Angles (Including Parallel Lines): Eduqas GCSE Foundation June 2025 Paper 2 Q16

Foundation June 2025 Paper 2 Q16

EduqasFoundationCurrent spec4 marksAngles (Including Parallel Lines)

16. \(PQRS\) is a parallelogram.
\(U\) is a point on the line \(PS\).
\(T\) is a point on the line \(PR\).

Diagram: parallelogram PQRS with diagonal PR; T on PR and U on PS joined by UT; angle QPR is 48 degrees, angle PTU is 53 degrees, angle SUT is 112 degrees and angle PSR is x

Diagram not drawn to scale

Show that \(x = 73^\circ\).
Give a reason for each step of your answer. [4]

Mark scheme

Scheme	Marks	Comments
\(P\hat{R}S\) = 48(°) (alternate angles are equal)	B1	Angles may be seen on the diagram. Must be correctly placed on the diagram or appropriately labelled
\(U\hat{T}R\) = 127(°) (angles on a straight line = 180°)	B1	Must be correctly placed on the diagram, appropriately labelled or unambiguously shown on the working lines
(\(x\) = ) 360 – 127 – 112 − 48 = 73(°) (angles in a quadrilateral = 360°)	B1
At least two of the reasons given above but must include alternate angles.	E1	Award E1 provided at least B1 B1 awarded
	(4)

Alternative

Alternative final two marks (assuming \(x = 73^\circ\))

Scheme	Marks	Comments
(\(Q\hat{P}R = P\hat{R}S =\) ) 360 – 127 – 112 − 73 = 48(°) (angles in a quadrilateral = 360°)	B1
At least two of the reasons given above but must include alternate angles.	E1	Award E1 provided at least B1 B1 awarded

Alternative method

Scheme	Marks	Comments
\(P\hat{U}T\) = 68(°) (angles on a straight line = 180°)	B1	Angles may be seen on the diagram. Must be correctly placed on the diagram, appropriately labelled or unambiguously shown on the working lines
\(U\hat{P}T\) = 59(°) (angles in a triangle = 180°)	B1	FT ‘their 68(°)’ Must be correctly placed on the diagram, appropriately labelled or unambiguously shown on the working lines
(\(x\) = 180 - \(U\hat{P}T\) - \(R\hat{P}Q\) = 73(°)) (\(x\) = ) 180 – 59 – 48 = 73(°) (allied/co-interior angles sum to 180°)	B1
At least two of the reasons given above but must include allied/co-interior angles	E1	Award E1 provided at least B1 B1 awarded

Alternative final two marks (using properties of a parallelogram)

Scheme	Marks	Comments
\(x\) = (360 – 2 × (59 + 48)) ÷ 2 = 73(°) (opposite angles in a parallelogram are equal and sum to 360°)	B1
At least two of the reasons given above but must include opposite angles in a parallelogram are equal and sum to 360°	E1	Award E1 provided at least B1 B1 awarded Allow pairs of angles in a parallelogram are equal

Note to markers
Allow other full and complete methods to prove that \(x\) = 73(°) with appropriate reasons e.g.
\(P\hat{R}S\) = 48(°) (alternate angles)
and \(P\hat{R}S + R\hat{P}S + P\hat{S}R\) = 48 + 59 + 73 = 180(°)
(angles in a triangle = 180°)