Geometry Formulas | Area, Perimeter, Volume & Shapes

Basic Shapes

Fundamental geometric shapes and their properties form the foundation of geometry.

Rectangle

\[\text{Area} = l \times w \] \[\text{Perimeter} = 2(l + w) \]

Where:

l = length
w = width

Example:

Find the area and perimeter of a rectangle with length 8 cm and width 5 cm.

\( \textbf{Area (A)} \)
\begin{align} \textbf{A} & = \text{length} \times \text{width} \\ \textbf{A} & = 8 \times 5 \\ \textbf{A} & = \boldsymbol{40\;cm^2} \end{align}

\( \textbf{Perimeter (P)} \)
\begin{align} \textbf{P} & = 2(\text{length} + \text{width})\\ \textbf{P} & = 2 \times (8 + 5) \\ \textbf{P} & = 2 \times (13) \\ \textbf{P} & = \boldsymbol{26\;cm} \end{align}

Square

\[\text{Area} = s^2\] \[\text{Perimeter} = 4s\]

Where:

s = side length

A square is a special rectangle where all sides are equal.

Example:

Find the area and perimeter of a square with side length 6 cm.

\( \textbf{Area (A)} \)
\begin{align} \textbf{A} & = s^2 \\ \textbf{A} & = 6^2 \\ \textbf{A} & = \boldsymbol{36\;cm^2} \end{align}

\( \textbf{Perimeter (P)} \)
\begin{align} \textbf{P} & = 4s\\ \textbf{P} & = 4 \times 6 \\ \textbf{P} & = \boldsymbol{24\;cm} \end{align}

Triangles

Triangles are three-sided polygons with many important properties and formulas.

Triangle Area

\[\text{Area} = \frac{1}{2} \times b \times h\] \[\text{Perimeter} = a + b + c \]

Where:

a, b, c = sides of triangle
h = height (perpendicular to base)

Example:

Find the area of a triangle with base = 8 cm and height = 6 cm.

\( \textbf{Area (A)} \)
\begin{align} \textbf{A} & = \frac{1}{2} \times b \times h \\ \textbf{A} & = \frac{1}{2} \times 8 \times 6 \\ \textbf{A} & = \frac{1}{2} \times 48 \\ \textbf{A} & = \boldsymbol{24\;cm^2} \end{align}

Pythagorean Theorem

The Pythagorean Theorem is one of the most famous theorems in mathematics. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

\[a^2 + b^2 = c^2\]

For right triangles only:

a, b = legs (shorter sides)
c = hypotenuse (longest side)

This theorem helps find the length of any side when you know the other two.

Example:

A ladder is leaning against a wall. The ladder is 13 feet long, and the base of the ladder is 5 feet from the wall. How high up the wall does the ladder reach.

\( \textbf{hypotenuse (c)} = 13 \;feet \)

\( \textbf{leg (a)} = 5 \;feet \)

\( \textbf{leg (b)} = \; ? \;feet \)

\( \textbf{Using} \)
\( \textbf{Pythagorean theorem} \) \begin{align} a^2 + b^2 & = c^2 \\ 5^2 + b^2 & = 13^2 \\ 25 + b^2 & = 169 \\ b^2 & = 169 - 25 = 144 \\ b & = \sqrt{144} \\ b & = \boldsymbol{12\;feet} \end{align}

Heron's Formula

Heron's Formula allows you to find the area of any triangle when you know the lengths of all three sides. Named after Hero of Alexandria, this formula is particularly useful when the height of the triangle is unknown.

\[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\] \[s = \frac{a+b+c}{2}\]

Where:

a, b, c = side lengths
s = semi-perimeter

Use this when you know all three side lengths but not the height.

Example:

Find the area of a triangular section with sides measuring 8 cm, 10 cm, and 12 cm.

