Foundation June 2019 Paper 3 Q28
28 The diagram shows a hexagon.
The hexagon has one line of symmetry.
\(FA = BC\)
\(EF = CD\)
Angle \(ABC = 117^\circ\)
Angle \(BCD = 2 \times\) angle \(CDE\)
Work out the size of angle \(AFE\).
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 162 supported | M1 | for method to find sum of the interior angles of a hexagon eg \((6 - 2) \times 180\ (= 720)\) oe OR for method to find sum of the interior angles of a pentagon, eg \((5 - 2) \times 180\ (= 540)\) OR for method to find angle \(AFC\) or \(BCF\), eg \((360 - 2 \times 117) \div 2\ (= 63)\) OR for dropping a perpendicular from \(A\) or \(B\) to \(ED\) with 90° marked on \(ED\) and 27° at the top |
| M1 | for method to use ratio 2 : 1 eg marks as \(2x\) and \(x\) or as \(x\) and \(\frac{1}{2}x\) on diagram OR for ([angle sum of hexagon] \({} - 2 \times 117) \div 6\ (= 81)\) oe or ([angle sum of hexagon] \({} \div 2 - 117) \div 3\ (= 81)\) oe or \(117 + 117 + 2x + 2x + x + x =\) [angle sum of hexagon] oe OR eg ([angle sum of pentagon] \({} - 117 - 180) \div 3\ (= 81)\) oe or \(117 + 180 + 2x + x =\) [angle sum of pentagon] oe | |
| M1 | for finding angle \(FED = 81\) or for finding angle \(CDE = 81\) OR for complete process to find angle \(AFE\) eg ([angle sum of hexagon] \({} - 2 \times 117) \div 6 \times 2\) oe OR ([angle sum of pentagon] \({} - 117 - 180) \div 3 \times 2\) oe | |
| C1 | for accurate working leading to angle \(AFE = 162\) |
Additional guidance
For the first M1: Must be a complete process that would lead to a figure of 720 if evaluated correctly.
For a pentagon there must be an indication that they have divided the hexagon into two halves.
63 may be shown on the diagram for angle \(AFC\) or angle \(BCF\)
For the second M1: Ratio must be used correctly if awarded for diagram
Award provided [angle sum of hexagon] is greater than 700 or [angle sum of pentagon] is greater than 500
Algebraic route needs to show both sides of the equation.
LHS of equation may be simplified.
For the third M1: This may be shown by solving a correct equation to find the value of \(x\).
For the C1: Award marks for 162 on the diagram with working and not contradicted by the answer line.
Award 0 marks for 162 without working.