Higher June 2022 Paper 2 Q11
11 A shape is made by joining a right-angled triangle to a rectangle.
Not drawn accurately
Work out the area of the shape. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(16^2\) or 256 and \(30^2\) or 900 | M1 | oe implied by 1156 |
| \(\sqrt{16^2 + 30^2}\) or \(\sqrt{256 + 900}\) or \(\sqrt{1156}\) or 34 | M1dep | oe eg \(\sqrt{16^2 + 30^2 - 2 \times 16 \times 30 \times \cos 90}\) |
| \(52 \times\) their 34 or 1768 | M1dep | oe if M1M0 their 34 can be any value other than 16, 30 or 52 dep on 1st M |
| \(0.5 \times 30 \times 16\) or 240 | M1 | oe eg \(0.5 \times 30 \times 16 \times \sin 90\) |
| 2008 | A1 | SC3 2248 |
| Alternative method 2 | ||
| \(\tan^{-1} \dfrac{16}{30}\) or [28, 28.1] or \(\tan^{-1} \dfrac{30}{16}\) or [61.9, 62] | M1 | oe may be on diagram |
| \(\dfrac{30}{\cos(\text{their } [28, 28.1])}\) or \(\dfrac{16}{\cos(\text{their } [61.9, 62])}\) or 34 | M1dep | oe eg \(\dfrac{16}{\sin(\text{their } [28, 28.1])}\) or \(30 \cos(\text{their } [28, 28.1]) + 16 \cos(\text{their } [61.9, 62])\) |
| \(52 \times\) their 34 or 1768 | M1dep | oe if M1M0 their 34 can be any value other than 16, 30 or 52 dep on 1st M |
| \(0.5 \times 30 \times 16\) or 240 | M1 | oe eg \(0.5 \times 30 \times 16 \times \sin 90\) |
| 2008 | A1 | SC3 2248 |
Additional guidance
Up to M4 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts
The 4th mark in Alts 1 and 2 is not dependent on any other marks
34 or 1768 or 240 may be on the diagram
SC3 is for using \(30 \times 16\) for the area of the triangle
Ignore units