Measuring two-dimensional space and the outer surfaces of three-dimensional objects — from rectangles and circles to prisms, cylinders, pyramids, and cones
This chapter develops the measurement of two-dimensional space and the outer surfaces of three-dimensional objects. Students who have worked through triangles, similarity, circles, quadrilaterals, and polygons have built a strong foundation of shape relationships. Now those shapes become objects to measure.
Area is not a formula. Area is a quantity — the amount of flat space enclosed by a boundary. Before any formula can be trusted, a student must understand what the formula is counting. This chapter insists on that understanding at every step: what does each formula measure, where does it come from, when does it apply, and what happens when the shape is not one of the standard cases?
Surface area extends that thinking into three dimensions. When a three-dimensional object is unfolded into its faces, each face is a flat region with an area. Surface area is the total area of all those faces. It is the quantity that determines how much material is needed to wrap a box, paint a cylinder, or manufacture a cone-shaped container.
A contractor is finishing a triangular dormer window on a house. The window has a base of 4 feet and a height of 3 feet. Below the window, a semicircular decorative arch extends downward with a diameter of 4 feet. The contractor needs to order enough specialty glass to fill both the triangle and the arch, and enough trim to outline the full perimeter of the combined shape. How do you find the total glass area and the total trim length? By the end of this chapter, problems like this will be routine.
Area and surface area appear throughout mathematics, science, and practical work. In construction and architecture, area determines how much flooring, roofing, wallboard, paint, or fabric is needed for a project. In physics, the cross-sectional area of a pipe or wire determines how much fluid or current it can carry. In engineering and manufacturing, surface area calculations determine material costs, thermal properties, drag forces, and structural needs. In biology and medicine, surface area governs how quickly substances are absorbed across membranes — the lungs, intestines, and blood vessels are enormous in surface area because of internal folding. This chapter is not an isolated collection of formulas. It is a systematic study of measurement in two dimensions and on the surfaces of three-dimensional objects.
Imagine tiling a flat floor with square tiles, each one foot on a side. The number of tiles required to cover the floor exactly — with no gaps and no overlaps — is the area of the floor in square feet. Each tile represents one square foot of space.
This is what area measures. Area is not the boundary of a shape. Area is not the length of the longest side. Area is the two-dimensional space enclosed inside the boundary of a flat figure.
When we say a rectangle has an area of 24 square inches, we mean that 24 non-overlapping unit squares of side length one inch could be arranged to fill the rectangle completely.
Length is measured in one-dimensional units: inches, feet, centimeters, meters. When we measure area, we are covering a two-dimensional region using squares. Each square has a length and a width, both measured in the same unit. A square that is 1 inch on each side has an area of 1 × 1 = 1 square inch, written 1 in². This is why area is always measured in squared units: in², ft², cm², m², and so on. An answer of "24" with no squared unit is incomplete.
Perimeter measures the length of the boundary. Area measures the space enclosed by that boundary. These are different quantities with different units. A rectangle 6 ft × 4 ft has perimeter 20 ft and area 24 ft². A square 5 ft on each side has perimeter 20 ft and area 25 ft². Both have the same perimeter but different areas.
Do not confuse perimeter and area. Area problems often require you to identify a height, which is a perpendicular distance — never a slant side. Perimeter adds up the actual side lengths. Height is used for area only.
Many area calculations involve π, square roots, or other irrational numbers. Leave answers in exact form unless the problem asks for a decimal approximation.
Exact: A = 9π cm²
Approximate: A ≈ 28.27 cm²
A rectangle is a quadrilateral with four right angles. Its area is the product of its base and its height. Because a rectangle has right angles, the height is simply the side perpendicular to the base.
A = bh where b is the base and h is the height.
A = bh = 14 × 11 = 154 ft²A parallelogram has two pairs of parallel sides. Its slant sides are not perpendicular to the base, so the height of a parallelogram must be measured perpendicular to the base, not along the slant side. If you cut a right triangle off one end of a parallelogram and move it to the other end, you form a rectangle with the same base and height. Therefore the area of the parallelogram equals the area of that rectangle.
Draw a parallelogram with a slanted left and right side. Draw a dashed perpendicular from the top side down to the base and label it h. Label the base b. Mark the slant sides clearly so the height is distinguished from the slant height.
A = bh where h is the perpendicular height, not the slant side.
A = bh = 9 × 5 = 45 cm²Never use the slant side as the height of a parallelogram. The height is always the perpendicular distance between the two parallel sides.
A triangle is half of a parallelogram. Take any parallelogram and draw one diagonal — you divide it into two congruent triangles. Each triangle has the same base and height as the parallelogram but half the area.
A = ½bh where h is the perpendicular distance from the base to the opposite vertex.
