Year 11 circle theorems worksheet answers to download as PDF

These Year 11 circle theorems worksheet answers give a full worked solution to all 20 questions of the circle theorems worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. \(AOB=2\times38=\) \(76^\circ\). Reason: the angle at the centre is twice the angle at the circumference.
2. \(ACB=130\div2=\) \(65^\circ\). Reason: the angle at the circumference is half the angle at the centre.
3. \(3x-20=2x\), so \(x=20\). Check: \(AOB=40^\circ\) and \(ACB=20^\circ\), and \(40=2\times20\). The angle at the centre is twice the angle at the circumference, so \(AOB=2\times ACB\).
Answer to Question 2
1. \(ACB=\) \(90^\circ\). Reason: the angle in a semicircle is a right angle.
2. \(ABC=180-90-34=\) \(56^\circ\) (angles in a triangle add to \(180^\circ\)).
3. \(MA=MC\) (radii), so triangle \(AMC\) is isosceles with base angles \(34^\circ\). Angle \(AMC=180-34-34=\) \(112^\circ\). (Check: this is \(2\times56^\circ\), twice angle \(ABC\).)
Answer to Question 3
1. \(ACD=\) \(42^\circ\). Reason: angles in the same segment are equal (both stand on chord \(AD\)). Name the chord each angle stands on: angle \(ABD\) and angle \(ACD\) both stand on \(AD\), so they are equal.
2. \(CBD=\) \(35^\circ\) (both \(CBD\) and \(CAD\) stand on chord \(CD\)). Angles subtended by the same arc at the circumference are equal.
3. \(BDC=\) \(28^\circ\) (both \(BDC\) and \(BAC\) stand on chord \(BC\)). So the three missing angles are \(42^\circ\), \(35^\circ\) and \(28^\circ\).
Answer to Question 4
1. Opposite angles of a cyclic quadrilateral add to \(180^\circ\): \(C=180-108=\) \(72^\circ\). Angles \(A\) and \(C\) are opposite each other in the cyclic quadrilateral.
2. \(D=180-85=\) \(95^\circ\). Angles \(B\) and \(D\) are also opposite, so they add up to \(180^\circ\).
3. \(2x+10+3x-20=180\), so \(5x-10=180\) and \(5x=190\), \(x=38\). \(B=2\times38+10=\) \(86^\circ\) and \(D=3\times38-20=\) \(94^\circ\). Check: \(86+94=180\). Opposite angles \(B\) and \(D\) add up to \(180^\circ\), so we add the two expressions and set the total equal to 180.
Answer to Question 5
1. The tangent meets the radius at \(90^\circ\), so angle \(OAT=90^\circ\). Then \(ATO=180-90-62=\) \(28^\circ\).
2. Tangents from a point are equal, so \(TA=TB\) and triangle \(TAB\) is isosceles: \(TAB=(180-50)\div2=\) \(65^\circ\).
3. \(OAB=90-65=\) \(25^\circ\). In the kite \(OATB\): \(AOB=360-90-90-50=\) \(130^\circ\). The angles of a quadrilateral add up to \(360^\circ\).
Answer to Question 6
1. \(ACB=\) \(62^\circ\). Reason: the angle between a tangent and a chord equals the angle in the alternate segment.
2. \(ABC=\) \(48^\circ\) (alternate segment theorem with angle \(QAC\)).
3. \(BAC=180-62-48=\) \(70^\circ\) (angles on a straight line at \(A\), or angles in triangle \(ABC\)).
Answer to Question 7
1. \(ACB=112\div2=\) \(56^\circ\) (angle at the centre is twice the angle at the circumference).
2. \(D\) is on the opposite arc, so \(ACBD\) is a cyclic quadrilateral. \(ADB=180-56=\) \(124^\circ\) (opposite angles add to \(180^\circ\)).
3. Angles in the same segment are equal: every point on the arc \(ACB\) is in the same segment and stands on the chord \(AB\), so each sees \(56^\circ\).
