Surface area of combination of solids — cone on cylinder, hemisphere on cone

Question

A solid is formed by placing a cone (height 8 cm, radius 6 cm) on top of a cylinder (height 10 cm, radius 6 cm). Find the total surface area of the combined solid.

(CBSE Class 10 pattern)

Solution — Step by Step

flowchart TD
    A["Combined Solid\n(Cone on Cylinder)"] --> B["Surfaces to count:"]
    B --> C["Curved surface\nof cylinder"]
    B --> D["Curved surface\nof cone"]
    B --> E["Base of cylinder\n(bottom circle)"]
    B --> F["DO NOT count:\njoint circle\n(where cone meets cylinder)"]

When two solids are joined, the circular face at the junction is hidden inside the combined solid. We must not include it.

Total SA = CSA of cylinder + CSA of cone + Area of bottom circle

l = \sqrt{h^2 + r^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}

CSA of cylinder = $2\pi r h = 2\pi \times 6 \times 10 = 120\pi$ cm²

CSA of cone = $\pi r l = \pi \times 6 \times 10 = 60\pi$ cm²

Base circle = $\pi r^2 = \pi \times 36 = 36\pi$ cm²

\text{Total SA} = 120\pi + 60\pi + 36\pi = 216\pi = \mathbf{678.86 \text{ cm}^2}

(Using $\pi = 22/7$ gives $216 \times 22/7 = 678.86$ cm²)

Why This Works

When solids are combined, some surfaces become internal and invisible. The strategy is always the same: add up all the exposed curved surfaces and subtract any flat faces that are hidden at the junction. For cone-on-cylinder, both have the same radius, so one circular face from the cylinder and the base of the cone overlap — neither is exposed.

Think of it like gift-wrapping: you only wrap the surfaces you can see from outside. The hidden joint needs no wrapping.

Alternative Method — For Hemisphere on Cylinder

If a hemisphere (radius $r$ ) sits on top of a cylinder (same radius $r$ , height $h$ ):

Total SA = CSA of cylinder + CSA of hemisphere + base circle

$= 2\pi rh + 2\pi r^2 + \pi r^2 = 2\pi rh + 3\pi r^2$

Again, the top circle of the cylinder and the flat face of the hemisphere are hidden.

In CBSE Class 10, a common mistake-prone variant: a toy shaped like a cone topped with a hemisphere (both with same radius). The total SA = CSA of cone + CSA of hemisphere. There is no flat circle anywhere because the hemisphere’s flat face covers the cone’s base, and the cone’s base covers the hemisphere’s flat face — both are hidden.

Common Mistake

The most common error: including the hidden circular face at the junction. Students compute $2\pi rh + 2\pi r^2 + \pi rl + \pi r^2$ (both circular faces of cylinder plus cone) instead of removing the hidden one. Remember: at every junction between two solids, one circular area disappears from each solid. Count only what you can see from outside.

Question

Solution — Step by Step

Why This Works

Alternative Method — For Hemisphere on Cylinder

Common Mistake

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