Circle Theorems: Eduqas GCSE Higher June 2017 Paper 1 Q15

Higher June 2017 Paper 1 Q15

EduqasHigherCurrent spec9 marksCircle Theorems Similar Shapes

15.

(a) PQRS is a cyclic quadrilateral of a circle, centre O.
\(S\hat{P}Q = 54^\circ\).
The ratio of \(O\hat{S}R\) to \(O\hat{Q}R\) is 2 : 1.

Diagram not drawn to scale

(i) Find \(S\hat{O}Q\).
Give a reason for your answer. [2]

(ii) Find \(S\hat{R}Q\). [1]

(iii) Calculate the size of \(O\hat{S}R\). [3]

(b) The diagram shows points A, B, C and D on the circumference of a circle.
AC and BD intersect at the point E.

Diagram not drawn to scale

Show clearly that triangle AEB is similar to triangle DEC.
Give a reason for each step of your answer. [3]

Mark scheme (a)

Scheme	Marks	Comments
(a)(i) \(S\hat{O}Q = 108^\circ\)	B1
Angle at centre (is twice angle at circumference)	E1	Do not allow eg ‘Angle SOQ = 2Angle SPQ’
(a)(ii) \(S\hat{R}Q = 126^\circ\)	B1	Ignore any reason if stated
(a)(iii) \((O\hat{Q}R =)\ \dfrac{360 - (108 + 126)}{3}\)	M1	or equivalent, FT ‘their 108’ and ‘their 126’ May be seen or implied in further work; may be in steps
\((O\hat{S}R =)\ \dfrac{360 - (108 + 126)}{3} \times 2\)	M1	FT ‘their 108’ and ‘their 126’
\(84^\circ\)	A1	CAO

Mark scheme (b)

Scheme	Marks	Comments
Any two of \(C\hat{D}B = C\hat{A}B\) (same segment) or equivalent \(A\hat{C}D = A\hat{B}D\) (same segment) or equivalent \(D\hat{E}C = A\hat{E}B\) (vertically opposite)	B2	B1 for \(C\hat{D}B = C\hat{A}B\) same segment or equivalent or for \(A\hat{C}D = A\hat{B}D\) same segment or equivalent or for two pairs of angles eg \(C\hat{D}B = C\hat{A}B\), \(A\hat{C}D = A\hat{B}D\) with no reason/invalid reasons stated
Correct explanation e.g. ‘The triangles have the same angles (and so are similar).’	E1	May be stated or implied earlier.

Scheme	Marks	Comments
\(\mathrm{AE} \times \mathrm{EC} = \mathrm{BE} \times \mathrm{ED}\) or states ‘intersecting chords theorem’	B1
\(\dfrac{\mathrm{AE}}{\mathrm{DE}} = \dfrac{\mathrm{BE}}{\mathrm{CE}}\) or equivalent ratios	B1
Correct explanation e.g. ‘Corresponding sides are in the same ratio so the triangles are similar.’ or eg extending the intersecting chords theorem \(\dfrac{\mathrm{AE}}{\mathrm{DE}} = \dfrac{\mathrm{BE}}{\mathrm{CE}} = \dfrac{\mathrm{AB}}{\mathrm{DC}}\) (so the triangles are similar)	E1

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