Area & Volume

Study Sheet

Area & Volume

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

Perimeter & Area of Polygons

Tip

Two rules to carry through everything:

  • 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 π≈3.14\pi \approx 3.14 and say so. Always keep the correct unit on your final answer.

Area formulas are built from the rectangle by cutting and rearranging: a parallelogram is a sheared rectangle, a triangle is half a parallelogram, a trapezoid is the average of its bases times the height. Volumes of prisms and cylinders are base area times height; pyramids and cones are one third of that; the sphere is 43πr3\tfrac43\pi r^3.

Concept
Formula Reference — Flat Shapes
b = 10h = 4
Slide the shaded triangle from the left end to the right: the parallelogram becomes a 10 by 4 rectangle.

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

  • Rectangle (length ℓ\ell, width ww):   P=2ℓ+2wP = 2\ell + 2w,   A=ℓwA = \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.

Every formula here comes from the rectangle by cutting and moving pieces, which never changes area. Cut a right triangle off one end of a parallelogram and slide it to the other end: you get a rectangle with the same base and the same perpendicular height, so A=bhA = bh. Two copies of a triangle make a parallelogram, so a triangle is 12bh\tfrac12 bh. Two copies of a trapezoid, one turned upside down, make a parallelogram with base b1+b2b_1 + b_2, so a trapezoid is 12(b1+b2)h\tfrac12(b_1 + b_2)h.

Example
Rectangle: perimeter and area
8 cm5 cm
40 unit squares fit inside.

For this rectangle, ℓ=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.
Side note
Where it goes wrong

Forgetting the 13\tfrac13 for pyramids and cones, or using the slant height where the perpendicular height belongs in a volume formula (slant height belongs to surface area).

Example
Triangle and parallelogram
64104
Heights are the dashed perpendiculars, not the slanted sides.

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
1064
Parallel sides 6 and 10, height 4.

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.

Try it
Quick check

A triangle has area 42 cm242\text{ cm}^2 and base 1212 cm. Find its height.

Show the answer

12⋅12⋅h=42\tfrac12 \cdot 12 \cdot h = 42, so 6h=426h = 42 and h=7h = \mathbf{7} cm.

Regular Polygons & Circles

Concept
Formula Reference — Regular Polygons and Circles
  • 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@@

Why 12aP\tfrac12 aP: lines from the center to the vertices cut a regular nn-gon into nn congruent isosceles triangles, each with a side ss as base and the apothem aa as height. Their total area is n⋅12sa=12a(ns)=12aPn \cdot \tfrac12 s a = \tfrac12 a (ns) = \tfrac12 aP. The circle formula is the same statement for a polygon with ever more, ever shorter sides.

Example
Regular hexagon using the apothem
6a ≈ 5.2
The apothem runs from the center to the middle of a side.

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

A=12(5.2)(36)=93.6 cm2.A = \tfrac{1}{2}(5.2)(36) = 93.6\text{ cm}^2.
Side note
Where it goes wrong

Using the radius (center to a vertex) as the apothem (center to the middle of a side). For the hexagon with side 66 the radius is 66 and the apothem is about 5.25.2; 12⋅6⋅36=108\tfrac12 \cdot 6 \cdot 36 = 108 overestimates the area.

Example
Circle: circumference, area, and a sector (with π≈3.14\pi \approx 3.14)
106
A circle of radius 10, and a quarter of a circle of radius 6.

A circle has radius r=10r = 10:

C=2πr≈2(3.14)(10)=62.8 cm,A=πr2≈3.14(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

90360⋅πr2≈14(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 12r⋅2πr=πr2\tfrac{1}{2}r\cdot 2\pi r = \pi r^2. Neat!

Try it
Quick check

A regular octagon has side 1010 cm and apothem about 12.0712.07 cm. Find its area.

Show the answer

Perimeter P=8⋅10=80P = 8 \cdot 10 = 80 cm. Area =12aP≈12(12.07)(80)=482.8 cm2= \tfrac12 aP \approx \tfrac12 (12.07)(80) = \mathbf{482.8\ cm^2}.

