Foundation June 2022 Paper 2 Q21
21 The diagram shows an 8-sided shape \(ABCDEFGH\).
Diagram NOT accurately drawn
\(HG = 28\) cm \(FG = 12\) cm \(AB = EF = 5\) cm
The height of the shape is 20 cm
\(CD\) is parallel to \(HG\)
The area of shape \(ABCDEFGH\) is 434 cm\(^2\)
Find the length of \(CD\).
(4)
| Scheme | Marks |
|---|---|
28 × 12 (=336) or 5 × 12 (= 60) or 18 × 12 (= 216) or 28 × 20 (=560) or \(\dfrac{1}{2}(CD + \text{``}{18}\text{''})\text{``}{8}\text{''}\) oe eg 72 +4\(CD\) [numbers in “ ” come from correct working] Check diagram for areas | M1 |
“336” + 0.5(“18”+ \(CD\))“8” = 434 oe eg 4(“18” + \(CD\)) = 98 or eg 0.5(“18” + \(CD\))”8” = “98” oe eg \(\dfrac{1}{2}(18 + CD) = 12.25\) or \(\text{``}{560}\text{''} - 2\left(0.5(5 + x)\text{``}{8}\text{''}\right) = 434\) oe (where \(x\) is horizontal from \(D\) to perp with \(AF\)) [numbers in “ ” come from correct working] | M1 |
eg (\(CD\) =) \(\dfrac{196 - 144}{8}\left(= \dfrac{52}{8}\right)\) or (\(CD\) =) \(\dfrac{98 - 72}{4}\left(= \dfrac{26}{4}\right)\) or (\(CD\) =) \(\dfrac{434 + 152 - 560}{4}\) or (\(CD\) =) 2 × 12.25 – 18 or 98 × 2(= 196), “196” ÷ 8(= 24.5), “24.5” – 18 | M1 |
| 6.5 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: For a correct method to find the area of a rectangle (may be seen as part calculation) or a correct expression for the area of the trapezium with numbers substituted.
Allow for other correct method to find area linked to this shape.
M1: correct use of their values from correct working for an equation involving \(CD\) (\(CD\) could be labelled with any letter)
M1: a correct process to solve a correct equation or a correct process to find \(CD\) using correct values
A1: oe