Draw three triangles side by side: an acute triangle with the height drawn inside; an obtuse triangle with the height drawn outside as a dashed extension; and a right triangle where the height is one of the legs. In each case label b and h and mark the right angle at the foot of the height.
A = ½ × 12 × 7 = ½ × 84 = 42 in²36 = ½(9)(h)
36 = 4.5h
h = 36 ÷ 4.5 = 8 ft
For obtuse triangles, you may need to extend the base line to draw the perpendicular height. The height falls outside the triangle, but the formula A = ½bh still applies correctly with that perpendicular distance.
A trapezoid has exactly one pair of parallel sides, called the bases and labeled b₁ and b₂. The height h is the perpendicular distance between the two parallel sides. The formula multiplies the average of the two bases by the height — the average base acts like the single base of a rectangle with the same height and the same area.
A = ½(b₁ + b₂)h
Derivation: place two identical trapezoids together to form a parallelogram with base b₁ + b₂ and height h. Each trapezoid is half that parallelogram.
A = ½(8 + 14)(5) = ½(22)(5) = ½(110) = 55 cm²A kite is a quadrilateral with two pairs of consecutive congruent sides. Its diagonals are perpendicular, and one diagonal bisects the other. The area uses the diagonal lengths.
A = ½d₁d₂ where d₁ and d₂ are the lengths of the two diagonals.
A = ½(10)(6) = 30 in²A rhombus is a parallelogram with four congruent sides. Its diagonals are perpendicular bisectors of each other. The area can be found two ways: the parallelogram formula (base × height) when the perpendicular height is known, or the diagonal formula when both diagonals are given. A rhombus is a special kite, which is why the diagonal formula is the same.
Method 1 (base and height): A = bh
Method 2 (diagonals): A = ½d₁d₂
Use the diagonal formula when both diagonals are given. Use base-height when the perpendicular height is given.
120 = ½(15)(d₂)
120 = 7.5d₂
d₂ = 16 ft
A regular polygon is both equilateral and equiangular. The apothem is the perpendicular distance from the center to the midpoint of any side — the shortest distance from the center to a side. By dividing the polygon into congruent triangles from the center, each with base s (one side) and height a (the apothem), the area of all n triangles gives the formula below.
Draw a regular hexagon with its center marked. Draw one line segment from the center perpendicular to a side, meeting the midpoint of that side. Label this segment a (the apothem). Label one side s. Show the triangle formed by two radii to adjacent vertices and the side s.
A = ½Pa where P is the perimeter and a is the apothem.
Derived by summing n triangles, each with area ½ · s · a. Since P = ns: total area = ½ · (ns) · a = ½Pa.
A = ½(48)(4√3) = ½(192√3) = 96√3 m²A = ½Pa = ½(35)(4.82) = ½(168.7) = 84.35 ft²The ratio of the circumference of any circle to its diameter is always the same constant: π ≈ 3.14159. Pi is irrational — its decimal expansion never repeats. The circumference is the perimeter of a circle. The area formula comes from rearranging thin pie-slice wedges into a near-rectangle whose width approaches πr and whose height is r, giving area πr · r = πr².
Circumference: C = 2πr or C = πd
Area: A = πr²
Always use the radius in A = πr². If given the diameter, divide by 2 first.
C = 2π(7) = 14π cm
A = π(7)² = 49π cm²
100π = πr²
r² = 100
r = 10 ft
d = 2r = 20 ft
The area formula uses the radius, not the diameter. If you use the diameter where the radius is required, you will get a result four times too large: π(2r)² = 4πr², not πr².
A sector of a circle is a region bounded by two radii and the arc between them — like a slice of pie. If the central angle is θ degrees, the sector is θ/360 of the full circle. This fraction applies equally to arc length and to area.
Arc length: ℓ = (θ/360) · 2πr
Sector area: A = (θ/360) · πr²
A = (1/3) · π(9)² = (1/3)(81π) = 27π in²
ℓ = (1/3) · 2π(9) = (1/3)(18π) = 6π in
48π = (θ/360) · π(12)²
48π = (θ/360) · 144π
48 = 144θ/360
48 = 0.4θ
θ = 120°
A composite figure is a region made by combining, removing, or overlapping simpler figures. Real floors, rooms, windows, signs, pools, and architectural features are rarely perfect rectangles or circles. They are combinations of familiar shapes.
Step 1: Identify the basic shapes that make up the composite figure.
Step 2: Decide whether you are adding regions (shapes combined) or subtracting regions (shapes cut out).
Step 3: Calculate the area of each piece separately.
Step 4: Add or subtract as needed.
Step 5: Check that units are correct and the answer is reasonable.
Draw three composite figures side by side: (1) an L-shaped region divisible into two rectangles, (2) a rectangle with a semicircle added on one end, and (3) a large rectangle with a circular hole cut from the interior. In each case, show the dividing lines or the shape being removed.
