Foundation January 2023 Paper 2 Q25
25 The diagram shows an isosceles triangle, with base length 24 cm.
Diagram NOT accurately drawn
The perimeter of the triangle is 54 cm.
Work out the area of the triangle.
(5)
| Scheme | Marks |
|---|---|
| (54 – 24) ÷ 2 (=15) [may be marked on diagram] | M1 |
| \(\text{“}15\text{”}^2 - (24 \div 2)^2\;(= 81)\) | M1 |
| [height =] \(\sqrt{\text{“}15\text{”}^2 - (24 \div 2)^2}\;(= 9)\) | M1 |
| \((24 \times \text{“}9\text{”}) \div 2\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 108 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: ft their “15” (if > 12)
M1: ft their “15” (if > 12)
M1: figures must be from correct working
A1: allow 107.9 – 108.1
| Scheme | Marks |
|---|---|
| (54 – 24) ÷ 2 (=15) [may be marked on diagram] | M1 |
or \(x = \cos^{-1}\left(\dfrac{\text{“}12\text{”}}{\text{“}15\text{”}}\right)(= 36.86...)\) or \(y = \sin^{-1}\left(\dfrac{24 \div 2}{\text{“}15\text{”}}\right)(= 53.13...)\) or \(A = \cos^{-1}\left(\dfrac{15^2 + 15^2 - 24^2}{2 \times 15 \times 15}\right)(= 106.2...)\) or \(B = \cos^{-1}\left(\dfrac{15^2 + 24^2 - 15^2}{2 \times 15 \times 24}\right)(= 36.8....)\) | M1 |
| or “12”tan”36.86…” (= 9) (allow 8.9… for these) “12” ÷ tan”53.13…” (= 9) or “15” × sin “36.86…” (= 9) or “15” × cos “53.13…” (= 9) | M1 |
| \((24 \times \text{“}9\text{”}) \div 2\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 108 | A1 |
Notes
M1: ft their “15” (if > 12)
[ using Hero’s formula S = 0.5 × 54 (= 27) and ]
27 × (27 – 24) × (27 – “15”) × (27 – “15”)
M1: ft their “15” (if > 12)
A1: allow 107.9 – 108.1
M2 for 0.5 × 24 × “15” × sin”36.86…” or 0.5 × “15” × “15” × sin(2 × “53.13…”) or 0.5 × “15” × “15” × sin(“106.2...”) or \(\sqrt{\text{“}27\text{”}(\text{“}27\text{”} - 24)(\text{“}27\text{”} - \text{“}15\text{”})(\text{“}27\text{”} - \text{“}15\text{”})}\)