Area & Volume

Study Sheet

Area & Volume

Flat shapes, solid shapes, and the formulas that measure them

Perimeter & Area of Polygons

Tip
b_1b_2h

Two rules to carry through everything:

  • [leftmargin=6mm]
  • Perimeter is a length, so its units are plain: cm\text{cm}, m\text{m}, in\text{in}. Area is a covering, so its units are squared: cm2\text{cm}^2. Volume is a filling, so its units are cubed: cm3\text{cm}^3.
  • Whenever a formula uses π\pi, we will round with π3.14\pi \approx 3.14 and say so. Always keep the correct unit on your final answer.

A trapezoid has two parallel bases.

Concept
Formula Reference --- Flat Shapes

Let bb be a base and hh the matching perpendicular height.

  • [leftmargin=6mm]
  • Rectangle (length \ell, width ww):   P=2+2wP = 2\ell + 2w,   A=wA = \ell w.
  • Triangle:   P=a+b+cP = a+b+c (add the three sides),   A=12bhA = \tfrac{1}{2}bh.
  • Parallelogram:   P=2a+2bP = 2a + 2b,   A=bhA = bh.
  • Trapezoid (parallel sides b1b_1 and b2b_2):   A=12(b1+b2)hA = \tfrac{1}{2}(b_1+b_2)\,h.

The height is always measured straight across, at a right angle to the base --- never along a slanted side.

Example
Rectangle: perimeter and area

For this rectangle, =8 cm\ell = 8\text{ cm} and w=5 cmw = 5\text{ cm}:

P=2(8)+2(5)=26 cm,A=8×5=40 cm2.P = 2(8) + 2(5) = 26\text{ cm}, \qquad A = 8 \times 5 = 40\text{ cm}^2.
Example
Triangle and parallelogram

Triangle with base b=6b = 6 and height h=4h = 4:

A=12(6)(4)=12 cm2.A = \tfrac{1}{2}(6)(4) = 12\text{ cm}^2.

Parallelogram with base b=10b = 10 and height h=4h = 4:

A=(10)(4)=40 cm2.A = (10)(4) = 40\text{ cm}^2.
Example
Trapezoid

The two parallel sides are b1=6b_1 = 6 and b2=10b_2 = 10; the height between them is h=4h = 4:

A=12(b1+b2)h=12(6+10)(4)=12(16)(4)=32 cm2.A = \tfrac{1}{2}(b_1 + b_2)\,h = \tfrac{1}{2}(6 + 10)(4) = \tfrac{1}{2}(16)(4) = 32\text{ cm}^2.
Tip

Tip: The 12\tfrac{1}{2} in the triangle and trapezoid formulas is not a coincidence --- a triangle is half of a parallelogram, and a trapezoid is the average of its two bases times the height.

Regular Polygons & Circles

Concept
Formula Reference --- Regular Polygons and Circles
  • [leftmargin=6mm]
  • Regular polygon (all sides and angles equal): the apothem aa is the distance from the center straight out to the middle of a side. With perimeter PP, @@BLOCK0@@
  • Circle with radius rr (and diameter d=2rd = 2r): @@BLOCK1@@
  • Sector (a “pizza slice” with central angle θ\theta degrees): @@BLOCK2@@
Example
Regular hexagon using the apothem

A regular hexagon has side 66, so P=6×6=36P = 6 \times 6 = 36. Its apothem is about a5.2a \approx 5.2:

A=12(5.2)(36)=93.6 cm2.A = \tfrac{1}{2}(5.2)(36) = 93.6\text{ cm}^2.
Example
Circle: circumference, area, and a sector (with π3.14\pi \approx 3.14)

A circle has radius r=10r = 10:

C=2πr2(3.14)(10)=62.8 cm,A=πr23.14(10)2=314 cm2.C = 2\pi r \approx 2(3.14)(10) = 62.8\text{ cm}, \qquad A = \pi r^2 \approx 3.14(10)^2 = 314\text{ cm}^2.

A quarter of a circle of radius 66 (so θ=90\theta = 90^\circ) has area

90360πr214(3.14)(36)=14(113.04)=28.26 cm2.\frac{90}{360}\cdot \pi r^2 \approx \tfrac{1}{4}(3.14)(36) = \tfrac{1}{4}(113.04) = 28.26\text{ cm}^2.
Tip

Remember: A=12aPA = \tfrac{1}{2}aP works for any regular polygon. A circle is like a regular polygon with endlessly many tiny sides --- its “apothem” is the radius and its “perimeter” is the circumference, and 12r2πr=πr2\tfrac{1}{2}r\cdot 2\pi r = \pi r^2. Neat!

