Menu

Geometric formulas are mathematical equations used to calculate the properties of various shapes and figures. These properties include area, perimeter, surface area, and volume. Mastering these formulas is essential for solving real-world problems in fields like engineering, architecture, design, and more.

1. Area:

\[

A = ab

\]

2. Perimeter:

\[

P = 2a + 2b

\]

Worked Example:

If \( a = 5 \) and \( b = 8 \), then:

\[

A = 5 \times 8 = 40

\quad

P = 2(5) + 2(8) = 10 + 16 = 26

\]

Â

1. Area:

\[

A = bh = ab \sin(\theta)

\]

2. Perimeter:

\[

P = 2a + 2b

\]

Worked Example:

If \( b = 6 \), \( h = 4 \), and \( \theta = 45^\circ \), then:

\[

A = 6 \times 4 = 24 \quad \text{or} \quad A = 6 \times 4 \times \sin(45^\circ) \approx 24

\]

Â

1. Area:

\[

A = \frac{1}{2}bh = \frac{1}{2}ab \sin(\theta)

\]

2. Perimeter:

\[

P = a + b + c

\]

Worked Example:

If \( b = 7 \), \( h = 5 \), then:

\[

A = \frac{1}{2} \times 7 \times 5 = 17.5

\quad \text{and} \quad P = 7 + 5 + 8 = 20

\]

Â

1. Area:

\[

A = \frac{1}{2}h(a + b)

\]

2. Perimeter:

\[

P = a + b + h \left( \csc(\theta) + \csc(\phi) \right)

\]

Worked Example:

For \( a = 6 \), \( b = 4 \), \( h = 3 \):

\[

A = \frac{1}{2} \times 3 \times (6 + 4) = 15

\quad \text{and} \quad P = 6 + 4 + 3(\csc(\theta) + \csc(\phi))

\]

Â

1. Area:

\[

A = \pi r^2

\]

2. Perimeter (Circumference):

\[

C = 2\pi r

\]

Worked Example:

If \( r = 3 \):

\[

A = \pi (3)^2 = 28.27 \quad \text{and} \quad C = 2\pi(3) = 18.85

\]

1. Area:

\[

A = \frac{1}{4} n b^2 \cot\left(\frac{\pi}{n}\right)

\]

2. Perimeter:

\[

P = n b

\]

Worked Example:

If \( n = 6 \) (a hexagon) and \( b = 4 \), then:

\[

A = \frac{1}{4} \times 6 \times 4^2 \times \cot\left(\frac{\pi}{6}\right) = 41.57

\quad \text{and} \quad P = 6 \times 4 = 24

\]

Â

1. Area:

\[

A = \frac{1}{2} r^2 \theta \quad (\theta \text{ in radians})

\]

2. Arc Length:

\[

s = r \theta

\]

Worked Example:

For \( r = 6 \) and \( \theta = \frac{\pi}{4} \):

\[

A = \frac{1}{2} \times 6^2 \times \frac{\pi}{4} = 14.14 \quad \text{and} \quad s = 6 \times \frac{\pi}{4} = 4.71

\]

Â

1. Radius:

\[

r = \frac{\sqrt{(s – a)(s – b)(s – c)}}{s}

\]

where \( s = \frac{a + b + c}{2} \) is the semiperimeter.

Worked Example:

If \( a = 7 \), \( b = 8 \), and \( c = 5 \), then:

\[

s = \frac{7 + 8 + 5}{2} = 10 \quad \text{so} \quad r = \frac{\sqrt{(10 – 7)(10 – 8)(10 – 5)}}{10} = 1.26

\]

Â

1. Radius:

\[

R = \frac{abc}{4 \sqrt{s(s – a)(s – b)(s – c)}}

\]

where \( s \) is the semiperimeter:

\[

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

\]

Worked Example:

For \( a = 6 \), \( b = 8 \), and \( c = 10 \):

\[

s = \frac{6 + 8 + 10}{2} = 12

\]

\[

R = \frac{6 \times 8 \times 10}{4 \sqrt{12(12 – 6)(12 – 8)(12 – 10)}} = 5

\]

Â

1. Radius:

\[

r = \frac{\sqrt{(s – a)(s – b)(s – c)}}{s}

\]

where \( s = \frac{a + b + c}{2} \) is the semiperimeter.