\( \textbf{Area (A)} \) \begin{align} \textbf{A} & = \sqrt{s(s-a)(s-b)(s-c)}, \;where\; s = \frac{a+b+c}{2} \\ \textbf{s} & = \frac{8 + 10 + 12}{2} \\ \textbf{s} & = \frac{30}{2} \\ \textbf{s} & = \boldsymbol{15\;cm} \\ \textbf{A} & = \sqrt{15(15 - 8)(15 - 10)(15 - 12)} \\ \textbf{A} & = \sqrt{1575} \\ \textbf{A} & \approx \boldsymbol{39.69\;cm^2} \end{align}

Circles

A Circle is the set of all the points in a plane that are equidistant from a given point. There are a few particularly notable features of a circle.

Center - The given point from which all points of the circle are equidistant. Circles are often named by their center point.
Radius - A segment that connects the center and any point on the circle. Its length is usually represented algebraically by r.
Diameter - A segment whose endpoints are on the circle and that passes through the center. Its length is usually represented algebraically by d.
Circumference - The perimeter of a circle, usually represented algebraically by C.

Circle Formulas

\[\text{Area} = \pi r^2\] \[\text{Circumference/Perimeter} = 2\pi r = \pi d\]

Where:

r = radius
d = diameter = 2r
\( \boldsymbol{\pi} \) ≈ 3.14159

Example:

Find the area and circumference of a circle with radius = 5 cm

\( \textbf{Area (A)} \)
\begin{align} \textbf{A} & = \pi r^2 \\ \textbf{A} & = \pi \times 5^2 \\ \textbf{A} & = \pi \times 25 \\ \textbf{A} & = 25\pi \\ \textbf{A} & \approx \boldsymbol{78.54 {cm}^2} \\ \end{align}

\( \textbf{Circumference (C)} \)
\begin{align} \textbf{C} & = 2\pi r \\ \textbf{C} & = 2\pi \times 5 \\ \textbf{C} & = 10\pi \\ \textbf{C} & \approx \boldsymbol{31.42\;cm} \end{align}

Sector and Arc

A sector of a circle is formed when a circle is divided using two radii.
Each sector has an angle between where the two radii meet at the centre of the circle. The sector with an angle less than 180 degrees is called a minor sector and the sector with an angle greater than 180 degrees is called a major sector. If the central angle formed equals 180 degrees, the two sectors would be semicircles.

An arc of a circle is a part of the circle, with all its points on the circle. It is a curve that is a part of its circumference.

\begin{gathered} \text{When}\;\boldsymbol{\theta}\;\text{in}\;\textbf{degrees} \end{gathered} \begin{align} \text{Area of a Sector (A)} & = \frac{\theta}{360^o} \times \pi r^2 \\ \text{Length of Arc AB (L)} & = \frac{\theta}{360^o} \times 2\pi r \\ \\ \end{align} \begin{gathered} \text{When}\;\boldsymbol{\theta}\;\text{in}\;\textbf{radians} \end{gathered} \begin{align} \text{Area of a Sector (A)} & = \frac{\theta}{2} \times r^2 \\ \text{Length of Arc AB (L)} & = r\times\theta \end{align}

Where:

θ = central angle in degrees/radian
A = area of sector
L = arc length
\( \boldsymbol{\pi} \approx \) 3.14159

Example:

Find the area of the sector of a circle with radius of 7 cm and the angle of the sector is \(40^o\)?

\( \textbf{Area of a Sector (A)} \) \begin{align} \textbf{A} & = \frac{\theta}{360^o} \times \pi r^2 \\ \textbf{A} & = \frac{40^o}{360^o} \times \pi \times 7^2 \\ \textbf{A} & = \frac{1}{9} \times \pi \times 49\\ \textbf{A} & = \frac{49}{9} \times \pi \\ \textbf{A} & \approx \boldsymbol{17.10\;cm^2} \\ \end{align}

\( \textbf{Length of Arc AB (L)} \) \begin{align} \textbf{L} & = \frac{\theta}{360^o} \times 2\pi r \\ \textbf{L} & = \frac{40^o}{360^o} \times 2\pi \times 7 \\ \textbf{L} & = \frac{14}{9} \times \pi \\ \textbf{L} & \approx \boldsymbol{4.89\;cm} \end{align}

Quadrilaterals

Four-sided polygons with various special properties and formulas.

Parallelogram

A parallelogram has opposite sides that are parallel and equal. The area is calculated using the base and perpendicular height, not the slant height.