A₁ (large rectangle) = 20 × 15 = 300 ft²
A₂ (missing corner) = 8 × 7 = 56 ft²
A = 300 − 56 = 244 ft²
A_rectangle = 4 × 6 = 24 ft²
A_semicircle = ½π(2)² = ½(4π) = 2π ft²
A_total = (24 + 2π) ft²
A_square = 10² = 100 in²
A_circle = π(3)² = 9π in²
A_remaining = (100 − 9π) in²
Label each region before computing. In a composite figure with three or more parts, it is easy to lose track of which area goes where. A quick sketch with region labels prevents most errors.
A prism is a three-dimensional figure with two congruent, parallel polygonal bases connected by rectangular lateral faces. The surface area of a prism is the total area of all its faces: both bases and all lateral faces.
A net is a flat diagram showing all faces of a three-dimensional figure unfolded into a plane. The net of a rectangular prism looks like a cross: a center rectangle surrounded by four more rectangles — two of each dimension.
Draw the net of a rectangular prism. Label the two base rectangles (top and bottom) with dimensions l × w. Label the four lateral face rectangles: two with dimensions l × h and two with dimensions w × h. Show fold lines between adjacent faces.
SA = 2(lw + lh + wh)
General Prism: SA = 2B + Ph where B = base area, P = base perimeter, h = prism height.
The lateral surface of a prism, when unrolled, forms a rectangle with width = perimeter of base and height = prism height.
SA = 2(lw + lh + wh)
= 2(10×6 + 10×4 + 6×4)
= 2(60 + 40 + 24)
= 2(124) = 248 cm²
B = ½(6)(3√3) = 9√3 ft²
P = 3 × 6 = 18 ft
SA = 2(9√3) + (18)(10)
= 18√3 + 180 ft²
A cylinder has two congruent circular bases connected by a curved lateral surface. When the lateral surface is unrolled, it becomes a rectangle. The rectangle's width equals the circumference of the base circle (2πr) and its height equals the cylinder height h.
Draw a cylinder and beside it its net: two circles and a rectangle. Label the rectangle's width as 2πr (the circumference) and its height as h. Label the circles with radius r. Show the fold lines connecting each circle to the rectangle's long edges.
SA = 2πr² + 2πrh
2πr² = both circular bases. 2πrh = lateral surface.
Lateral surface only: SA_lateral = 2πrh
SA = 2π(5)² + 2π(5)(12)
= 2π(25) + 2π(60)
= 50π + 120π = 170π m²
Some problems ask for total surface area (both bases plus lateral). Others ask for lateral surface area only (no bases). Read the problem carefully before choosing the formula.
A pyramid has a polygonal base and triangular lateral faces that meet at a single apex. The slant height ℓ is the distance from the apex to the midpoint of a base edge, measured along a lateral face — not the vertical height h of the pyramid.
Draw a square pyramid. Label the vertical height h from the apex down to the center of the base. Label the slant height ℓ from the apex down the face to the midpoint of a base edge. Show clearly that h and ℓ are different lengths. Mark the right angle where h meets the base.
SA = B + ½Pℓ
B = base area. P = base perimeter. ℓ = slant height.
Each triangular lateral face has area ½ · (side) · ℓ. Adding all n faces gives ½Pℓ.
Finding slant height: ℓ = √(h² + a²) where a = apothem of the base.
B = 8² = 64 ft²
P = 4 × 8 = 32 ft
SA = 64 + ½(32)(10) = 64 + 160 = 224 ft²
ℓ = √(h² + a²) = √(4² + 3²) = √(16 + 9) = √25 = 5 m
B = 6² = 36 m²
P = 4 × 6 = 24 m
SA = 36 + ½(24)(5) = 36 + 60 = 96 m²
A cone has a circular base and a lateral surface tapering to a single apex. The slant height ℓ is the distance from the apex to any point on the base edge along the lateral surface. The lateral surface, when unrolled, forms a sector of a circle whose arc length equals the base circumference 2πr — giving lateral area πrℓ.
Draw a right cone. Label the apex at the top. Label vertical height h from apex to center of base. Label radius r of the base. Label slant height ℓ from apex to base edge. Show that h, r, and ℓ form a right triangle with the right angle at the center of the base.
SA = πr² + πrℓ
πr² = circular base. πrℓ = lateral surface.
Slant height: ℓ = √(h² + r²)
SA = π(6)² + π(6)(10) = 36π + 60π = 96π cm²ℓ = √(12² + 5²) = √(144 + 25) = √169 = 13 inSA = π(5)² + π(5)(13) = 25π + 65π = 90π in²Real surface area problems require reading carefully to decide which faces to include. A cardboard box with no lid omits the top face. A pipe open at both ends uses only the lateral surface. A tank sitting on the ground needs the lateral surface and top base painted, but not the bottom. Identifying which faces matter is as important as calculating them correctly.