Composite (Compound) Figures

Concept
Break It Apart, Then Add or Subtract

A composite figure is made of familiar shapes joined together. To find its area:

  • [leftmargin=6mm]
  • Cut the figure into shapes you know (rectangles, triangles, half-circles, …).
  • Find each piece's area.
  • Add the pieces that are present; subtract any region that is cut out.

For perimeter, trace the outside edge only, and do not count any inner cut lines.

Example
A rectangle topped with a half-circle

This shape is a 6×46 \times 4 rectangle with a semicircle of radius 33 sitting on top. Add the two areas (π3.14\pi \approx 3.14):

Arect=6×4=24,Asemi=12πr212(3.14)(3)2=14.13.A_{\text{rect}} = 6 \times 4 = 24, \qquad A_{\text{semi}} = \tfrac{1}{2}\pi r^2 \approx \tfrac{1}{2}(3.14)(3)^2 = 14.13.
Atotal=24+14.13=38.13 cm2.A_{\text{total}} = 24 + 14.13 = 38.13\text{ cm}^2.
Tip

Tip: When a piece is removed (a hole, a bite, a corner cut off), you subtract its area. When a piece is added on, you add it. Always ask: “Is this region part of my shape, or missing from it?”

Surface Area of Solids

Concept
Formula Reference --- Surface Area
rh

Surface area (SA) is the total area of all the outside faces --- the amount of wrapping paper needed. It is always in square units.

  • [leftmargin=6mm]
  • Prism (base area BB, base perimeter PP, height hh):   SA=2B+PhSA = 2B + Ph.
  • Cube (edge ss):   SA=6s2SA = 6s^2 (six equal square faces).
  • Cylinder (radius rr, height hh):   SA=2πr2+2πrhSA = 2\pi r^2 + 2\pi r h.
  • Pyramid (base area BB, base perimeter PP, slant height \ell):   SA=B+12PSA = B + \tfrac{1}{2}P\ell.
  • Cone (radius rr, slant height \ell):   SA=πr2+πrSA = \pi r^2 + \pi r \ell.
  • Sphere (radius rr):   SA=4πr2SA = 4\pi r^2.

The slant height \ell runs up the slanted face of a pyramid or cone --- it is longer than the straight-up height.

A cylinder.

Example
Rectangular prism

A box measures =4\ell = 4, w=3w = 3, h=2h = 2. Add the areas of all six faces (opposite faces match):

SA=2(w+h+wh)=2(43+42+32)=2(12+8+6)=52 cm2.SA = 2(\ell w + \ell h + w h) = 2(4\cdot3 + 4\cdot2 + 3\cdot2) = 2(12 + 8 + 6) = 52\text{ cm}^2.
Example
Cylinder (with π3.14\pi \approx 3.14)

A can has radius r=3r = 3 and height h=10h = 10. The surface is two circular ends plus the wrapper:

SA=2πr2+2πrh2(3.14)(9)+2(3.14)(3)(10)=56.52+188.4=244.92 cm2.SA = 2\pi r^2 + 2\pi r h \approx 2(3.14)(9) + 2(3.14)(3)(10) = 56.52 + 188.4 = 244.92\text{ cm}^2.
Example
Pyramid and cone (using slant height \ell)

Square pyramid, base side 66 (so B=36B = 36, P=24P = 24), slant height =5\ell = 5:

SA=B+12P=36+12(24)(5)=36+60=96 cm2.SA = B + \tfrac{1}{2}P\ell = 36 + \tfrac{1}{2}(24)(5) = 36 + 60 = 96\text{ cm}^2.

Cone, radius r=3r = 3, slant height =5\ell = 5 (π3.14\pi \approx 3.14):

SA=πr2+πr3.14(9)+3.14(3)(5)=28.26+47.1=75.36 cm2.SA = \pi r^2 + \pi r \ell \approx 3.14(9) + 3.14(3)(5) = 28.26 + 47.1 = 75.36\text{ cm}^2.
Example
Sphere (with π3.14\pi \approx 3.14)

A ball has radius r=6r = 6:

SA=4πr24(3.14)(6)2=4(3.14)(36)=452.16 cm2.SA = 4\pi r^2 \approx 4(3.14)(6)^2 = 4(3.14)(36) = 452.16\text{ cm}^2.
Tip

Remember: For a prism, PhPh is the “label around the can” (the sides), and 2B2B is the two ends. For a pyramid or cone, use the slant height \ell for the slanted faces --- save the straight-up height for volume.

