Angles in Polygons: Eduqas GCSE Higher June 2019 Paper 2 Q12

Higher June 2019 Paper 2 Q12

12. Ben draws an irregular pentagon.
The interior angles of the pentagon he has drawn are all less than 180°.

Ben attempts to express the interior angles of his pentagon using algebra.
His expressions are

\[x^\circ,\quad (x + 40)^\circ,\quad (2x - 30)^\circ,\quad 3(x - 40)^\circ\quad \text{and}\quad 3x^\circ.\]

Show that Ben is incorrect. [6]

Mark scheme

Scheme	Marks	Comments
\(x + x + 40 + 2x - 30 + 3x - 120 + 3x\)	B1	\(3(x - 40) = 3x - 120\) may be seen in later working
(Interior angle sum) \(3 \times 180(^\circ)\) or alternative FULL method	M1
\(540(^\circ)\)	A1
\(10x - 110 = 540\) or \(10x = 540 + 110\) or \(10x = 650\)	M1	FT ‘their \(10x - 110\)’ \(= n\) where \(n \geqslant 360\)
\(x = 65(^\circ)\)	A1	CAO
\((65(^\circ), 105(^\circ), 100(^\circ), 75(^\circ),)\ \mathbf{195}(^\circ)\) (so this angle is greater than 180°)	E1	FT provided similar outcome
	(6)