SA_lateral = 2π(8)(30) = 480π ft²
A_top = π(8)² = 64π ft²
SA_total = 480π + 64π = 544π ft²
SA = 2(lw + lh + wh)
= 2(12×8 + 12×5 + 8×5)
= 2(96 + 60 + 40) = 2(196) = 392 in²
SA_lateral = πrℓ = π(4)(9) = 36π cm² ≈ 113.1 cm²The height must be the perpendicular distance between the parallel sides. The slant side is always longer, so using it produces an area that is too large. Draw the perpendicular height and use that value.
Area is always measured in squared units. Writing "36 cm" instead of "36 cm²" is an incomplete and incorrect answer. Every area calculation must end with a squared unit.
The formula A = πr² uses the radius. If the diameter is 10, then r = 5, and A = π(5)² = 25π. Using the diameter instead of the radius gives an answer four times too large.
A triangle is half a parallelogram. Always include the ½: A = ½(8)(5) = 20. The same applies to kite and rhombus diagonal formulas: A = ½d₁d₂.
That computes the perimeter, not the area. Area requires the perpendicular height: A = ½(6)(3√3) = 9√3. Perimeter and area use different formulas and have different units.
Surface area uses the slant height ℓ, not the vertical height h. Find the slant height first: ℓ = √(h² + r²) = √(144 + 25) = 13. Then SA = π(5)² + π(5)(13) = 90π.
Total surface area includes both circular bases: SA = 2π(4)² + 2π(4)(10) = 32π + 80π = 112π. Only use the lateral formula when the problem explicitly asks for lateral surface area only.
Convert all measurements to the same unit before computing. Either convert 2 ft to 24 inches: A = 24 × 18 = 432 in², or convert 18 in to 1.5 ft: A = 2 × 1.5 = 3 ft².
| Figure | Formula | Notes |
|---|---|---|
| Rectangle | A = bh | h is perpendicular to b |
| Parallelogram | A = bh | h is perpendicular height, not slant side |
| Triangle | A = ½bh | h is perpendicular to base |
| Trapezoid | A = ½(b₁ + b₂)h | b₁ and b₂ are the two parallel sides |
| Kite | A = ½d₁d₂ | d₁ and d₂ are the diagonals |
| Rhombus | A = bh or A = ½d₁d₂ | Use whichever is given |
| Regular Polygon | A = ½Pa | P = perimeter; a = apothem |
| Circle | A = πr² | Use radius, not diameter |
| Sector | A = (θ/360)·πr² | θ = central angle in degrees |
| Arc length | ℓ = (θ/360)·2πr | θ = central angle in degrees |
| Figure | Formula | Notes |
|---|---|---|
| Rectangular Prism | SA = 2(lw + lh + wh) | All six faces |
| General Prism | SA = 2B + Ph | B = base area; P = base perimeter |
| Cylinder | SA = 2πr² + 2πrh | Lateral: 2πrh only |
| Regular Pyramid | SA = B + ½Pℓ | ℓ = slant height |
| Cone (total) | SA = πr² + πrℓ | ℓ = slant height |
| Cone (lateral) | SA = πrℓ | No circular base |
| Slant height (cone) | ℓ = √(h² + r²) | Pythagorean Theorem |
| Slant height (pyramid) | ℓ = √(h² + a²) | a = apothem of base |
Answer each question in complete sentences.
Identify the basic shapes, then find the total area. Leave π in exact answers unless told to approximate.
Define variables, show all work, and include units in every answer.
Each problem contains an error. Identify it, explain what went wrong, and give the correct answer.
What is wrong? What additional information is needed? What is the correct approach?
Identify the error and write the correct area.
What is the error? What is the correct area?
What is missing from this surface area calculation? What is the correct total SA?
What part of the surface area is missing? Write the complete correct answer.
Identify the error in applying the sector formula and write the correct area.
What is wrong with the triangle area? What is the correct total area?
Find the actual slant height, then compute the correct total surface area.
Chapter 11 prepared you for two closely related topics in Chapter 12. Volume is the three-dimensional counterpart of area: just as area measures the flat space inside a boundary, volume measures the three-dimensional space inside an enclosed solid. The base area B you computed for surface area becomes the starting point for volume formulas: V = Bh for prisms and cylinders, V = ⅓Bh for pyramids and cones.
Scaling reveals how area and volume change when a figure is enlarged or reduced. When all lengths of a figure are multiplied by a scale factor k, all areas scale by k² and all volumes scale by k³. This is why a box twice as long, wide, and tall holds eight times the volume, not twice — and why it needs four times the wrapping paper, not twice. Keep the formulas from Chapter 11 handy throughout Chapter 12.