Foundation June 2024 Paper 2 Q23
23 The diagram shows a 6-sided shape \(ABCDEF\)
Diagram NOT accurately drawn
\(AF = 21\) cm \(\qquad CD = 15\) cm \(\qquad AB = FE = 13\) cm
The perpendicular height of the shape is \(h\) cm
\(CD\) is parallel to \(AF\)
The area of the shape is 390 cm²
Work out the value of \(h\)
(4)
| Scheme | Marks |
|---|---|
| eg 13 × 21 (=273) or 21 × \(h\) (= 21\(h\)) or 0.5(15 + 21) × \(y\) oe or 15(\(h\) – 13) or 2 ×½(3(\(h\) – 13)) or ½(13 + \(h\)) ×3 (= 19.5 + 1.5\(h\)) or 15 × \(h\) (= 15\(h\)) | M1 |
| eg 390 – “273” (= 117) or 13 × 21 and 0.5(15 + 21)(\(h\) – 13) oe or 13 × 21 and 0.5(15 + 21)\(y\) oe or 21\(h\) and 2 ×½(3(\(h\) – 13)) oe or 13 × 21 and 15(\(h\) – 13) and 2 ×½(3(\(h\) – 13)) oe or 2 × ½(13 + \(h\)) ×3 and 15 × \(h\) | M1 |
| “117” ÷ (0.5 × (15 + 21))(= 6.5) or \(\tfrac{1}{2}(15 + 21) \times y = \text{``}{117}\text{''}\) or 273 + 18(\(h\) – 13) = 390 or 15(\(h\) – 13) + 2 ×½(3(\(h\) – 13)) =”117” oe or 2 × ½(13 + \(h\)) ×3 + 15\(h\) = 390 Typical equations here simplify to : 18\(y\) = 117, 18\(h\) – 234 = 117, 18\(h\) + 39 = 390, 18\(h\) = 351 | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 19.5 | A1oe |
| (4) | |
| (4 marks) |
Notes
M1: A correct calculation for an area linked to the shape. (\(h\) – 13) might be written as \(x\) or \(y\) etc: this is acceptable (even allow \(h\))
[allow without brackets for this mark only]
M1: For considering the area of all parts of the shape (parts need not be added or subtracted for the whole shape)
(where \(y\) = height of \(BCDE\) )
(\(h\) – 13) might be written as \(x\) or \(y\) etc: this is acceptable (even allow \(h\))
[correct use of brackets]
M1: A correct calculation to find height of trapezium or height of shape or a correct equation involving height of trapezium or height of shape
or
6.5
(\(h\) – 13) might be written as \(x\) or \(y\) etc: this is acceptable (even allow \(h\))
[correct use of brackets]
A1oe: eg \(\dfrac{39}{2}\)