Perimeter, area, and volume

Learning objectives

Choose the correct measurement and units; identify perpendicular heights; decompose composite figures; distinguish exposed surface area from volume; recover missing dimensions; apply linear, square, and cubic scale factors correctly.

Choose the quantity before the formula

Perimeter measures the boundary of a plane figure. Area measures a two-dimensional region. Surface area totals the exposed two-dimensional faces or curved surfaces of a solid. Volume measures three-dimensional space. Similar-looking questions can require very different calculations depending on which quantity is requested.

Plane figure	Area	What the variables mean
Rectangle	$A = lw$	Adjacent side lengths l , w. Perimeter is $2l + 2w.$
Square	$A = s 2$	Side length s. Perimeter is 4s.
Triangle	$\frac{1}{2}$	Base b and its perpendicular height h.
Parallelogram	$A = bh$	Base and perpendicular height, not a slanted side.
Trapezoid	$\frac{1}{2}$	Parallel bases b₁, b₂ and perpendicular distance h between them.
Circle or disk	$A = πr 2$	Radius r. Boundary circumference is 2πr.

Why perpendicular height matters. Base times height measures the amount of space between parallel lines. Tilting a parallelogram’s side while keeping its base and height fixed does not change its area. A slanted side is generally longer than the height and therefore cannot simply replace it.

You can often recover an unknown height by drawing a right triangle. In an isosceles triangle, an altitude from the apex splits the base in half. In a right circular cone, the radius, perpendicular height, and slant height form a right triangle. The height used in a volume formula is perpendicular to the base plane.

Composite figures: add regions, trace boundaries

Break a complicated region into familiar nonoverlapping pieces, or subtract a missing region from a larger simple figure. Use addition and subtraction of areas only for the portions actually present. For perimeter, trace the outside boundary: an internal dividing line is not part of the perimeter. A circular ring, or annulus, with outer radius R and inner radius r has area $π(R 2 - r 2).$ This is not $π(R - r) 2 :$ the difference of squares is not the square of a difference. A rectangular frame of uniform border width w loses 2w from each outer dimension when you determine its inner rectangle.

Solids: volume and exposed surface area

For a prism, let B denote the area of one base and h the perpendicular distance between bases. Then $V = Bh.$ A cylinder has a circular base, so $V = πr 2 h.$ A pyramid or cone has one-third the volume of a prism or cylinder with the same base area and perpendicular height: $V = Bh/3.$

Solid	Volume	Total surface area when closed
Rectangular prism	lwh	$2(lw + l h + wh)$
Right circular cylinder	$πr 2 h$	$2πr 2 + 2πrh$
Right circular cone	$\frac{1}{3}$	$πr 2 + πrs,$ where s is slant height
Sphere	$\frac{4}{3}$	$4πr 2$
Pyramid	$\frac{1}{3}$	Base area plus the areas of its triangular faces

For a right prism with base perimeter P, the lateral area is Ph and total closed surface area is $2B + Ph.$ For a cylinder, unrolling the curved surface produces a rectangle of width 2πr and height h, explaining its lateral area 2πrh. A hemisphere has volume $\frac{2}{3}$ and curved surface area $2πr 2 .$ If its circular base is exposed, add another $πr 2$ for a total of $3πr 2 .$ An open-top cylinder similarly omits one circular base from its surface area.

Left: cone height h differs from slant height s. Right: a pyramid’s face altitude also differs from h.

Shared faces disappear from the outside

When two solids are joined, add their volumes if their interiors do not overlap. For exposed surface area, omit the faces where they touch. Adding the total surface areas of both separate solids generally counts hidden internal faces twice. Likewise, a hollow object may require an inner surface, an outer surface, or both; read what is to be painted, covered, or measured.

Scaling, percentages, and units

For similar figures with positive scale factor k,

$length ratio = k,$ $area ratio = k 2,$ $volume ratio = k 3 .$

Thus a known area ratio gives a length ratio by a square root, and a known volume ratio gives a length ratio by a cube root. An area ratio of q gives a volume ratio of $q 3/2$ for similar solids. For a p% increase in every length, use $k = 1 + p/100.$ The area percent increase is $100(k 2 - 1)%$ and the volume percent increase is $100(k 3 - 1)%.$ Do not multiply the original percent by 2 or 3: the changes compound across dimensions.

Similarity requires uniform scaling. If a cylinder’s radius changes by a factor a and its height by a factor b, its volume changes by a²b, not automatically by a³. A rectangular box whose three dimensions change by factors a, b, c has volume factor abc.

Units also scale by dimension. Since $1 m = 100 cm,$

$1 m 2 = 10,000 cm 2,$ $1 m 3 = 1,000,000 cm 3 .$

Convert all input lengths into one unit before multiplying. A length conversion factor must be squared for area and cubed for volume. If an idealized solid is melted and recast without material loss, volume is conserved; surface area generally is not.

