Circle theorems – Year 11 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(ACB=118\div2=\) \(59^\circ\), because the angle at the centre is twice the angle at the circumference (1 each) Angle \(ACB\) stands on the same arc \(AB\) as the angle at the centre, so it is half of \(118^\circ\).
   2. \(ACB=90^\circ\) (angle in a semicircle), so \(ABC=180-90-37=\) \(53^\circ\) (1 each) The angles of triangle \(ABC\) add up to \(180^\circ\).

   In (b), the theorem name must be stated for the reason mark.

2. Answer to Question 2 (4 marks)

   1. \(4x+5+2x+25=180\) (1), \(6x=150\) (1), \(x=25\) (1) Angles \(A\) and \(C\) are opposite angles of a cyclic quadrilateral, so they add up to \(180^\circ\). Collect the terms: \(6x+30=180\), then subtract 30 and divide by 6.
   2. \(105^\circ\) (1) Substitute \(x=25\) into \(4x+5\): \(4\times25+5=105\). Check: angle \(C=2\times25+25=75^\circ\) and \(105+75=180\).

   Do not set the two angles equal: \(4x+5=2x+25\) gives \(x=10\), which is wrong because opposite angles of a cyclic quadrilateral are supplementary, not equal.

3. Answer to Question 3 (4 marks)

   1. Angle \(OAT=90^\circ\) (1), \(TA^2=17^2-8^2=225\) (1), \(TA=\) 15 cm (1) The tangent \(TA\) is perpendicular to the radius \(OA\), so triangle \(OAT\) has a right angle at \(A\) and \(OT=17\) cm is the hypotenuse. Subtract the squares, then take the square root.
   2. \(\sin(ATO)=\dfrac{8}{17}\), so \(ATO=\) \(28.1^\circ\) (1) \(OA=8\) cm is opposite angle \(ATO\) and \(OT=17\) cm is the hypotenuse, so we use sine and then the inverse sine key.

4. Answer to Question 4 (4 marks)

   1. The angle in the alternate segment is \(64^\circ\) (1) and the angle at the centre is twice this (1): \(AOB=\) \(128^\circ\) (1) By the alternate segment theorem, the angle between the tangent and the chord \(AB\) equals any angle in the opposite segment.
   2. \((180-128)\div2=\) \(26^\circ\) (1) Triangle \(OAB\) is isosceles because \(OA=OB\) are radii.

   Check (b): \(26^\circ+26^\circ+128^\circ=180^\circ\). Another route: angle \(OAT=90^\circ\) because the tangent is perpendicular to the radius, so \(OAB=90-64=26^\circ\).

5. Answer to Question 5 (4 marks)

   1. The half-chord is 15, the distance from the centre to the chord is \(r-5\) and Pythagoras gives the result (1)
   2. \(r^2=225+r^2-10r+25\) (1), \(10r=250\) (1), \(r=25\) m (1) The \(r^2\) terms cancel, which leaves a linear equation in \(r\).

   In (a), draw the radius to one end of the chord: the radius is the hypotenuse, half of the 30 m chord is one side, and the perpendicular from the centre to the chord, \(r-5\), is the other. Check (b): \(15^2+20^2=225+400=625=25^2\).

Total : 20 marks