DWG NO. 11.1 — Lesson 1 of 5

Surface Area of Prisms and Cylinders

Unit 11: Surface Area and Volume · ~15–30 min

Objective

Find the surface area of a prism or cylinder by unfolding it into a net and adding the areas of its bases and lateral faces.

Unfolding a solid: the net

A three–dimensional solid has a surface made of flat or curved faces. If you cut the solid along its edges and flatten it out, you get a net — a two–dimensional pattern showing every face at once. Surface area is just the total area of that net, so the strategy for any prism or cylinder is the same: find the area of each face, then add them up.

Every prism and cylinder splits naturally into two kinds of surface:

Bases — the two congruent, parallel faces that give the solid its name (two rectangles for a rectangular prism, two circles for a cylinder).
Lateral surface — everything connecting the two bases (the rectangular side faces of a prism, or the curved side of a cylinder, which unrolls into a single rectangle).

Surface area formulas

Let B be the area of one base, P be the perimeter of the base, h be the height of the solid (the distance between the bases), and r be the radius of a circular base.

Solid	Lateral area	Total surface area
Prism	L.A. = Ph	S.A. = 2B + Ph
Cylinder	L.A. = 2πrh	S.A. = 2πr² + 2πrh

The lateral area of a cylinder comes from the same idea as a prism: unrolled, the curved side becomes a rectangle whose width is the base's circumference (2πr) and whose height is h — so L.A. = (2πr)(h).

A cylinder unfolds into two circular bases plus a rectangle of width 2πr and height h.

Worked Example 1 · Rectangular prism

ProblemFind the surface area of a rectangular prism with length 8 cm, width 5 cm, and height 4 cm.

1.The base is the 8 × 5 rectangle, so B = 8(5) = 40 cm² and the base perimeter is P = 2(8) + 2(5) = 26 cm.

2.Lateral area: Ph = 26(4) = 104 cm².

3.Total: S.A. = 2B + Ph = 2(40) + 104 = 80 + 104 = 184 cm².

Surface area = 184 cm²

Worked Example 2 · Cylinder

ProblemA can has radius 3 in and height 9 in. Find its surface area in terms of π, then to the nearest tenth.

1.Base area: πr² = π(3)² = 9π, so two bases contribute 18π.

2.Lateral area: 2πrh = 2π(3)(9) = 54π.

3.Total: S.A. = 18π + 54π = 72π ≈ 226.2 in².

Surface area = 72π in² ≈ 226.2 in²

Guided practice

1.Find the surface area of a cube with side length 6 ft.
Show answer
A cube has 6 congruent square faces. Each face: 6² = 36 ft². S.A. = 6(36) = 216 ft².
2.A triangular prism has a base that is a right triangle with legs 6 cm and 8 cm (hypotenuse 10 cm), and a height (length) of 12 cm. Find its surface area.
Show answer
Base area B = ½(6)(8) = 24 cm². Base perimeter P = 6 + 8 + 10 = 24 cm. Lateral area = Ph = 24(12) = 288 cm². S.A. = 2(24) + 288 = 48 + 288 = 336 cm².
3.A cylinder has diameter 10 m and height 5 m. Find its surface area in terms of π.
Show answer
Radius r = 5 m. Bases: 2π(5)² = 50π. Lateral: 2π(5)(5) = 50π. S.A. = 50π + 50π = 100π m².
4.Explain why the lateral area formula Ph works the same way for any prism, no matter the base shape.
Show answer
Every lateral face is a rectangle whose width is one side of the base and whose height is the prism's height h. Adding up all those rectangle areas gives (sum of base sides) × h = Ph, regardless of how many sides the base has.
5.Two cylinders have the same height but Cylinder B has twice the radius of Cylinder A. Does Cylinder B have exactly twice the surface area of Cylinder A? Explain.
Show answer
No. Surface area depends on r² in the base terms and on r in the lateral term, so doubling r does not scale every term by the same factor — the base areas quadruple while the lateral area only doubles. The overall surface area increases by somewhere between 2× and 4×, not exactly 2×.

← Unit 11 Unit 11 Overview Lesson 11.2 →