Answer to Question 8
1. \(AB\) is a diameter, so the angle in a semicircle is \(90^\circ\). So the cut and the two halves of the diameter make a right-angled triangle with the right angle at \(C\).
2. Pythagoras: \(BC^2=30^2-18^2=900-324=576\), so \(BC=\) 24 cm. The diameter \(AB=30\) cm is the hypotenuse because it is opposite the right angle.
3. Area \(=\tfrac12\times18\times24=\) 216 cm². The two shorter sides are perpendicular, so they are the base and the height of the triangle.
4. \(360\div8=\) \(45^\circ\). The 8 equal slices share the full turn of \(360^\circ\) at the centre.
Answer to Question 9
1. The radius \(OA\) meets the tangent \(TA\) at \(90^\circ\). A tangent is always perpendicular to the radius at the point where it touches the circle.
2. \(TA^2=10^2-6^2=100-36=64\), so \(TA=\) 8 m. \(OT=10\) m is the hypotenuse, because it is opposite the right angle at \(A\).
3. \(TB=\) 8 m: tangents drawn to a circle from the same point are equal in length.
4. \(\sin(ATO)=\dfrac{6}{10}=0.6\), so \(ATO=\) \(36.9^\circ\). In the right-angled triangle, \(OA\) is opposite angle \(ATO\) and \(OT\) is the hypotenuse, so we use sine.
Answer to Question 10
1. The line from the centre of a circle to the midpoint of a chord is perpendicular to the chord (equivalently, \(OA=OB\) makes triangle \(OAB\) isosceles and \(OM\) is its line of symmetry).
2. \(AM=8\) and \(OA=10\). \(OM^2=100-64=36\), so \(OM=\) 6 m.
3. \(\tfrac12\times16\times6=\) 48 m².
4. \(\sin(AOM)=\dfrac{8}{10}\), so \(AOM=53.13^\circ\) and \(AOB=2\times53.13=\) \(106.3^\circ\).
Answer to Question 11
1. \(O\) is \(r-4\) metres below the chord (the top of the arc is \(r\) above \(O\), and the chord is 4 m below the top).
2. The half-chord is 12 m, the distance from \(O\) to the chord is \(r-4\) and the radius is \(r\), so \(r^2=12^2+(r-4)^2\).
3. \(r^2=144+r^2-8r+16\), so \(8r=160\) and \(r=20\) m.
4. \(20-4=\) 16 m below the water.
Answer to Question 12
1. \(ABCD\) is a cyclic quadrilateral and opposite angles add to \(180^\circ\).
2. \(6x+30=180\), so \(6x=150\) and \(x=25\). Collect the \(x\) terms and the numbers, then divide by 6.
3. \(DAB=4\times25+5=\) \(105^\circ\) and \(BCD=2\times25+25=\) \(75^\circ\). Check: \(105+75=180\).
4. \(ADC=180-100=\) \(80^\circ\), since \(B\) and \(D\) are opposite angles of a cyclic quadrilateral.
Answer to Question 13
1. The ship's path is a tangent and the radius to the point of contact is perpendicular to it.
2. \(SP^2=13^2-5^2=144\), so \(SP=\) 12 km. The distance \(OS=13\) km is the hypotenuse, so we subtract the squares.
3. \(\sin(OSP)=\dfrac{5}{13}\), so \(OSP=\) \(22.6^\circ\). \(OP\) is opposite angle \(OSP\) and \(OS\) is the hypotenuse, so we use sine.
4. \(\dfrac{12}{24}=0.5\) hours \(=\) 30 minutes. Time \(=\) distance \(\div\) speed, then multiply by 60 to change hours into minutes.
Answer to Question 14
1. \(AOB=2\times40=\) \(80^\circ\) (angle at the centre is twice the angle at the circumference).
2. \(40^\circ\): angles in the same segment are equal.