\( \text{Area} = \text{base(b)} \times \text{height(h)} \) \[ \text{Perimeter} = 2 \times (a + b) \]

Where:

h = height
b = base
a = width

Example:

Find the area of a parallelogram with base 12 cm and height 7 cm.

\( \textbf{Area (A)} \) \begin{align} \textbf{A} & = \text{base(b)} \times \text{height(h)} \\ \textbf{A} & = 12 \times 7 \\ \textbf{A} & = \boldsymbol{84\;cm^2} \end{align}

Trapezoid

A trapezoid (or trapezium) has exactly one pair of parallel sides called bases. The area formula averages the two bases and multiplies by the height.

\[ \text{Area} = \frac{1}{2}(a + b) \times h \] \[ \text{Perimeter} = \text{Sum of all sides} \]

Where:

a, b = parallel bases
h = height

Example:

Find the area of a trapezoid with parallel sides 8 cm and 12 cm, and height 6 cm.

\( \textbf{Area (A)} \) \begin{align} \textbf{A} & = \frac{1}{2}(a + b) \times h \\ \textbf{A} & = \frac{1}{2}\times(8 + 12) \times 6 \\ \textbf{A} & = \frac{1}{2}\times(20) \times 6 \\ \textbf{A} & = \boldsymbol{60\;cm^2} \end{align}

Rhombus

A rhombus is a quadrilateral where all four sides are equal in length. Its diagonals are perpendicular bisectors of each other, creating four congruent right triangles.

\[ \text{Area} = \frac{1}{2} \times a \times b \] \[ \text{Perimeter} = 4 \times s \]

Where:

s = side
a, b = diagonals

Example:

The lengths of the two diagonals of a rhombus are 18 cm and 12 cm. Find the area of the rhombus.

\( \textbf{Area (A)} \) \begin{align} \textbf{A} & = \frac{1}{2} \times a \times b \\ \textbf{A} & = \frac{1}{2}\times18 \times 12 \\ \textbf{A} & = \boldsymbol{108\;cm^2} \end{align}

3D Shapes (Solids)

Three-dimensional shapes have volume and surface area formulas.

Cube

A cube is a three-dimensional shape with six square faces, twelve edges, and eight vertices. All edges of a cube are equal in length, making it a regular polyhedron.

\[\text{Volume} = s^3 \] \[\text{Total Surface Area} = 6s^2 \]

Where:

s = length of one side (edge) of the cube

Example:

Find the volume and surface area of a cube with side length 4 cm.

\( \textbf{Volume (V)} \) \begin{align} \textbf{V} & = s^3 \\ \textbf{V} & = 4^3 \\ \textbf{V} & = \boldsymbol{64\;cm^3} \end{align}

\( \textbf{Total Surface Area (SA)} \) \begin{align} \textbf{SA} & = 6s^2 \\ \textbf{SA} & = 6\times4^2 \\ \textbf{SA} & = 6\times 16 \\ \textbf{SA} & = \boldsymbol{96\;cm^2} \end{align}

Cuboid (Rectangular Prism)

A cuboid, also known as a rectangular prism, is a three-dimensional shape with six rectangular faces. It has three different dimensions: length, width, and height.

\[\text{Volume} = l \times w \times h \] \[\text{Curved Surface Area} = 2(lw + lh) \] \[\text{Total Surface Area} = 2(lw + lh + wh)\]

Where:

l, w, h = length, width, height

Example:

Find the volume and surface area of a rectangular prism with dimensions 6 cm, 4 cm and 3 cm.

\( \textbf{Volume (V)} \) \begin{align} \textbf{V} & = l \times w \times h \\ \textbf{V} & = 6 \times 4 \times 3 \\ \textbf{V} & = \boldsymbol{72\;cm^3} \end{align}

\( \textbf{Total Surface Area (SA)} \) \begin{align} \textbf{SA} & = 2(lw + lh + wh) \\ \textbf{SA} & = 2(6\times4 + 6\times3 + 4\times3) \\ \textbf{SA} & = 2(24 + 18 + 12) \\ \textbf{SA} & = 2\times 54 \\ \textbf{SA} & = \boldsymbol{108\;cm^2} \end{align}

Sphere

A sphere is perfectly round with every point on its surface equidistant from the center. It has the largest volume-to-surface-area ratio of any 3D shape.

\[\text{Volume} = \frac{4}{3}\pi r^3\] \[\text{Surface Area} = 4\pi r^2\]

Where:

r = radius
\( \boldsymbol{\pi} \approx \) 3.14159

Example:

Find the volume and surface area of sphere with radius 3 cm.