Volume of Solids

Concept
Formula Reference --- Volume
rh

Volume is how much space fills the inside --- always in cubic units.

  • [leftmargin=6mm]
  • Prism or cylinder:   V=BhV = Bh, where BB is the base area and hh the height. itemize
  • Rectangular prism: V=whV = \ell w h.   Cube: V=s3V = s^3.   Cylinder: V=πr2hV = \pi r^2 h.

Pyramid or cone:   V=13BhV = \tfrac{1}{3}Bh (exactly one-third of the matching prism or cylinder).

  • Cone: V=13πr2hV = \tfrac{1}{3}\pi r^2 h.

Sphere:   V=43πr3V = \tfrac{4}{3}\pi r^3. itemize

A cylinder.

Example
Prism, cylinder, cone, and sphere (π3.14\pi \approx 3.14)

Box 4×3×24 \times 3 \times 2:   V=wh=432=24 cm3.V = \ell w h = 4\cdot3\cdot2 = 24\text{ cm}^3.

Cylinder r=3r = 3, h=10h = 10:   V=πr2h3.14(9)(10)=282.6 cm3.V = \pi r^2 h \approx 3.14(9)(10) = 282.6\text{ cm}^3.

Cone r=3r = 3, h=4h = 4:   V=13πr2h13(3.14)(9)(4)=13(113.04)=37.68 cm3.V = \tfrac{1}{3}\pi r^2 h \approx \tfrac{1}{3}(3.14)(9)(4) = \tfrac{1}{3}(113.04) = 37.68\text{ cm}^3.

Sphere r=6r = 6:   V=43πr343(3.14)(216)=43(678.24)=904.32 cm3.V = \tfrac{4}{3}\pi r^3 \approx \tfrac{4}{3}(3.14)(216) = \tfrac{4}{3}(678.24) = 904.32\text{ cm}^3.

Example
Pyramid: one-third of its box

A square pyramid has base 6×66 \times 6 (so B=36B = 36) and height h=10h = 10:

V=13Bh=13(36)(10)=120 cm3.V = \tfrac{1}{3}Bh = \tfrac{1}{3}(36)(10) = 120\text{ cm}^3.

A full prism with that base and height would hold 360 cm3360\text{ cm}^3 --- the pyramid is exactly one-third of it.

Tip

Remember: A pyramid or cone always holds 13\tfrac{1}{3} of the prism or cylinder that shares its base and height. If you forget the 13\tfrac{1}{3}, your answer will be three times too big!

Scaling Dimensions & Real-World Problems

Concept
What Happens When You Resize a Shape

If you multiply every length of a figure by the same factor kk:

  • [leftmargin=6mm]
  • Perimeter (a length) is multiplied by kk.
  • Area is multiplied by k2k^2.
  • Volume is multiplied by k3k^3.

So doubling the sides (k=2k=2) makes area 4×4\times and volume 8×8\times bigger.

Example
Scaling a cube

A cube with edge 22 has volume 23=82^3 = 8. Triple every edge to 66: the new volume is 63=2166^3 = 216. That is 216÷8=27216 \div 8 = 27 times as large --- and sure enough, k3=33=27k^3 = 3^3 = 27.

Example
A real-world cost problem

A rectangular floor is 5 m5\text{ m} by 4 m4\text{ m}. Carpet costs $3\$3 per square meter. The area is 5×4=20 m25 \times 4 = 20\text{ m}^2, so the cost is

20×$3=$60.20 \times \$3 = \$60.
Tip

Tip: Real-world word problems are just formula problems in a costume. Read carefully to spot which measurement you need --- area for painting or carpeting, volume for filling or capacity --- then round money sensibly and keep the units.

Going Deeper: Advanced Area & Volume

Concept
Heron's Formula --- Area from the Three Sides
bh

Sometimes you know a triangle's three sides but not its height. Heron's formula finds the area directly. First compute the semiperimeter (half the perimeter):

s=a+b+c2.s = \frac{a+b+c}{2}.

Then the area is

A=s(sa)(sb)(sc).A = \sqrt{\,s\,(s-a)(s-b)(s-c)\,}.

This always works for a valid triangle, and it never needs a perpendicular height.

Half of a rectangle with the same base and height.

Example
Worked example: Heron's formula on a 13--14--15 triangle

A triangle has sides a=13a = 13, b=14b = 14, c=15c = 15. First the semiperimeter:

s=13+14+152=422=21.s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21.

