Higher June 2023 Paper 3 Q16
16
Not drawn accurately
In this right-angled triangle,
\(a = 16\) cm
\(a : c = 4 : 5\)
Work out the area of the triangle. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – using Pythagoras’ theorem or 3, 4, 5 triangle | ||
| \(16 \div 4 \times 5\) or 20 (cm) or identifies triangle as 3, 4, 5 | M1 | oe length of \(c\) may be on diagram |
| \(\sqrt{(\text{their } 20)^2 - 16^2}\) or \(\sqrt{400 - 256}\) or \(\sqrt{144}\) or \(4 \times 3\) | M1dep | |
| 12 (cm) | A1 | length of \(b\) may be on diagram |
| 96 | A1ft | ft \(\dfrac{1}{2} \times 16 \times\) their 12 with M2 awarded |
| Alternative method 2 – using trigonometry and \(\tfrac{1}{2}ab\sin C\) formula | ||
| \(16 \div 4 \times 5\) or 20 (cm) | M1 | oe length of \(c\) may be on diagram |
| \(\cos^{-1}\left(\dfrac{16}{20}\right)\) or 36.8(…) or 36.9 | M1dep | angle between sides \(a\) and \(c\) |
| \(\dfrac{1}{2} \times 16 \times 20 \times \sin(\text{their } 36.8(\ldots))\) | M1dep | dep on M2 |
| 96 | A1 | |
Additional guidance
| \(\dfrac{1}{2} \times 16 \times 12 \times \sin 90\) | M1M1M1 |