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:

  • 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.

This works because area is additive: if a region is split into pieces that do not overlap, its area is the sum of theirs. The only care needed is that the pieces really do not overlap and really do fill the figure — which is why it pays to draw the cut lines (dashed) before computing anything.

Example
A rectangle topped with a half-circle
643
The dashed line splits the shape into a rectangle and a semicircle.

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

Arect=6×4=24,Asemi=12πr2≈12(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.
Example
A square with a hole
103
Shaded: the tile; the circle is the hole.

A 10 cm10\text{ cm} square tile has a circular hole of radius 3 cm3\text{ cm} cut from its center. Find the area that remains (π≈3.14\pi \approx 3.14).

The hole is missing from the shape, so subtract it:

A=102−π(3)2≈100−3.14(9)=100−28.26=71.74 cm2.A = 10^2 - \pi(3)^2 \approx 100 - 3.14(9) = 100 - 28.26 = 71.74\text{ cm}^2 .
Side note
Where it goes wrong

Counting inner cut lines in the perimeter. For the rectangle-plus-semicircle, the dashed line across the top of the rectangle is not part of the boundary; the perimeter is 6+4+4+12(2π⋅3)≈23.426 + 4 + 4 + \tfrac12(2\pi \cdot 3) \approx 23.42, not that plus 66.

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?”

Try it
Quick check

An L-shape is a 1010 cm by 88 cm rectangle with a 44 cm by 33 cm rectangle cut from one corner. Find its area and its perimeter.

Show the answer

Area: 80−12=68 cm280 - 12 = \mathbf{68\ cm^2}. Perimeter: the notch removes 4+34 + 3 of the outer edge but adds its two inner edges, 4+34 + 3, back, so the perimeter equals the full rectangle's: 2(10+8)=362(10 + 8) = \mathbf{36} cm.

Surface Area of Solids

Concept
Formula Reference — Surface Area
4 × 24 × 34 × 34 × 23×23×2
Net of the 4 by 3 by 2 box: six rectangles, total 52.

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

  • 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 ℓ\ell):   SA=B+12PℓSA = B + \tfrac{1}{2}P\ell.
  • Cone (radius rr, slant height ℓ\ell):   SA=πr2+πrℓSA = \pi r^2 + \pi r \ell.
  • Sphere (radius rr):   SA=4πr2SA = 4\pi r^2.

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

Surface area is the area of the net — the solid unfolded flat. A prism unfolds into its two bases and a band of rectangles whose combined width is the base perimeter, hence 2B+Ph2B + Ph. A cylinder's band is one rectangle, 2πr2\pi r wide (the circumference) and hh tall. A pyramid unfolds into its base and triangles whose height is the slant height ℓ\ell; a cone's side unrolls into a sector whose area works out to πrℓ\pi r \ell.

Example
Rectangular prism
432
Three pairs of opposite faces: 12, 8 and 6.

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

SA=2(ℓw+ℓh+wh)=2(4⋅3+4⋅2+3⋅2)=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 π≈3.14\pi \approx 3.14)
310
Two circles of radius 3 and a wrapper 2π⋅\pi \cdot3 wide and 10 tall.

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

SA=2πr2+2πrh≈2(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 ℓ\ell)
6l = 53l = 5
Slant heights run down the faces; both solids here are 4 tall.

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

SA=B+12Pℓ=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 ℓ=5\ell = 5 (π≈3.14\pi \approx 3.14):

SA=πr2+πrℓ≈3.14(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 π≈3.14\pi \approx 3.14)
6
Radius 6: the surface is four great circles' worth, 4π⋅\pi \cdot36.

A ball has radius r=6r = 6:

SA=4πr2≈4(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.
Side note
Where it goes wrong

Leaving out faces or counting a hidden one. A closed can has two circular ends (2πr22\pi r^2); an open-top can has one. Read whether the solid is closed before adding up the net.