Now the differences are sa=8s-a = 8, sb=7s-b = 7, sc=6s-c = 6, so

A=21876=7056=84 cm2.A = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84\text{ cm}^2.
Concept
From Coordinates: the Shoelace Formula & Pick's Theorem

When a polygon is given by the coordinates of its corners, two tools find its area without any height at all.

Shoelace formula. List the vertices in order (going around once) as (x1,y1),(x2,y2),,(xn,yn)(x_1,y_1),(x_2,y_2),\dots,(x_n,y_n), then loop back to the first. Multiply “down the diagonals” and subtract:

A=12i=1n(xiyi+1xi+1yi).A = \tfrac{1}{2}\left| \sum_{i=1}^{n} \big(x_i\,y_{i+1} - x_{i+1}\,y_i\big) \right|.

It is called the shoelace because the cross-multiplications criss-cross like laces.

Pick's theorem. If every corner sits on a grid (lattice) point, count II, the interior grid points, and BB, the grid points on the boundary. Then

A=I+B21.A = I + \frac{B}{2} - 1.

Two very different counts, one exact area --- and they must agree.

Example
Worked example: shoelace and Pick's theorem agree

Take the lattice triangle with corners (0,0)(0,0), (4,0)(4,0), and (2,3)(2,3).

Shoelace (list the corners, wrap back to the start):

A=12(x1y2x2y1)+(x2y3x3y2)+(x3y1x1y3).A = \tfrac{1}{2}\big| (x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2) + (x_3 y_1 - x_1 y_3) \big|.
A=12(0040)+(4320)+(2003)=120+12+0=6 units2.A = \tfrac{1}{2}\big| (0\cdot 0 - 4\cdot 0) + (4\cdot 3 - 2\cdot 0) + (2\cdot 0 - 0\cdot 3) \big| = \tfrac{1}{2}|0 + 12 + 0| = 6\text{ units}^2.

Pick's theorem. The boundary points are the 33 corners plus the 33 extra grid points along the bottom edge from (0,0)(0,0) to (4,0)(4,0) --- the slanted edges pass through no grid points --- so B=6B = 6. The interior points are (1,1),(2,1),(3,1),(2,2)(1,1),(2,1),(3,1),(2,2), so I=4I = 4:

A=I+B21=4+621=4+31=6 units2.A = I + \frac{B}{2} - 1 = 4 + \frac{6}{2} - 1 = 4 + 3 - 1 = 6\text{ units}^2.

Both methods give 66.

Concept
Combined Solids --- Deriving SA and Volume

A combined (composite) solid is glued from familiar solids: a cylinder capped by a hemisphere, an ice-cream cone (cone ++ hemisphere), a house (prism ++ pyramid), and so on. Treat it like a composite figure, but in 33D:

  • [leftmargin=6mm]
  • Volume: add the volumes of the separate pieces. Nothing is lost where they join.
  • Surface area: add only the surfaces that are exposed on the outside. The circle (or face) where two pieces meet is inside the solid --- do not count it.

So for a cone stacked on a hemisphere of the same radius, the exposed surface is the cone's slant side πr\pi r \ell plus the hemisphere's curved half 2πr22\pi r^2 --- neither flat base circle is counted, because they are hidden at the join.

Example
Worked example: cone on a hemisphere (π3.14\pi \approx 3.14)

A solid is a cone (radius r=3r = 3, height h=4h = 4, so slant height =5\ell = 5) sitting on a hemisphere of the same radius r=3r = 3.

Volume == cone ++ hemisphere:

V=13πr2h+23πr313(3.14)(9)(4)+23(3.14)(27)=37.68+56.52=94.2 cm3.V = \tfrac{1}{3}\pi r^2 h + \tfrac{2}{3}\pi r^3 \approx \tfrac{1}{3}(3.14)(9)(4) + \tfrac{2}{3}(3.14)(27) = 37.68 + 56.52 = 94.2\text{ cm}^3.

Surface area == cone's slant side ++ hemisphere's curved half (the shared circle is hidden):

SA=πr+2πr2(3.14)(3)(5)+2(3.14)(9)=47.1+56.52=103.62 cm2.SA = \pi r \ell + 2\pi r^2 \approx (3.14)(3)(5) + 2(3.14)(9) = 47.1 + 56.52 = 103.62\text{ cm}^2.
Concept
Cross-Sections, Nets, and the Scaling Laws

Cross-section: the flat shape you see when you slice straight through a solid. Slicing a cylinder horizontally gives a circle; slicing it vertically through the middle gives a rectangle. Slicing a cube can give a square, a rectangle, or even a triangle or hexagon, depending on the angle.

