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Geometry & measures · Perimeter, area & volume

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Surface area of prisms, cylinders, cones & spheres

Surface area as the area of the unfolded net — counting a prism's faces without missing any, why a cylinder's curved surface is 2πrh, cones and the slant height, spheres, and the hemisphere trap.

Geometry & measures · Perimeter, area & volume

Surface area of prisms, cylinders, cones & spheres

Surface area as the area of the unfolded net — counting a prism's faces without missing any, why a cylinder's curved surface is 2πrh, cones and the slant height, spheres, and the hemisphere trap.

Why it works

Surface area is the area of the net. Imagine unfolding the solid flat and finding the area of every face, then adding. That single picture prevents the two standard errors: missing a hidden face, and mixing surface area (cm²) up with volume (cm³).

Prisms: two ends plus a wrap. A prism's surface is its two identical cross-sections plus a rectangle for each edge of the cross-section, all of length equal to the prism's length. For a triangular prism that is 22 triangles +3+ 3 rectangles — count them off on the net so none escape.

Cylinder: the label peels into a rectangle. Unroll the curved surface of a cylinder and you get a rectangle whose height is hh and whose width is the distance around the circle — the circumference 2πr2\pi r. So

curved surface=2πrh,total=2πrh+2πr2\text{curved surface} = 2\pi r h, \qquad \text{total} = 2\pi r h + 2\pi r^2

(the wrap plus the two circular ends). An open-topped tank or a pipe drops one or both circles — read the context before adding 2πr22\pi r^2 automatically.

Cone: the slant height does the work. The curved surface of a cone is πrl\pi r l, where ll is the slant height — the distance from rim to apex along the surface (this formula is printed in the question). If you are given the perpendicular height hh instead, the radius, height and slant make a right-angled triangle: l=r2+h2l = \sqrt{r^2 + h^2} by Pythagoras. Total surface area of a closed cone =πrl+πr2= \pi r l + \pi r^2.

Sphere: 4πr24\pi r^2 (also given in the question). The hemisphere trap: half a sphere's surface is 2πr22\pi r^2 of curve — but cutting the sphere exposed a flat circular face of area πr2\pi r^2. A solid hemisphere's total surface area is

2πr2+πr2=3πr2,2\pi r^2 + \pi r^2 = 3\pi r^2,

not 2πr22\pi r^2. Composite solids follow the same logic everywhere: add up only the faces that are actually exposed, and leave out any face where two pieces are glued together.