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 ℓ\ell for the slanted faces — save the straight-up height for volume.

Try it
Quick check

Find the surface area of a closed cylinder with radius 22 and height 55. Leave π\pi in the answer.

Show the answer

Ends: 2π(2)2=8π2\pi(2)^2 = 8\pi. Wrapper: 2π(2)(5)=20π2\pi(2)(5) = 20\pi. Total 28π≈87.96\mathbf{28\pi} \approx 87.96.

Volume of Solids

Concept
Formula Reference — Volume

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

  • Prism or cylinder:   V=BhV = Bh, where BB is the base area and hh the height. itemize
  • Rectangular prism: V=ℓwhV = \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

V=BhV = Bh for prisms and cylinders because the solid is a stack of identical slices, each a copy of the base; hh layers of BB unit cubes. Pyramids and cones taper to a point, and the slice at each level is a scaled copy of the base with the scale shrinking steadily to 00; adding those shrinking slices gives exactly one third of the full stack (see Why It Works). Cavalieri's principle says only the slice areas matter, so a slanted (oblique) prism or cylinder has the same BhBh.

Example
Prism, cylinder, cone, and sphere (π≈3.14\pi \approx 3.14)
4×3×2r 3, h 10r 3, h 4r 6
The four solids of the example, on one scale.

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

Cylinder r=3r = 3, h=10h = 10:   V=πr2h≈3.14(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=13πr2h≈13(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=43πr3≈43(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
610
The pyramid sits inside the 6 by 6 by 10 prism with the same base and height.

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.

Side note
Where it goes wrong

Using the slant height in a volume. For a cone of radius 33 and slant height 55, the height is 44 (a 33-44-55 triangle), and V=13π(9)(4)=12πV = \tfrac13\pi(9)(4) = 12\pi — not 13π(9)(5)=15π\tfrac13\pi(9)(5) = 15\pi.

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!

Try it
Quick check

A cylinder and a cone both have radius 44 and height 99. Find both volumes, leaving π\pi in the answers.

Show the answer

Cylinder: π(4)2(9)=144π\pi(4)^2(9) = \mathbf{144\pi}. Cone: one third of that, 48π\mathbf{48\pi}.

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:

  • 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.

Count the factors of kk: a length is one length, so it scales by kk; an area is a product of two lengths, each scaled, so k⋅kk \cdot k; a volume multiplies three, k3k^3. The cube in the example shows it literally — tripling the edge fits 33 small cubes along each of the three directions.

Example
Scaling a cube
26
Triple the edge: 3 ×\times 3 ×\times 3 = 27 small cubes fill the big one.

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.

Side note
Where it goes wrong

Tripling the volume when the edges triple. The new cube is not three small cubes but 2727; the paint for its surface is 99 times as much, not 33 times.

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.

Try it
Quick check

Every length of a solid is multiplied by 44. By what factor do its surface area and volume change? If the original volume was 5 cm35\text{ cm}^3, what is the new volume?

Show the answer

Surface area ×42=16\times 4^2 = \mathbf{16}; volume ×43=64\times 4^3 = \mathbf{64}. New volume 5⋅64=320 cm35 \cdot 64 = \mathbf{320\ cm^3}.

Going Deeper: Advanced Area & Volume

Concept
Heron's Formula — Area from the Three Sides
1413151259
A check on Heron: the height to the side 14 is 12 (triangles 5-12-13 and 9-12-15).

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 (s−a)(s−b)(s−c) .A = \sqrt{\,s\,(s-a)(s-b)(s-c)\,}.

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

Where it comes from: the Law of Cosines gives cos⁡C\cos C from the three sides, the identity sin⁡2C=1−cos⁡2C\sin^2 C = 1 - \cos^2 C gives sin⁡C\sin C, and the area is 12absin⁡C\tfrac12 ab\sin C. Factoring the result as a difference of squares twice produces s(s−a)(s−b)(s−c)s(s - a)(s - b)(s - c). For the 1313-1414-1515 triangle the figure confirms it the ordinary way: the height to the base 1414 is 1212, and 12⋅14⋅12=84\tfrac12 \cdot 14 \cdot 12 = 84.