15 worked examples

6.01. Perimeter supplies dimensions; dimensions supply area

A rectangle has perimeter 54. Its length is 3 more than twice its width. Find its area.

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Recognize the structure. Use the perimeter to solve for one dimension before applying the area formula.

Work it through. Let the width be w, so the length is $2w + 3.$ Then

$2w + 2(2w + 3) = 54 6w + 6 = 54 w = 8.$

The length is 19, giving area $8(19) = 152.$

Answer: 152 square units.

Avoid the trap. Perimeter and area use different operations. Dividing 54 by 4 assumes a square, a condition not given here.

6.02. Recover the perpendicular height of a parallelogram

A parallelogram has base 20 and an adjacent slanted side 13. A perpendicular from the upper-left vertex to the base meets the base 5 units from its lower-left endpoint. Find the area.

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Recognize the structure. The slanted side is a hypotenuse, not the height.

Work it through. The altitude h, the 5-unit base segment, and the 13-unit slanted side form a right triangle:

$h 2 + 5 2 = 13 2 h = 12.$

Thus $A = bh = 20(12) = 240.$

Answer: 240 square units.

Avoid the trap. $20(13) = 260$ uses the slanted side instead of the perpendicular height. Draw the height explicitly before substituting.

6.03. A trapezoid uses the average of its bases

A trapezoid has parallel bases of lengths 8 and 18 and perpendicular height 6. Find its area.

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Recognize the structure. The trapezoid formula multiplies its height by the average of the parallel bases.

Work it through. Compute $\frac{1}{2}$

Answer: 78 square units.

Key takeaway. The average base is 13, which lies between 8 and 18. The area therefore lies between $8(6) = 48$ and $18(6) = 108,$ consistent with the result.

6.04. Removing a corner changes area but can preserve perimeter

A 4-by-3 rectangle is removed from one corner of a 12-by-10 rectangle, with the removed sides aligned to the original sides. Find the remaining area and perimeter.

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Recognize the structure. Subtract the missing area, but trace the new outside boundary for perimeter.

Work it through. The area is $12(10) - 4(3) = 120 - 12 = 108.$ The new boundary lengths are 12, 7, 4, 3, 8, 10, so the perimeter is 44.

Answer: Area 108 square units; perimeter 44 units.

Key takeaway. The removed outer pieces of lengths 4 and 3 are replaced by inner-facing boundary pieces of the same lengths. This corner cut preserves perimeter; other cuts, such as an inward notch, may not.

6.05. Area of a circular path

A circular garden has radius 7 meters. A path of uniform width 2 meters surrounds the garden. What is the area of the path?

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Recognize the structure. The path is the outer disk minus the garden; its outer radius is $7 + 2 = 9.$

Work it through. Compute $A path = π(9) 2 - π(7) 2 = π(81 - 49) = 32π.$

Answer: 32π square meters.

Avoid the trap. $π(2) 2$ is the area of a disk of radius 2, not a ring of width 2. The inner and outer radii must each be squared before subtracting.

6.06. A semicircle changes the outside boundary

A semicircle of diameter 6 is attached externally along a 6-unit side of a 10-by-6 rectangle. What is the perimeter of the combined figure?

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Recognize the structure. The shared diameter is internal and must not be counted.

Work it through. The exposed straight edges total $10 + 10 + 6 = 26.$ The semicircle has radius 3, so its curved edge is half a circumference: $πr = 3π.$

Answer: $26 + 3π$ units.

Avoid the trap. Adding the rectangle’s full perimeter and the semicircle’s full perimeter counts the shared diameter twice. Both copies must be removed from that sum.

6.07. Volume and surface area measure different things

A closed rectangular prism has length 8, width 5, and height 6. Find its volume and total surface area.

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Recognize the structure. Volume multiplies three dimensions; surface area adds the six face areas.

Work it through. The volume is $V = 8(5)(6) = 240.$ Opposite faces occur in matching pairs, so

$S = 2(8 \cdot 5 + 8 \cdot 6 + 5 \cdot 6) = 2(40 + 48 + 30) = 236.$

Answer: Volume 240 cubic units; surface area 236 square units.

Key takeaway. The numerical values happen to be close, but they describe different quantities and have different units. Never compare a volume directly with an area as though one must be larger.

6.08. A cylinder’s diameter is not its radius

A right circular cylinder has diameter 10 and volume 200π. What is its height?

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Recognize the structure. Convert the diameter to radius before using $V = πr 2 h.$

Work it through. The radius is 5. Therefore

$200π = π(5) 2 h = 25πh h = 8.$

Answer: 8 units.

Avoid the trap. Substituting 10 as the radius would make the base area four times too large and the recovered height four times too small.

6.09. A cone gives slant height instead of volume height

A right circular cone has radius 6 and slant height 10. Find its volume.

