L5 · Surface Area of Spheres and Composite Objects

Learning goal

“I can find the surface area of a sphere (4πr²) and of a closed hemisphere (3πr²), and the surface area of a composite object by adding only the parts that are actually exposed - a face where two solids join is counted zero times.”

Sphere — a perfectly round solid; every point on its surface is the radius r from the centre. It has no net you can draw flat without stretching - its area formula is simply 4πr².

Hemisphere — half a sphere. OPEN (a bowl): curved surface only, 2πr². CLOSED (with its flat circular face): 2πr² + πr² = 3πr².

Composite object — a solid built from two or more known solids joined together (a silo: cylinder + hemisphere; a pencil: cylinder + cone).

Hidden face — the part of a surface where two solids touch. It is inside the composite - exposed to nothing - so it is subtracted from BOTH solids that share it (counted zero times total).

Worked example

A sphere has radius 6 cm. Find its surface area. Step 1: SA = 4πr² = 4π x 36. Step 2: SA = 144π ≈ 452.39 cm2. (Memory anchor: a sphere's area is exactly FOUR great circles' worth - 4 x πr².)

Check yourself — explain it

A closed hemisphere (a dome with a floor) has radius 6 cm. Its TOTAL surface area is...

Paired problem — example still visible

Closed hemisphere, r = 5 cm -> curved 2π x 25 = 50π; flat circle 25π; total 75π ≈ 235.62 cm2

Faded completion — last step is yours

"Sphere, r = 7 cm: SA = 4π x ___ = ___π ≈ ___ cm2" -> 49; 196π ≈ 615.75 cm2

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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Grade 10 — Mathematics 10C › C10U02 Surface Area & Volume

L5 · Surface Area of Spheres and Composite Objects