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 s−a=8s-a = 8, s−b=7s-b = 7, s−c=6s-c = 6, so

A=21⋅8⋅7⋅6=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=12∣∑i=1n(xi yi+1−xi+1 yi)∣.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+B2−1.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
xy−11234−1123
Solid dots: B = 6 boundary points. Open dots: I = 4 interior points.

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∣(x1y2−x2y1)+(x2y3−x3y2)+(x3y1−x1y3)∣.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∣(0⋅0−4⋅0)+(4⋅3−2⋅0)+(2⋅0−0⋅3)∣=12∣0+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+B2−1=4+62−1=4+3−1=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:

  • 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 πrℓ\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 (π≈3.14\pi \approx 3.14)
345
The shared circle (dashed) is inside the solid, so it is not surface.

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

Volume == cone ++ hemisphere:

V=13πr2h+23πr3≈13(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=πrℓ+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.
Side note
Where it goes wrong

Counting the hidden circle. Adding the base of the cone, πr2≈28.26\pi r^2 \approx 28.26, gives 131.88 cm2131.88\text{ cm}^2 — but that disk is glued to the hemisphere inside the solid, and no paint reaches it. The surface is 103.62 cm2103.62\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
16 × 412 × 810 × 10
Three pens with perimeter 40: areas 64, 96 and 100.

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 length ℓ\ell and width ww satisfy ℓ+w=20\ell + w = 20, so try a few:

ℓ (m)w (m)Area=ℓw (m2)164641557512896119991010100\begin{array}{c|c|c} \ell\ (\text{m}) & w\ (\text{m}) & \text{Area} = \ell w\ (\text{m}^2) \\ \hline 16 & 4 & 64 \\ 15 & 5 & 75 \\ 12 & 8 & 96 \\ 11 & 9 & 99 \\ 10 & 10 & \mathbf{100} \end{array}

The area is largest when ℓ=w=10\ell = w = 10 — a square. Algebra shows why: with ℓ=10+d\ell = 10 + d and w=10−dw = 10 - d, the area is (10+d)(10−d)=100−d2(10 + d)(10 - d) = 100 - d^2, which is largest when d=0d = 0. (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.

Try it
Quick check

Find the surface area of a sphere with radius 10 cm10\text{ cm} (π≈3.14\pi \approx 3.14).

Work it on paper before revealing the answer.

Show the answer

Answer: SA=4(3.14)(100)=1256 cm2SA = 4(3.14)(100) = 1256\text{ cm}^2.

Why It Works

Theorem
The reason behind the rule
One of six pyramids that meet at the cube's center; each has a face as its base.

Why 13\tfrac13? Join the center of a cube of edge ss to all eight corners. This cuts the cube into six identical pyramids, one on each face, each with base area s2s^2 and height s2\tfrac{s}{2} (center to face). Each is one sixth of the cube: s36=13⋅s2⋅s2\tfrac{s^3}{6} = \tfrac13 \cdot s^2 \cdot \tfrac{s}{2} — exactly 13\tfrac13 base ×\times height. Cavalieri's principle (solids with equal cross-sections at every height have equal volumes) then extends the factor 13\tfrac13 to every pyramid and cone. The sphere's 43πr3\tfrac43\pi r^3 comes from comparing its cross-sections with a cylinder minus a cone.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

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 12⋅6⋅4=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 π⋅22⋅5=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(s−a)(s−b)(s−c),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=6⋅3⋅2⋅1=6A=\sqrt{6\cdot 3\cdot 2\cdot 1}=6.

Why it works. Start from A=12absin⁡CA=\tfrac12 ab\sin C, replace sin⁡C\sin C with 1−cos⁡2C\sqrt{1-\cos^2C}, and substitute cos⁡C\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.