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Area, Perimeter and Volume with Algebra

Turn lengths involving algebra into perimeter, area and volume equations, choose valid dimensions, and keep exact answers where possible.

Translate the shape into an equation

Start by deciding whether the question concerns a boundary length, an area, a surface area or a volume. Label the dimensions, choose the formula and substitute the algebraic lengths before simplifying. A diagram need not be to scale.

Use consistent units. Perimeter is measured in units such as cm, area in cm² and volume in cm³. After solving, reject values that make any required length zero or negative. Keep surds and multiples of \pi exact unless a rounded answer is requested.

Useful area and volume formulas

Rectangle area

lw

Parallelogram area

bh

Triangle area

\frac12bh

Trapezium area, for parallel sides a,b:

\frac12(a+b)h

Circle area

\pi r^2

Circle circumference

2\pi r

Heights must be perpendicular.

Prism volume

\text{cross-sectional area}\times\text{length}

Cylinder volume

\pi r^2h

Pyramid volume

\frac13\times\text{base area}\times h

Cone volume

\frac13\pi r^2h

Sphere volume

\frac43\pi r^3

Cylinder curved area

2\pi rh

Cone curved area, with slant height l:

\pi rl

Sphere surface area

4\pi r^2

Add any exposed flat faces separately.

Worked example 1

A rectangle has side lengths (3x+2) cm and (x-1) cm and perimeter 42 cm. Find x and the area.

  1. 2(3x+2)+2(x-1)=42, so 8x+2=42

  2. Your turn. Find x

    x=5, giving sides 17 cm and 4 cm.

  3. Answer: x=5 and area 17\times4=68 cm². Both side lengths are positive.

Worked example 2

A trapezium has parallel sides (x+2) cm and (3x-2) cm, perpendicular height x cm and area 72 cm². Find its dimensions.

  1. \frac12[(x+2)+(3x-2)]x=72, which simplifies to 2x^2=72

  2. Your turn. Find the positive value of x

    x^2=36, so x=6. The value -6 would give negative lengths.

  3. Answer: parallel sides 8 cm and 16 cm, height 6 cm. Check: \frac12(8+16)6=72 cm².

Worked example 3

A rectangle of width 2x cm and height (x+3) cm has a semicircle attached along its top edge, with that edge as the diameter. Find expressions for the outer perimeter and total area, then evaluate the area at x=2.

  2. The outer perimeter is 2(x+3)+2x+\pi x=(4+\pi)x+6 cm. Count two vertical sides, the bottom edge and the semicircular arc.
  3. Total area =2x(x+3)+\frac12\pi x^2 cm². The shared diameter does not remove any area.

  4. Your turn. At x=2, what is the rectangular part’s area in cm²?

    2(2)(2+3)=20 cm².

  5. Answer: perimeter (4+\pi)x+6 cm; area 2x(x+3)+\frac12\pi x^2 cm²; at x=2, area 20+2\pi cm².

Worked example 4

A closed cylinder has radius x cm and height 4x cm. Its volume is 108\pi cm³. Find its dimensions and total surface area.

  1. \pi x^2(4x)=108\pi, so 4x^3=108 after cancelling the non-zero factor \pi

  2. Your turn. Find x, given x^3=27

    The radius is 3 cm and the height is 12 cm.

  3. Total surface area includes both ends: 2\pi r^2+2\pi rh=2\pi(3^2)+2\pi(3)(12)=90\pi cm².
  4. Answer: radius 3 cm, height 12 cm and total surface area 90\pi cm².

Worked example 5

A cone has radius 3x cm and perpendicular height 4x cm, where x>0. Its total surface area, including its base, is 216\pi cm². Find x and its volume.

  2. The slant height is l=\sqrt{(3x)^2+(4x)^2}=5x. Total surface area is \pi(3x)(5x)+\pi(3x)^2=24\pi x^2

  3. Your turn. Solve 24\pi x^2=216\pi for positive x

    x^2=9, so x=3

  4. The radius is 9 cm and perpendicular height is 12 cm. Volume =\frac13\pi(9^2)(12)=324\pi cm³.
  5. Answer: x=3 and volume 324\pi cm³. Surface area uses the slant height; volume uses the perpendicular height.

Worked example 6

A right square-based pyramid has base side x cm and perpendicular height 6 cm. The apex is directly above the base centre and its volume is 32 cm³. Find its total surface area exactly.

  1. \frac13 x^2(6)=32, so x^2=16 and x=4 cm.

  3. Your turn. The face height is s. Calculate s^2=6^2+2^2

    s=\sqrt{40}=2\sqrt{10} cm.

  4. The base area is 16 cm². Each triangular face has area \frac12(4)(2\sqrt{10})=4\sqrt{10} cm². There are four such faces.
  5. Answer: total surface area 16+16\sqrt{10} cm². The face height is not the perpendicular height of the pyramid.

Worked example 7

A sphere of radius 3x cm has the same volume as a cylinder of radius 2x cm and height 9 cm, where x>0. Find x and the common volume.

  1. \frac43\pi(3x)^3=\pi(2x)^2(9), so 36\pi x^3=36\pi x^2

  2. Your turn. Divide by 36\pi x^2, which is non-zero because x>0. Find x

    x=1

  3. Answer: x=1 and common volume 36\pi cm³. The algebraic value x=0 is excluded by the positive-length condition.

Common mistakes

  • Using the diameter as the radius. Halve the diameter before using a radius formula.
  • Counting internal edges in a perimeter. Trace only the outside boundary of the composite shape.
  • Counting the wrong surface faces. Include all exposed faces, but exclude covered or joined faces.
  • Confusing slant and perpendicular heights. Match the height to the formula: volume needs the perpendicular height.
  • Forgetting to square the coefficient. Square every factor: (3x)^2=9x^2.
  • Accepting impossible lengths or wrong units. Check lengths are positive; use squared units for area and cubed units for volume.

Now try it: Area, Perimeter and Volume with Algebra practice questions

More on this topic: Area, Perimeter and Volume with Algebra worksheet with full solutions