Foundation June 2024 Paper 1R Q23
23 The diagram shows a trapezium, \(ABCD\)
Diagram NOT accurately drawn
\(DAB\) and \(ADC\) are right angles.
\(AD = 15\) cm \(\qquad DC = 14\) cm
The area of the trapezium is 360 cm²
Work out the perimeter of the trapezium.
(6)
| Scheme | Marks |
|---|---|
\(\dfrac{14 + AB}{2} \times 15 = 360\) oe or \(360 - 14 \times 15\ (= 150)\) oe | M1 |
| \(AB = 34\) or \(MB = 20\) (where \(M\) is point on \(AB\) such that \(MC\) is perpendicular to \(AB\)) | A1 |
| \((CB^2 =)\ 15^2 + 20^2\ (= 625)\) or \((CB^2 =)\ 15^2 + MB^2\) | M1 |
\((CB =)\ \sqrt{15^2 + 20^2}\ (= 25)\) or \((CB =)\ \sqrt{15^2 + MB^2}\) | M1 |
| 14 + 15 + “34” + “25” oe or 14 + 15 + 14 + \(MB\) + \(CB\) oe | M1ft |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 88 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for setting up an equation using the area of the trapezium or method to find the area of the triangle
A1: could be seen on diagram
M1: allow use of their \(MB\)
M1: allow use of their \(MB\)
M1ft: (dep on previous two M marks) for a method to find the perimeter of the trapezium, allow use of their \(MB\) and \(CB\)
A1: cao