Surface Areas and Volumes for Math Olympiad: Complete Guide

Why This Matters

Surface areas and volumes of 3D solids are tested extensively in Math Olympiads. Combined solids, conversion problems, and frustum calculations push your spatial reasoning and formula application skills.

For Class 9-10 students, competition problems often involve solids that are combinations of basic shapes — a hemisphere on a cone, a cylinder with conical ends, requiring careful decomposition.

Best Strategy

Common Pitfalls

Mistakes:

• TSA vs CSA — Total Surface Area includes bases; Curved Surface Area does not.
• Combined solid surface area — Do NOT include the surfaces where solids are joined.
• Slant height vs height — For cones, slant height $l = \sqrt{h^2 + r^2}$. CSA uses slant height, volume uses height.
• Unit conversion — $1 m^3 = 1000000 cm^3 = 1000$ litres.

Practice Questions

Try these!

Question 1

A solid is formed by joining a hemisphere (radius 7 cm) to a cylinder (radius 7 cm, height 10 cm). Find the total surface area.

Solution: TSA = CSA of cylinder + CSA of hemisphere + one base of cylinder
$= 2\pi(7)(10) + 2\pi(7)^2 + \pi(7)^2 = 140\pi + 98\pi + 49\pi = 287\pi \approx 901.6$ sq cm.

Question 2

A cone of height 24 cm and radius 6 cm is melted to form a sphere. Find the sphere's radius.

Solution: Volume of cone = $\frac{1}{3}\pi(6)^2(24) = 288\pi$
$\frac{4}{3}\pi r^3 = 288\pi$, $r^3 = 216$, $r = 6$ cm.

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