Foundation June 2023 Paper 1 Q20
20 The diagram shows a shaded shape \(AEBCD\) made by removing triangle \(AEB\) from rectangle \(ABCD\)
Diagram NOT accurately drawn
\(AE\) = 7.2 cm \(BE\) = 5.4 cm \(BC\) = 6 cm angle \(AEB\) = 90°
Work out the perimeter of the shaded shape.
(4)
| Scheme | Marks |
|---|---|
| \(7.2^2 + 5.4^2\) (= 81) | M1 |
| \(\sqrt{7.2^2 + 5.4^2}\) (= 9) | M1 |
| \(7.2 + 5.4 + 6 + \text{``}{9}\text{''} + 6\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 33.6 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for correct first step using Pythagoras
M1: for complete Pythagoras method to find length of \(AB\)/\(DC\) check the diagram for sight of 9, \(DC\) marked as 9 implies M2
M1: for a complete method to find the perimeter
A1: oe
If using trig:
M1 for reaching one step from the length of \(AB\) eg \((EAB =) \tan^{-1}\left(\dfrac{5.4}{7.2}\right)\) (= 36.8…) and \(\sin(\text{``}{36.8...}\text{''}) = \dfrac{5.4}{AB}\)
M1 for complete method to find the length of \(AB\)/\(DC\) eg \(\dfrac{5.4}{\sin(\text{``}{36.8...}\text{''})}\) (= 9)