Complete List of Mensuration Formulas for 2D and 3D Shapes (PDF Download)

Mensuration is a core branch of mathematics focused on geometric properties and their measurements. It encompasses essential parameters such as length, area, volume, and surface area. By studying mensuration, you can master the precise calculation of both 2D and 3D geometric figures.

Mensuration Formulas

Using the correct mensuration formulas allows you to accurately calculate the area, perimeter, volume, and surface area of various complex shapes. Before diving into the formulas, it is crucial to distinguish between 2D and 3D geometry. This guide is designed for school students as well as candidates preparing for competitive exams, where mensuration is a high-yield topic. For aspirants of SSC CGL, SSC CHSL, SSC MTS, SSC CPO, Delhi Police Constable, or Bank Exams, understanding these concepts is vital for scoring in the Quantitative Aptitude section. These resources are also perfect for Class 8, 9, and 10 students looking to strengthen their mathematical foundations.

To ensure clarity, we have categorized the following mensuration formulas into dedicated tables for 2D and 3D shapes. Explore these references to simplify your calculations.

Mensuration Formulas for 2-D Figures

Areas of 2D figures are calculated in square units, while perimeters are measured in linear units. Below is a comprehensive list of formulas for shapes including squares, triangles (scalene, isosceles, equilateral, right), trapeziums, parallelograms, rhombi, and circles. Study these formulas carefully to improve your problem-solving speed.

Mensuration Formulas for 2-Dimensional Figures
Shape	Area	Perimeter
Circle	πr²	2 π r
Square	(side)²	4 × side
Rectangle	length × breadth	2 (length + breadth)
Scalene Triangle	√[s(s−a)(s−b)(s−c)], Where, s = (a+b+c)/2	a+b+c (sum of sides)
Isosceles Triangle	½ × base × height	2a + b (sum of sides)
Equilateral Triangle	(√3/4) × (side)²	3 × side
Right Angled Triangle	½ × base × hypotenuse	A + B + hypotenuse, where the hypotenuse is √A²+B²
Parallelogram	base × height	2(l+b)
Rhombus	½ × diagonal1 × diagonal2	4 × side
Trapezium	½ h(sum of parallel sides)	a+b+c+d (sum of all sides)

Mensuration Formulas for 3-D Figures

For 3D figures, the total surface area is the sum of the curved (lateral) surface area and the areas of the top and bottom bases. While surface areas are expressed in square units, volume is measured in cubic units. Access our organized list of formulas for 3D objects like cones, cylinders, cubes, cuboids, and spheres below.

Mensuration Formulas for 3-Dimensional Figures
Shape	Area	Curved Surface Area (CSA)/  Lateral Surface Area (LSA)	Total Surface Area (TSA)
Cone	(1/3) π r² h	π r l	πr (r + l)
Cube	(side)³	4 (side)²	6 (side)²
Cuboid	length × breadth × height	2 height (length + breadth)	2 (lb +bh +hl)
Cylinder	π r² h	2π r h	2πrh + 2πr²
Hemisphere	(2/3) π r³	2 π r²	3 π r²
Sphere	4/3πr³	4πr²	4πr²

Differences between 2-Dimensional and 3-Dimensional Figures

Geometry relies on the distinction between 2D (flat) and 3D (spatial) figures. Review the table below to grasp the fundamental differences and characteristics of these shapes.

2-Dimensional Figures	3-Dimensional Figures
As the name suggests, a 2D shape means that it will have only 2 dimensions which are length and breadth.	Here, a 3D shape means that this figure will have 3 dimensions, which are length, breadth, and height.
For 2D shapes, we can calculate 2 things i.e. area and perimeter.	For 3D shapes, we can calculate their volume, total surface area, and curved/lateral surface area.
2D shapes are flat as they do not have depth and also, these cannot be held physically because of the lack of depth.	3D shapes contain a depth so they can be held physically and are not flat like 2D shapes.
Example: Square, Rectangle, Triangle, Circle, etc.	Example: Cone, Cylinder, Sphere, Cube, Prism, Pyramid, etc.

Mensuration Formulas PDF

To master these calculations, download our comprehensive Mensuration Formulas PDF provided below for offline practice and revision.

Mensuration Formulas PDF 1- Click to Download

Mensuration Formulas PDF 2- Click to Download

Mensuration Formulas Examples

Below, we detail the essential mensuration math formulas for every 2D and 3D geometric figure.

1. Cube Formula

2. Cuboid Formula

3. Prism Formula

4. Sphere Formula

5. Hemisphere Formula

6. Pyramids Formula

7. Right Circular Cone Formula

8. Right Circular Cylinder Formula

9. Equilateral Triangle Formula

Important Terms Related to Mensuration Formula

To effectively apply these formulas, you must understand these core measurement concepts:

● Perimeter: The total length of the boundary of a closed figure, measured in linear units like meters (m) or centimeters (cm).

● Area: The measure of the surface enclosed within a 2D boundary, expressed in square units (m², cm²).

● Volume: The total space occupied by a 3D object, measured in cubic units (m³, cm³).