Worked Example:

If \( a = 6 \), \( b = 7 \), and \( c = 8 \):

\[

s = \frac{6 + 7 + 8}{2} = 10.5 \quad \text{so} \quad r = \frac{\sqrt{(10.5 – 6)(10.5 – 7)(10.5 – 8)}}{10.5} = 1.87

\]

Â

1. Radius:

\[

R = \frac{abc}{4 \sqrt{s(s – a)(s – b)(s – c)}}

\]

where \( s \) is the semiperimeter:

\[

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

\]

Worked Example:

For \( a = 5 \), \( b = 12 \), and \( c = 13 \):

\[

s = \frac{5 + 12 + 13}{2} = 15

\]

\[

R = \frac{5 \times 12 \times 13}{4 \sqrt{15(15 – 5)(15 – 12)(15 – 13)}} = 6.5

\]

Â

1. Area:

\[

A = \frac{1}{2} n r^2 \sin\left(\frac{2\pi}{n}\right)

\]

2. Perimeter:

\[

P = 2nr\sin\left(\frac{\pi}{n}\right)

\]

Worked Example:

If \( n = 6 \) (hexagon) and \( r = 4 \):

\[

A = \frac{1}{2} \times 6 \times 4^2 \times \sin\left(\frac{2\pi}{6}\right) = 41.57

\quad \text{and} \quad P = 2 \times 6 \times 4 \times \sin\left(\frac{\pi}{6}\right) = 24

\]

Â

1. Area of Shaded Part:

\[

A = \frac{1}{2} r^2 (\theta – \sin(\theta))

\]

Worked Example:

If \( r = 10 \) and \( \theta = \frac{\pi}{3} \):

\[

A = \frac{1}{2} \times 10^2 \times \left(\frac{\pi}{3} – \sin\left(\frac{\pi}{3}\right)\right) = 35.44

\]

Â

1. Area:

\[

A = \frac{1}{2} n r^2 \sin\left(\frac{2\pi}{n}\right)

\]

2. Perimeter:

\[

P = 2nr \sin\left(\frac{\pi}{n}\right)

\]

Worked Example:

For a hexagon (\(n = 6\)) inscribed in a circle of radius \(r = 4\):

\[

A = \frac{1}{2} \times 6 \times 4^2 \times \sin\left(\frac{2\pi}{6}\right) \approx 41.57

\]

\[

P = 2 \times 6 \times 4 \times \sin\left(\frac{\pi}{6}\right) = 24

\]

Regular Polygon of \(n\) Sides Circumscribing a Circle of Radius \(r\)

1. Area:

\[

A = \frac{1}{2}nr^2 \cot\left(\frac{\pi}{n}\right)

\]

2. Perimeter:

\[

P = 2nr\tan\left(\frac{\pi}{n}\right)

\]

Worked Example:

For a regular pentagon (\(n = 5\)) circumscribing a circle of radius \(r = 3\):

\[

A = \frac{1}{2} \times 5 \times 3^2 \times \cot\left(\frac{\pi}{5}\right) \approx 43.01

\]

\[

P = 2 \times 5 \times 3 \times \tan\left(\frac{\pi}{5}\right) = 21.84

\]

Segment of Circle of Radius \(r\)

1. Area of the Shaded Part:

\[

A = \frac{1}{2}r^2(\theta – \sin(\theta))

\]

Worked Example:

For a circle with radius \(r = 5\) and central angle \(\theta = \frac{\pi}{4}\):

\[

A = \frac{1}{2} \times 5^2 \times \left(\frac{\pi}{4} – \sin\left(\frac{\pi}{4}\right)\right) \approx 8.18

\]

Ellipse of Semi-major Axis \(a\) and Semi-minor Axis \(b\)

1. Area:

\[

A = \pi ab

\]

2. Perimeter (Approximation):

\[

P \approx \pi \left( 3(a + b) – \sqrt{(3a + b)(a + 3b)} \right)

\]

Worked Example:

