IB DP Math MAA SL : IB Style Mock Exams – Set 5 Paper 2
Question
The diagram shows a solid hemisphere with centre \( A(6,-1,-3) \). Point \( B(4,-5,-9) \) lies on the curved surface.
(a) Find \( AB \), the radius of the hemisphere.
(b) Hence, find the total surface area of the solid hemisphere.
(b) Hence, find the total surface area of the solid hemisphere.
▶️ Answer/Explanation
(a)
Using the 3D distance formula:
\( AB = \sqrt{(6-4)^2 + (-1+5)^2 + (-3+9)^2} \)
\( = \sqrt{2^2 + 4^2 + 6^2} = \sqrt{4 + 16 + 36} = \sqrt{56} \)
\( AB = 2\sqrt{14} \) (exact) or \( \approx 7.48 \) (to 3 s.f.).
(b)
Total surface area of a solid hemisphere:
Curved surface area \( = 2\pi r^2 \), base area \( = \pi r^2 \)
Total \( = 3\pi r^2 \).
Using \( r = 2\sqrt{14} \):
\( r^2 = 56 \)
\( \text{Surface area} = 3\pi \times 56 = 168\pi \) (exact) \( \approx 528 \) (to 3 s.f.).

