4MA0 Foundation January 2018 Paper 1R Q15

15 \(ABCD\) is a parallelogram.

Diagram NOT accurately drawn

Angle \(DCB = 124^\circ\)
\(E\) is the point on \(DC\) such that \(AE = DE\).
Angle \(AEC = x^\circ\)

Work out the value of \(x\).

(4)

Mark scheme

Scheme	Marks
\(\angle ADE = 180 - 124\;(= 56)\) or \(\angle ADE = \dfrac{360 - 2 \times 124}{2}\;(= 56)\)	M1
\(\angle DAE = \angle ADE = \text{‘}56\text{’}\)	M1
\(\angle AEC = 2 \times \text{‘}56\text{’}\)	M1
112	A1
	(4)

M1: for 2 × ‘56’ or for \(\angle AED\) = 180 – 2 × ‘56’ (=68) and \(\angle AEC\) = 180 – ‘68’