3. \(ARB=180-40=\) \(140^\circ\) (opposite angles of a cyclic quadrilateral).
4. Triangle \(OAB\) is isosceles with apex angle \(80^\circ\). The perpendicular from \(O\) bisects \(AB\): half-width \(3.66\) m and half-angle \(40^\circ\). \(r=\dfrac{3.66}{\sin40^\circ}=\) 5.69 m.
Answer to Question 15
1. Triangle \(OAB\) is isosceles (\(OA=OB\)), so \(OAB=(180-100)\div2=\) \(40^\circ\).
2. The tangent is perpendicular to the radius, so \(OAT=90^\circ\) and \(TAB=90-40=\) \(50^\circ\).
3. Quadrilateral \(OATB\): \(ATB=360-90-90-100=\) \(80^\circ\).
4. Tangents from a point are equal: \(TA=TB\). \(OATB\) has two pairs of equal adjacent sides (\(OA=OB\) and \(TA=TB\)), so it is a kite.
Answer to Question 16
1. \(ACB=\) \(64^\circ\) (alternate segment theorem).
2. \(AOB=2\times64=\) \(128^\circ\) (angle at the centre is twice the angle at the circumference).
3. \(OA=OB\) so triangle \(OAB\) is isosceles: \(OAB=(180-128)\div2=\) \(26^\circ\). The tangent-radius angle is \(TAB+OAB=64+26=90^\circ\), as required.
Answer to Question 17
1. The angle at the centre is twice the angle at the circumference, so the angle \(AOC\) standing on the arc \(ADC\) is \(2x\).
2. Angles around a point add to \(360^\circ\), so the other angle \(AOC\) (on the arc \(ABC\)) is \(360-2x\).
3. \(ADC\) stands on the arc \(ABC\), so \(ADC=\tfrac12(360-2x)=180-x\). Then \(ABC+ADC=x+(180-x)=180^\circ\). Opposite angles of a cyclic quadrilateral add to \(180^\circ\).
Answer to Question 18
1. \(ACB=\) \(90^\circ\) (angle in a semicircle).
2. \(ABC=180-90-28=\) \(62^\circ\).
3. \(ADB=90^\circ\) (angle in a semicircle), so \(BAD=180-90-35=\) \(55^\circ\).
4. \(CDB=CAB=28^\circ\) (angles in the same segment on chord \(CB\)). Then \(ADC=ADB-CDB=90-28=\) \(62^\circ\). This agrees with \(ABC=62^\circ\), because \(ADC\) and \(ABC\) are angles in the same segment on chord \(AC\).
Answer to Question 19
1. \(AB\) is a tangent to the smaller circle at \(M\), so \(OM\perp AB\). The perpendicular from the centre to a chord bisects the chord, so \(AM=MB\).
2. \(AM^2=13^2-5^2=144\), so \(AM=12\) and \(AB=\) 24 cm. In the right-angled triangle \(OMA\), the hypotenuse is the radius of the larger circle and \(OM\) is the radius of the smaller one.
3. Area \(=\pi\times13^2-\pi\times5^2=169\pi-25\pi=\) \(144\pi\) cm² (which is \(\pi\times AM^2\)).
Answer to Question 20
1. Angle \(ACB=90^\circ\) (semicircle), so \(AB^2=10^2+24^2=676\), \(AB=26\) and the radius is 13 cm.
2. Area of \(ABC=\tfrac12\times10\times24=120\). Also area \(=\tfrac12\times26\times h\), so \(13h=120\) and \(h=\dfrac{120}{13}=\) 9.23 cm.
3. The height of any triangle on base \(AB\) is at most the radius (13 cm), reached when \(C\) is at the top of the circle directly above \(O\). So the area is at most \(\tfrac12\times26\times13=169\) cm², i.e. \(13\times13\).

How to use these Year 11 circle theorems worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the circle theorems lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 11 circle theorems worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 11 circle theorems worksheet answers show every step, so nothing is left unexplained.