Intermediate June 2022 Unit 2 Q11
11.
In the diagram above, the area of triangle \(BCD\) is 60 cm2.
Calculate the length of \(CD\).
You must show all your working. [6]
| Scheme | Marks | Comments |
|---|---|---|
| (BC2 = ) 9·62 + 12·82 or equivalent | M1 | note: (BC2 =) 92·16 + 163·84 (ignore place values for M1) Award M1 for the correct values substituted into the Cosine rule. |
| (BC2 =) 256 or (BC =) √256 | A1 | |
| (BC =) 16 (cm) | A1 | Allow (BC =) ±16 (cm). FT from M1 for the correctly evaluated square root of 'their 256’ provided their answer > 12·8. |
| CD = 2 × 60 ÷ 16 or equivalent | M2 | FT ‘their derived BC’ OR ‘their stated 16’ (not derived) provided 12·8 < ‘their stated 16’ < 22·4. Award M1 for 60 = ½ × 16 × CD or equivalent. |
| (CD =) 7·5 (cm) | A1 | Allow M2A1 for a correct embedded answer BUT M2A0 if contradicted by CD ≠ 7·5 (cm). |
Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of ‘two-step’ method | M2 | A partial trigonometric method is M0. |
| (BC =) 16 (cm) | A1 | |
| CD = 2 × 60 ÷ 16 or equivalent | M2 | FT ‘their derived BC’ OR ‘their stated 16’ (not derived) provided 12·8 < ‘their stated 16’ < 22·4. Award M1 for 60 = ½ × 16 × CD or equivalent. |
| (CD =) 7·5 (cm) | A1 | Allow M2A1 for a correct embedded answer BUT M2A0 if contradicted by CD ≠ 7·5 (cm). |