IB DP Math MAA HL : IB Style Mock Exams – Set 4 Paper 2
Question
(a) Calculate the length of the segment \(VX\).
The square base of the pyramid has a side length of \(5 \text{ cm}\).
(b) Determine the length of the diagonal \(AC\).
(c) Find the angle formed between the lateral edge \([VC]\) and the base of the pyramid.
▶️ Answer/Explanation
(a)
Using the distance formula: \(VX = \sqrt{(-3-1)^2 + (4-7)^2 + (2-0)^2} = \sqrt{16 + 9 + 4} = \sqrt{29} \approx 5.39\).
\(\sqrt{29}\) cm (or \(5.39\) cm).
(b)
For a square of side \(5 \text{ cm}\), the diagonal is: \(AC = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2} \approx 7.07\).
\(5\sqrt{2}\) cm (or \(7.07\) cm).
(c)
Let \(\theta\) be the required angle. Triangle \(VXC\) is right‑angled at \(X\) (since \(VX\) is perpendicular to the base). \(XC = \frac{AC}{2} = \frac{5\sqrt{2}}{2} \approx 3.5355\).
Using \(\tan \theta = \frac{VX}{XC}\): \(\tan \theta = \frac{\sqrt{29}}{5\sqrt{2}/2} = \frac{2\sqrt{29}}{5\sqrt{2}}\). \(\theta = \arctan\!\left(\frac{2\sqrt{29}}{5\sqrt{2}}\right) \approx 56.7^\circ\) (or \(0.990\) rad).
\(56.7^\circ\) (or \(0.990\) rad).