● Curved/Lateral Surface area: The area of the side surfaces of a 3D figure, excluding the base and top, measured in square units.

● Total Surface area: The combined area of all surfaces of a 3D object, including the top, bottom, and lateral sides.

Mensuration Questions with Solutions

Q1. Find the area of a rhombus if its diagonals measure 9 cm and 7.2 cm.

Solution: Given d1 = 9 cm, d2 = 7.2 cm.

The area formula for a rhombus is ½ × (d1 × d2).

Area = ½ × 9 × 7.2 cm² = 32.4 cm².

Note: In these formulas, π is typically 22/7 or 3.14, 'r' represents the radius, and 'h' denotes height.

Q2. A rectangular paper (width 10 cm) is rolled to form a cylinder with a radius of 18 cm. Find the cylinder's volume. (Take π=22/7)

Solution: Rolling the rectangle along its width means the paper's width (10 cm) becomes the cylinder's height.

Height (h) = 10 cm, Radius (r) = 18 cm.

Volume (V) = πr²h = (22/7) × (18)² × 10

V ≈ 10,182.85 cm³ (approximately 10,183 cm³).

The volume of the cylinder is approximately 10,183 cm³.

Q3. For a circle with a 21 cm radius, calculate the circumference and area. (Use π = 22/7)

Solution: Circumference = 2πr = 2 × (22/7) × 21 = 132 cm.

Area = πr² = (22/7) × (21)² = 22 × 3 × 21.

Area of the circle = 1386 cm².

Q4. A cuboidal box has dimensions 30 cm × 25 cm × 20 cm. Find its total surface area.

Solution: Here, l = 30 cm, b = 25 cm, h = 20 cm.

Total Surface Area = 2(lb + bh + hl)

= 2(30 × 25 + 25 × 20 + 20 × 30)

= 2(750 + 500 + 600) = 3700 cm².

Q5. A 11 cm × 4 cm rectangular sheet is rolled to form a cylinder with height 4 cm. Find the volume of the cylinder.

Solution: The length of the sheet (11 cm) becomes the circumference of the cylinder's base, and the width (4 cm) is the height.

Circumference (2πr) = 11 cm

2 × (22/7) × r = 11

r = 1.75 cm (or 7/4 cm).

Volume = πr²h = (22/7) × (7/4)² × 4 = 38.5 cm³.

Q6. Tangents at points A and B of a circle (center O) meet at P. If ∠AOB = 120° and AP = 6 cm, calculate the area of ΔAPB.

Solution:

Given ∠AOB = 120°, AP = 6 cm, and ∠OAP = 90°.

∠AOP = 60°.

In ΔAOP: tan(60°) = AP/OA

√3 = 6/OA

OA = 2√3 cm.

Area ΔAOP = ½ × OA × AP = ½ × 2√3 × 6 = 6√3 cm².

Continue solving geometrically to reach final result.

Following the geometric properties, the area ΔAPB = 9√3 cm².

Calculation steps for ΔAMP...

Trigonometric analysis concludes...

Sin(60°) = AM/OA...

Resulting in AM = 3 cm.

AM = 3 cm.

OM = √3 cm.

Area ΔAOM = ½ × √3 × 3 = 1.5√3 cm².

Area ΔAOM = 1.5√3 cm².

Area ΔAMP = Area ΔAOP - Area ΔAOM

Area ΔAMP = 6√3 - 1.5√3 = 4.5√3 cm².

Area ΔAPB = 2 × Area ΔAMP.

Area ΔAPB = 2 × 4.5√3 = 9√3 cm².

Q7. If the total surface area of a hemisphere is 166.32 cm², what is its curved surface area?

Solution: Total Surface Area (TSA) = 3πr² = 166.32 cm².

Curved Surface Area (CSA) = 2πr².

Since CSA is 2/3 of TSA:

CSA = (2/3) × 166.32

CSA = 110.88 cm².

Q8. G is the centroid of a triangle with area 72 cm². What is the area of ΔBDG?

Solution: The area of the triangle formed by the centroid and two vertices is 1/3 of the total area.

A median divides the triangle into two equal parts.

Area of ΔBDG = 1/6 of total area (ΔABC).

= (1/6) × 72

= 12 cm².

Q9. Perimeters of similar triangles ABC and PQR are 156 cm and 46.8 cm. If BC = 19.5 cm, find QR.

Solution: Ratio of perimeters = Ratio of corresponding sides.

156 / 46.8 = 19.5 / QR

QR = (46.8 × 19.5) / 156

QR = 5.85 cm.

Q10. A metal sphere (radius 10 cm) is recast into spheres of radius 2 cm. How many small spheres are formed?

Solution: Volume = 4/3 πr³.

Total Volume = n × Volume of small sphere.

4/3 π (10)³ = n × 4/3 π (2)³

1000 = n × 8 ⇒ n = 125.

Conclusion

Mensuration can be challenging due to the vast number of formulas, but consistent practice makes it manageable. By revising these formulas daily and working through problems, you can master the topic. We hope this guide helps you succeed in your exams. Stay tuned for more educational content.