Higher November 2024 Paper 2 Q20
20 \(VABC\) is a solid pyramid.
\(ABC\) is an equilateral triangle.
\(M\) is the midpoint of \(AB\).
\(F\) is the point on \(MC\) such that \(MF : FC = 1 : 2\)
The vertex \(V\) is vertically above \(F\).
\(VA = VB = VC\)
\(VF = 8\) cm Angle \(VCM = 52^\circ\)
Work out the side length of the equilateral triangle \(ABC\).
Give your answer correct to 1 decimal place. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 10.8 | P1 | for process to find \(FC\), eg \(\tan 52 = \dfrac{8}{FC}\) (\(FC = 6.25(028..)\)) |
| P1 | for process that will lead to side length of \(ABC\), eg \(\sin 60 = \dfrac{\text{``}6.25\text{''} \times 1.5}{BC}\) or \(\cos 30 = \dfrac{\text{``}6.25\text{''} \times 1.5}{BC}\) or \((\text{``}6.25\text{''} \times 1.5)^2 + (0.5x)^2 = x^2\) oe | |
| A1 | for answer in range 10.8 to 10.83 |