For an ellipse with semi-major axis \(a = 6\) and semi-minor axis \(b = 4\):

\[

A = \pi \times 6 \times 4 = 75.40

\]

\[

P \approx \pi \left(3(6 + 4) – \sqrt{(3 \times 6 + 4)(6 + 3 \times 4)}\right) \approx 31.42

\]

Â

1. Volume:

\[

V = \frac{1}{3} \pi r^2 h

\]

2. Lateral Surface Area:

\[

A = \pi r \sqrt{r^2 + h^2}

\]

Worked Example:

For a cone with radius \(r = 3\) and height \(h = 5\):

\[

V = \frac{1}{3} \pi \times 3^2 \times 5 = 47.12

\]

\[

A = \pi \times 3 \times \sqrt{3^2 + 5^2} \approx 51.42

\]

Sphere of Radius \(r\)

1. Volume:

\[

V = \frac{4}{3} \pi r^3

\]

2. Surface Area:

\[

A = 4\pi r^2

\]

Worked Example:

For a sphere with radius \(r = 6\):

\[

V = \frac{4}{3} \pi \times 6^3 = 904.32

\]

\[

A = 4\pi \times 6^2 = 452.39

\]

Â

1. Volume:

\[

V = \frac{1}{3} \pi h (r_1^2 + r_1r_2 + r_2^2)

\]

2. Lateral Surface Area:

\[

A = \pi (r_1 + r_2) \sqrt{(r_1 – r_2)^2 + h^2}

\]

Worked Example:

For a frustum with \(r_1 = 4\), \(r_2 = 2\), and \(h = 5\):

\[

V = \frac{1}{3} \pi \times 5 \times (4^2 + 4 \times 2 + 2^2) = 150.80

\]

\[

A = \pi \times (4 + 2) \times \sqrt{(4 – 2)^2 + 5^2} \approx 87.96

\]

Torus with Inner Radius \(a\) and Outer Radius \(b\)

1. Volume:

\[

V = \frac{\pi^2}{4} (b – a)(b + a)^2

\]

2. Surface Area:

\[

A = 2\pi^2 (b^2 – a^2)

\]

Worked Example:

For a torus with inner radius \(a = 3\) and outer radius \(b = 6\):

\[

V = \frac{\pi^2}{4} \times (6 – 3) \times (6 + 3)^2 \approx 1,579.14

\]

\[

A = 2\pi^2 \times (6^2 – 3^2) \approx 592.91

\]

Â

**In Education**

- Geometric formulas are fundamental in mathematics education, helping students develop spatial reasoning and problem-solving skills.

**In Engineering and Architecture**

- Engineers and architects use geometric formulas to design structures, calculate material requirements, and ensure stability and efficiency in their projects.

**In Everyday Life**

- From determining the amount of paint needed for a room to calculating the volume of a swimming pool, geometric formulas have practical everyday applications.

– Practice Regularly: Consistent practice helps reinforce your understanding of different formulas.

– Understand the Concepts: Instead of memorizing formulas, comprehend the principles behind them.

– Use Visual Aids: Diagrams and drawings can help visualize problems and solutions.

– Apply Real-World Examples: Relating formulas to real-life situations enhances retention and understanding.

– Utilize Online Resources: Leverage educational websites, tutorials, and interactive tools for additional practice and explanations.

Q1: What is the most important geometric formula to memorize?

A1: While all geometric formulas are important, the area and perimeter formulas for basic shapes like triangles, circles, rectangles, and squares are fundamental and widely used.

Q2: How can I easily remember geometric formulas?

A2: Understanding the derivation of formulas, practicing regularly, and applying them to real-world scenarios can significantly aid in memorization.

Q3: Are geometric formulas used in other subjects besides math?

A3: Yes, geometric formulas are integral in fields such as physics, engineering, architecture, computer graphics, and even art and design.

Q4: How do I choose which geometric formula to use in a problem?

A4: Identify the shape involved and the property you need to calculate (area, perimeter, volume). Match these with the corresponding formula for that shape.

Q5: Can geometric formulas be applied to irregular shapes?

A5: Yes, by decomposing irregular shapes into regular ones, calculating each part’s area or volume, and then summing them up.