DCA alt ,∠s AB//AD)
      = 40^o
   3. DAB = DAC + BAC = 40 + 40 = 80^o
      BCD = 180^o – 80^o
      = 100^o
4. c) (ii) Radius = 2.3 ±0.1cm
   Name of QPR : Escribed circle
   
   
5. 1. ∠ACB = 10^o (∠s subtended by chord AB)
   2. ∠AOD = 160^o (∠at centre line at circumference)
   3. ∠CAB = 40^o (∠s subtended by chord AB)
   4. ∠ABC = 130^o ( Opposite ∠s of cyclic quadrilateral)
   5. ∠AXB = 60^o(sum angle of triangle
6. 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
7.  
   1. ∠RST = 180^o – 46^o Opposite angel in cyclic quadrilateral
      = 134^o 
   2. ∠SUT = 180o – 46^o – 27^o (Sum of angles in a traingle QRU)
      = 180^o – 173^o = 7^o
   3. ∠ROT = 2 x 46^o (angle substended by chord RT at the centre
      = 92^o
   4. ∠PST = 180^o – 37^o – 48^o – 53^o
      Sum of angles in a triangle PST
   5. Reflex ∠SOP = (2x 37^o) + 2x 42^o) = 158^o
      Angle subtended chord at centres is twice angle at circle
8. ∠POQ = 80^o
   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  x 2.645²) 
   = 4.884cm²
   Area of the shaded segment = (4.884 – 3.445) 
   = 1.439cm²
9. 1. ∢BDC = 90^o -33^o, 3rd angle of ∆BCD
      = 57^o, ∡BCD = 90.
      ∡ADC = ∡ADB + ∡BDC
      = 48^o + 57^o = 105^o
   2. Consider ∆ BCE
      ∡ AEB is an exterior opposite angle
      ∴ ∡ AEB = 33^o + 48^o = 81^o