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Recognize the structure. First recover the perpendicular height from the right-triangle cross-section.

Work it through. The perpendicular height satisfies $h 2 + 6 2 = 10 2,$ so $h = 8.$ Then

$\frac{1}{3}$

Answer: 96π cubic units.

Avoid the trap. The slant height 10 belongs in the cone’s lateral-area formula, not directly in its volume formula. The volume formula uses the perpendicular height 8.

6.10. A square pyramid’s face height

A square pyramid has base side length 10 and perpendicular height 12. Its apex is directly above the center of the base. What is its total surface area, including the base?

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Recognize the structure. The triangular faces use slant height from the apex to a base edge’s midpoint.

Work it through. The center-to-midpoint distance in the base is 5, so the face altitude is

$\sqrt{12^{2} + 5^{2}}$

Each triangular face has area $(1/2)(10)(13) = 65.$ There are four such faces, plus the square base:

$S = 4(65) + 10 2 = 260 + 100 = 360.$

Answer: 360 square units.

Avoid the trap. Using 12 as the triangular face height understates the lateral area. The face lies on a slant, while the given height is perpendicular to the base plane.

6.11. Recover a sphere’s radius, then its surface area

A sphere has volume 288π. What is its surface area?

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Recognize the structure. Volume determines r³; surface area requires r².

Work it through. From the volume formula,

$\frac{4}{3}$

Then $S = 4πr 2 = 4π(36) = 144π.$

Answer: 144π square units.

Avoid the trap. Take a cube root when solving $r 3 = 216.$ A square root would mix the dimensions of volume and area.

6.12. Exposed area of a joined cylinder and hemisphere

A closed solid consists of a cylinder of radius 3 and height 8, topped by a hemisphere of radius 3. The flat face of the hemisphere exactly covers the cylinder’s top. Find the total exposed surface area, including the cylinder’s bottom.

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Recognize the structure. List only the surfaces exposed to the outside.

Work it through. The cylinder’s curved area is $2πrh = 48π.$ The hemisphere’s curved area is $2πr 2 = 18π.$ The cylinder’s bottom contributes $πr 2 = 9π.$ The joined top/base faces are internal.

Answer: $48π + 18π + 9π = 75π$ square units.

Avoid the trap. Adding the two separate solids’ closed surface areas counts the shared circular region twice. Neither copy is exposed.

6.13. A 20 percent length increase is not a 60 percent volume increase

A solid is enlarged to a similar solid so that every corresponding length increases by 20%. By what percent does the volume increase?

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Recognize the structure. Convert the percent increase into a scale factor, then cube that factor.

Work it through. The length factor is $k = 1.20.$ The volume factor is

$k 3 = (1.20) 3 = 1.728.$

The new volume is 172.8% of the original, so the increase is 72.8%.

Answer: 72.8%.

Key takeaway. The corresponding area increase would be $(1.2 2 -1)100% = 44%.$ Percentage changes compound across dimensions rather than simply adding 20% for each dimension.

6.14. Surface-area ratio determines volume ratio

Two similar solids have surface areas 100 and 225 square units. The smaller solid has volume 80 cubic units. What is the larger solid’s volume?

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Recognize the structure. Take a square root to recover the length scale factor, then cube it for volume.

Work it through. If k is the smaller-to-larger length factor, then

$\frac{225}{100}$

Thus $k 3 = 27/8,$ and the larger volume is $80(27/8) = 270.$

Answer: 270 cubic units.

Avoid the trap. Multiplying 80 by 225/100 uses an area ratio for a volume. The common length factor is the bridge between them.

6.15. Conserve volume and reconcile units

A solid metal cylinder has radius 6 centimeters and height 0.12 meter. It is melted and recast into solid spheres of radius 3 centimeters, with no material loss. How many complete spheres can be made?

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Recognize the structure. Convert the height to centimeters, then divide the total volume by the volume of one sphere.

Work it through. The cylinder’s height is 12 centimeters. Its volume is $π(6) 2 (12) = 432π$ cubic centimeters. Each sphere has volume $\frac{4}{3}$

The number of spheres is $432π/(36π) = 12.$

Answer: 12 spheres.

Avoid the trap. This is a recasting problem, not a packing problem: only volume conservation matters. Surface area changes, and the original cylinder need not physically fit the spheres inside it.

Mastery check and error prevention

Identify the unit type before calculating. Circle the perpendicular height in a figure. Trace each exposed boundary or surface rather than relying on a remembered picture. For scaling questions, write the length factor k first and label the requested quantity as one-, two-, or three-dimensional.

When a result seems unreasonable, test an easier special case. If the radius of a cylinder doubles while its height stays fixed, its volume should multiply by 4; if every dimension doubles, volume should multiply by 8. This distinction reveals whether you have assumed similarity without justification.