Net: a solid unfolded flat, showing every face at once. A cube's net is 66 squares; a cylinder's net is two circles plus a rectangle whose width equals the circumference 2πr2\pi r. A net is the fastest way to see why a surface-area formula adds up --- the total area of the net is the surface area.

Scaling: multiply every length by a factor kk and lengths grow by kk, areas by k2k^2, and volumes by k3k^3. A subtle consequence: the surface-area-to-volume ratio scales like k2/k3=1/kk^2/k^3 = 1/k, so bigger objects have relatively less surface --- which is why large animals hold heat and small ones lose it fast.

Example
Worked example: an area optimization

A farmer has 40 m40\text{ m} of fencing for a rectangular pen. Which rectangle encloses the most area? With perimeter fixed at 4040, the width ww and length \ell satisfy +w=20\ell + w = 20, so try a few:

The area is largest when =w=10\ell = w = 10 --- a square. Among all rectangles of a given perimeter, the square always wins. (The same pattern holds in 33D: of all rectangular boxes with a fixed surface area, the cube holds the most volume.)

Tip

Big picture: Every advanced tool here is a shortcut around a missing height. Heron uses the three sides; the shoelace and Pick's theorem use coordinates; combined solids reuse the basic formulas piece by piece. And whenever a length is scaled by kk, remember the ladder: length k\to k, area k2\to k^2, volume k3\to k^3.

Formulas, Proofs & Tips

Tip
Area formulas
Arect=bh,A=12bh,Apar=bh,Atrap=12(b1+b2)h,A=πr2A_{\text{rect}}=bh,\quad A_{\triangle}=\tfrac12 bh,\quad A_{\text{par}}=bh,\quad A_{\text{trap}}=\tfrac12(b_1+b_2)h,\quad A_{\odot}=\pi r^{2}

What it means. Every one of these is really "base times height", adjusted.

Example. A triangle with base 66 and height 44 has area 1264=12\tfrac12\cdot6\cdot4=12.

Why it works. A parallelogram becomes a rectangle when you cut a triangle off one end and slide it to the other, so its area is bhbh. A triangle is half a parallelogram (two copies make one), giving 12bh\tfrac12 bh. Two copies of a trapezoid form a parallelogram of base b1+b2b_1+b_2, giving 12(b1+b2)h\tfrac12(b_1+b_2)h.

Tip. The height must be perpendicular to the base, not a slanted side. In an obtuse triangle the height can fall outside the triangle.

Tip
Volume and surface area
Vprism=Bh,Vcyl=πr2h,Vcone=13πr2h,Vsph=43πr3,Ssph=4πr2V_{\text{prism}}=Bh,\quad V_{\text{cyl}}=\pi r^2h,\quad V_{\text{cone}}=\tfrac13\pi r^2h,\quad V_{\text{sph}}=\tfrac43\pi r^3,\quad S_{\text{sph}}=4\pi r^2

What it means. Prisms and cylinders are "base area times height"; a cone is one third of its cylinder.

Example. A cylinder with r=2r=2, h=5h=5 has volume π225=20π\pi\cdot2^2\cdot5=20\pi.

Why it works. A prism stacks copies of its base hh high, giving BhBh. Filling a cone and pouring it into the matching cylinder takes exactly three cones — the 13\tfrac13 factor, which calculus confirms by integrating the cross-sections.

Tip. Watch the units: area is squared (cm2\text{cm}^2), volume is cubed (cm3\text{cm}^3).

Tip
Heron's formula
A=s(sa)(sb)(sc),s=a+b+c2A=\sqrt{s(s-a)(s-b)(s-c)},\qquad s=\frac{a+b+c}{2}

What it means. The area of a triangle from its three sides alone — no angle or height needed.

Example. Sides 3,4,53,4,5: s=6s=6, so A=6321=6A=\sqrt{6\cdot 3\cdot 2\cdot 1}=6.

Why it works. Start from A=12absinCA=\tfrac12 ab\sin C, replace sinC\sin C with 1cos2C\sqrt{1-\cos^2C}, and substitute cosC\cos C from the Law of Cosines. The algebra factors into the four bracketed terms.

Tip. ss is the SEMI-perimeter — half the perimeter. Forgetting the halving is the usual slip.