\( \textbf{Volume (V)} \) \begin{align} \textbf{V} & = \frac{4}{3}\pi r^3 \\ \textbf{V} & = \frac{4}{3}\pi \times 27 \\ \textbf{V} & = 36\pi \\ \textbf{V} & \approx \boldsymbol{113.10\;cm^3} \end{align}

\( \textbf{Total Surface Area (SA)} \) \begin{align} \textbf{SA} & = 4\pi r^2 \\ \textbf{SA} & = 4\pi\times 3^2 \\ \textbf{SA} & = 4\pi\times 9 \\ \textbf{SA} & = 36\pi \\ \textbf{SA} & \approx \boldsymbol{113.10\;cm^2} \end{align}

Cylinder

A cylinder has two parallel circular bases connected by a curved surface. The volume is found by multiplying the base area by the height.

\[\text{Volume} = \pi r^2 h\] \[\text{Curved Surface Area} = 2\pi r h \] \[\text{Total Surface Area} = 2\pi\times r(r + h) \]

Where:

r = radius of base
h = height
\( \boldsymbol{\pi} \approx \) 3.14159

Example:

Find the volume and total surface area of a cylinder with radius 4 cm and height 10 cm.

\( \textbf{Volume (V)} \) \begin{align} \textbf{V} & = \pi r^2 h \\ \textbf{V} & = \pi \times (4)^2 \times 10 \\ \textbf{V} & = \pi \times 16 \times 10 \\ \textbf{V} & = 160\pi \\ \textbf{V} & \approx \boldsymbol{502.65\;cm^3} \end{align}

\( \textbf{Total Surface Area (SA)} \) \begin{align} \textbf{SA} & = 2\pi\times r(r + h) \\ \textbf{SA} & = 2\pi\times 4 \times (4 + 10) \\ \textbf{SA} & = 2\pi\times 4 \times 14 \\ \textbf{SA} & = 2\pi\times 56 \\ \textbf{SA} & = 112\pi \\ \textbf{SA} & \approx \boldsymbol{351.86\;cm^2} \end{align}

Cone

A cone has a circular base and tapers smoothly to an apex. The slant height is the distance from the apex to any point on the base circumference.

\[\text{Volume} = \frac{1}{3}\pi r^2 h\] \[\text{Curved Surface Area} = \pi rl\] \[\text{Total Surface Area} = \pi r (l + r) \] \[\text{Slant Height} = \sqrt{r^2 + h^2} \]

Where:

r = radius of base
h = height
l = slant height
\( \boldsymbol{\pi} \approx \) 3.14159

Example:

Find the volume of a cone with radius 3 cm and height 8 cm.

\( \textbf{Volume (V)} \) \begin{align} \textbf{V} & = \pi r^2 h \\ \textbf{V} & = \frac{1}{3}\pi \times (3)^2 \times 8 \\ \textbf{V} & = \frac{1}{3}\pi \times 9 \times 8 \\ \textbf{V} & = \frac{1}{3}\pi \times 72 \\ \textbf{V} & = 24\pi \\ \textbf{V} & \approx \boldsymbol{75.40\;cm^3} \end{align}

\( \textbf{Slant Height (l)} \) \begin{align} \textbf{l} & = \sqrt{r^2 + h^2} \\ \textbf{l} & = \sqrt{3^2 + 8^2} \\ \textbf{l} & = \sqrt{73} \\ \textbf{l} & \approx \boldsymbol{8.54\;cm} \end{align}

Coordinate Geometry

Geometry using coordinate systems and algebraic methods.

Distance Formula

\[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]

Distance between two points (x₁, y₁) and (x₂, y₂)

Midpoint Formula

\[M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\]

Midpoint between two points (x₁, y₁) and (x₂, y₂)