[**Latest] Mensuration Basic to Advanced Maths Formulas for 2D and 3D Shapes (With PDF)

 

the basic and advanced mensuration formulas for 2D and 3D shapes

 The topic of mensuration revolves around some limited concepts, it becomes very easy to solve these problems if you remember some basic formulas. Hence, we have compiled some important Mensuration Formulas for SSC and Banking in PDF.

 

[**Latest] Mensuration Basic to Advanced Maths Formulas for 2D and 3D Shapes (With PDF)

You can easily print it and affix it in your study room. Just give it a glance when you are solving mensuration problems and you will automatically remember it.

**:2D Shapes:**

**Area:

• Square: A = side²
• Rectangle: A = length × width
• Triangle: A = 1/2 × base × height
• Circle: A = π × radius²
• Parallelogram: A = base × height
• Trapezoid: A = 1/2 × (base1 + base2) × height

**Perimeter:

• Square: P = 4 × side
• Rectangle: P = 2 × (length + width)
• Triangle: P = side1 + side2 + side3
• Circle: P = 2 × π × radius

**:3D Shapes:**

**Volume:

• Cube: V = side³
• Cuboid: V = length × width × height
• Cylinder: V = π × radius² × height
• Cone: V = 1/3 × π × radius² × height
• Sphere: V = 4/3 × π × radius³

**Surface Area:

• Cube: SA = 6 × side²
• Cuboid: SA = 2(lw + wh + lh)
• Cylinder: SA = 2πr(r + h)
• Cone: SA = πr² + πrl (l = slant height)
• Sphere: SA = 4πr²

**Lateral Surface Area:

• Cuboid: LSA = 2(lh + wh)
• Cylinder: LSA = 2πrh
• Cone: LSA = πrl

**Additional Formulas:

1. Pythagorean Theorem (for right triangles): a² + b² = c²
2. Area of a sector of a circle: A = (θ/360) × πr²
3. Volume of a frustum of a cone: V = 1/3 × πh (r1² + r2² + r1r2)

**:2D Shapes:**

**Area:

• Rhombus: A = (diagonal1 × diagonal2) / 2
• Kite: A = (diagonal1 × diagonal2) / 2
• Regular Polygon (n sides): A = (n × side²) / (4 × tan(π/n))
• Equilateral Triangle: A = (√3 / 4) × side²

**Perimeter:

• Rhombus: P = 4 × side
• Kite: P = 2 × (side1 + side2)
• Regular Polygon (n sides): P = n × side

**:3D Shapes:**

**Surface Area:

• Rectangular Pyramid: SA = base area + (1/2) × perimeter × slant height
• Triangular Pyramid: SA = base area + (1/2) × perimeter × slant height
• Prism: SA = 2 × base area + perimeter × height

**Volume:

• Regular Pyramid: V = (1/3) × base area × height
• Ellipsoid: V = (4/3) × π × a × b × c (where a, b, c are semi-axes)

**:Formulas for Special Cases:**

• Sector area of a circle: A = (θ/360) × πr²
• Arc length of a circle: L = (θ/360) × 2πr
• Central angle of a sector: θ = (Arc length / r) × 360°

**:2D Shapes:**

**Area:

• Annulus (Ring): A = π × (outer radius² - inner radius²)
• Sector of a Circle: A = (θ/360) × π × radius² (where θ is the angle in degrees)

**Perimeter:

• Annulus (Ring): P = 2 × π × (outer radius + inner radius)

**:3D Shapes:**

**Volume:

• Torus (Ring-shaped solid): V = 2π² × R × r² (where R is the distance from the center of the tube to the center of the torus, and r is the radius of the tube)

**Surface Area:

• Torus (Ring-shaped solid): SA = 4π² × R × r (where R is the distance from the center of the tube to the center of the torus, and r is the radius of the tube)

**:Formulas for Special Cases:**

• Surface area of a spherical cap: SA = 2πrh
• Volume of a spherical cap: V = (1/3)πh²(3R - h) (where R is the radius of the sphere, h is the height of the cap)

**:Key Points:**

1. Always use consistent units for measurements (e.g., cm, m, in).
2. Visualize the shapes to understand their dimensions and relationships.
3. Practice applying the formulas to various problems to solidify your understanding.
4. Refer to reliable sources for more complex shapes and formulas as needed.

**:Key Features:**

These formulas are particularly relevant when dealing with more specialized shapes like tori, annuli, and spherical caps. They offer solutions for scenarios involving these unique geometries.

✓Name: Mensuration Basic to Advanced Maths Formulas for 2D and 3D Shapes (With PDF) 

✓Type: pdf 

✓Size: Limited 

✓Useful For: competitive exams 

**:Conclusion:**

These formulas can be particularly useful for solving problems involving more complex shapes and situations. Understanding these allows for more diverse problem-solving in geometry and mathematics!