Higher June 2025 Paper 4 Q7

OCRHigherCurrent spec6 marksAngles in Polygons

7 The diagram shows a regular pentagon and part of another regular polygon meeting at point P.
One of the angles at point P is 92°.

Not to scale

(a) Show that the size of one interior angle of a regular pentagon is 108°. [2]

(b) Find the number of sides of the other regular polygon. [4]

Mark scheme (a)

Answer	Marks	Part marks and guidance
\(360 \div 5\ [= 72]\) oe \(180 - 72\ [= 108]\)	M1 M1		Accept any correct method but they must not start with 108 e.g \(108 \times 5 = 540\) scores M0
OR	OR
\((5 - 2) \times 180\) [= 540] oe	M1	allow to split into 3 triangles or into a quadrilateral and a triangle so 360 + 180 [= 540]	condone missing brackets
\(540 \div 5\) [= 108]	M1	condone \(540 \div 108 = 5\) with 5 is the number of sides [of a pentagon]	Note : \(\frac{(5 - 2) \times 180}{5}\) scores M1 M1

Mark scheme (b)

Answer	Marks	Part marks and guidance
18	4	M1 for \(360 - 108 - 92\) or 160 oe	Angles may be on diagram
18	4	M2 for \(\dfrac{360}{180 - \textit{their}\,160}\) oe or M1 for \(180 -\) their160 may be implied by 20	Alternative method 1: M1 for 160 as before M2 for \(180n -\) their\(160n = 360\) or M1 for \(\dfrac{(n - 2)180}{n} =\) their160 oe Alternative method 2: M2 for \(92 - 72\) or 20 M1dep for \(\dfrac{360}{\textit{their}\,20}\) dep. on M2 Trials: M1 for \(360 - 108 - 92\) or 160 M1 for each different value for the number of sides and the correct internal angle calculated to a maximum of M2 see appendix for values

Formula = \(180 - \frac{360}{n}\) where \(n\) is the number of sides