Formula sheet 2026–27

Geometry formula sheet

Areas and perimeters of plane figures, and $A_B$, $A_L$, $A_T$ and volume of solids. Also as a PDF to print.

Plane figures: area and perimeter

$A$: area · $P$: perimeter (the sum of all the sides).

Square

$A = c^2$

$P = 4c$

Rectangle

$A = b \cdot h$

$P = 2b + 2h$

Triangle

$A = \dfrac{b \cdot h}{2}$

$P = \text{sum of the 3 sides}$

Rhombus

$A = \dfrac{D \cdot d}{2}$

$P = 4c$

Parallelogram

$A = b \cdot h$

$P = \text{sum of the sides}$

Trapezium

$A = \dfrac{(B + b) \cdot h}{2}$

$P = \text{sum of the sides}$

Regular polygon

$P = n \cdot c$

$A = \dfrac{P \cdot a}{2}$

$n$: number of sides · $a$: apothem · $r$: radius. In a regular hexagon the side equals the radius, $c = r$, and the apothem comes from Pythagoras: $a^2 = r^2 - \left(\tfrac{c}{2}\right)^2$.

Circle

$A = \pi r^2$

$L = 2\pi r$

$L$: length of the circumference

Circular sector

$A = \dfrac{\pi r^2 \cdot \alpha}{360^\circ}$

$L_{\text{arc}} = \dfrac{2\pi r \cdot \alpha}{360^\circ}$

$P = L_{\text{arc}} + 2r$

Annulus

$A = \pi \left(R^2 - r^2\right)$

$P = 2\pi R + 2\pi r$

Pythagoras' theorem

$a^2 = b^2 + c^2$

In a right triangle, $a$ is the hypotenuse.

Solids: area and volume

$A_B$: base area · $A_L$: lateral area · $A_T$: total area · $P_B$: perimeter of the base · $V$: volume.

Prism

$A_B = \text{area of the base polygon}$

$A_L = P_B \cdot h$

$A_T = 2A_B + A_L$

$V = A_B \cdot h$

Cube with edge $c$: $A_T = 6c^2$, $V = c^3$.

Pyramid

$A_B = \text{area of the base polygon}$

$A_L = n \cdot A_{\triangle}$

$A_L = \dfrac{P_B \cdot a_p}{2}$

$A_T = A_B + A_L$

$V = \dfrac{A_B \cdot h}{3}$

The side faces are $n$ equal triangles. 1st way: $A_L = n \cdot A_{\triangle}$, with $A_{\triangle} = \tfrac{c \cdot a_p}{2}$ (base $c$, side of the polygon; height $a_p$, slant height). 2nd way: since $n \cdot c = P_B$, altogether $A_L = \tfrac{P_B \cdot a_p}{2}$. If the base is a regular polygon, $A_B = \tfrac{P_B \cdot a_b}{2}$ ($a_b$: apothem of the base). Pythagoras: $a_p^2 = h^2 + a_b^2$.

Cylinder

$A_B = \pi r^2$

$A_L = 2\pi r \cdot h$

$A_T = 2A_B + A_L$

$V = A_B \cdot h = \pi r^2 h$

Cone

$A_B = \pi r^2$

$A_L = \pi r \cdot g$

$A_T = A_B + A_L$

$V = \dfrac{A_B \cdot h}{3} = \dfrac{\pi r^2 h}{3}$

$g$: slant height, $g^2 = h^2 + r^2$.

Sphere

$A = 4\pi r^2$

$V = \dfrac{4}{3}\pi r^3$

No base: just the total area. Hemisphere: $V = \tfrac{2}{3}\pi r^3$.

Capacity and units

$1\ \text{dm}^3 = 1\ \text{L}$

$1\ \text{m}^3 = 1000\ \text{dm}^3 = 1000\ \text{L}$

$1\ \text{cm}^3 = 1\ \text{mL}$

Areas: each unit step is $\times 100$. Volumes: each step is $\times 1000$.

Notice the difference: prisms and cylinders have two bases ($A_T = 2A_B + A_L$); pyramids and cones have only one ($A_T = A_B + A_L$), and their volume is a third of that of the prism or cylinder with the same base and height.