Measurement Formulas are the mathematical expressions used to calculate length, area, volume, perimeter, and surface area of two-dimensional and three-dimensional geometric shapes. These formulas form the backbone of NCERT Mathematics from Class 6 through Class 12, and they appear regularly in CBSE board examinations. JEE Main and NEET also test measurement concepts in coordinate geometry and physics problems. This article covers every essential measurement formula, a complete formula sheet, three progressive solved examples, CBSE exam tips, and JEE/NEET applications.
Key Measurement Formulas at a Glance
Quick reference for the most important measurement formulas across 2D and 3D shapes.
- Area of Rectangle: \( A = l \times b \)
- Area of Circle: \( A = \pi r^2 \)
- Perimeter of Circle (Circumference): \( C = 2\pi r \)
- Volume of Cube: \( V = a^3 \)
- Volume of Sphere: \( V = \dfrac{4}{3}\pi r^3 \)
- Total Surface Area of Cylinder: \( TSA = 2\pi r(r + h) \)
- Lateral Surface Area of Cone: \( LSA = \pi r l \)
What are Measurement Formulas?
Measurement Formulas are standardised algebraic expressions that allow students to quantify geometric properties such as area, perimeter, volume, and surface area. The study of measurement begins in NCERT Class 6 (Chapter 10: Mensuration) and extends through Class 8, Class 10, and Class 12. At each level, the shapes become more complex and the formulas more sophisticated.
In simple terms, measurement is the process of assigning a numerical value to a physical quantity using a defined unit. For geometric figures, this means finding how much space a shape occupies (area or volume) or how long its boundary is (perimeter or circumference).
Measurement activities include direct measurement using rulers and scales, as well as indirect measurement using formulas. The indirect method is far more powerful. It lets us calculate the area of a football field or the volume of a water tank without physically filling it. NCERT textbooks introduce these formulas progressively, ensuring students build a strong conceptual foundation before tackling competitive exams like JEE Main and NEET.
All Measurement Formulas use the constant \( \pi \approx 3.14159 \) for curved shapes and standard algebraic variables for straight-edged shapes.
Measurement Formulas — Expression and Variables
The most fundamental measurement formula is for the area of a rectangle, which underpins nearly all other area calculations:
\[ A = l \times b \]
where \( A \) is area, \( l \) is length, and \( b \) is breadth. Every other measurement formula is a variation or extension of this core idea.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( A \) | Area | Square metres (m²) |
| \( V \) | Volume | Cubic metres (m³) |
| \( P \) | Perimeter | Metres (m) |
| \( C \) | Circumference | Metres (m) |
| \( r \) | Radius | Metres (m) |
| \( h \) | Height | Metres (m) |
| \( l \) | Length / Slant height | Metres (m) |
| \( b \) | Breadth | Metres (m) |
| \( a \) | Side of square / cube | Metres (m) |
| \( \pi \) | Pi (constant) | Dimensionless (≈ 3.14159) |
| \( TSA \) | Total Surface Area | Square metres (m²) |
| \( LSA \) | Lateral Surface Area | Square metres (m²) |
How Measurement Formulas Are Derived
Most area formulas derive from the rectangle formula \( A = l \times b \). A parallelogram is essentially a sheared rectangle, giving \( A = b \times h \). A triangle is exactly half a parallelogram, giving \( A = \frac{1}{2} b h \). A circle’s area formula \( A = \pi r^2 \) comes from dividing the circle into infinitely thin sectors and rearranging them into a rectangle of length \( \pi r \) and breadth \( r \). Volume formulas extend area into the third dimension: the volume of a prism equals its base area multiplied by its height, \( V = A_{base} \times h \).
Measurement Formulas for 2D Objects
Two-dimensional shapes have length and breadth but no depth. Their key measurements are area and perimeter.
Rectangle: A rectangle has length \( l \) and breadth \( b \).
\[ A_{rectangle} = l \times b \qquad P_{rectangle} = 2(l + b) \]
Square: A special rectangle where all sides equal \( a \).
\[ A_{square} = a^2 \qquad P_{square} = 4a \]
Triangle: For a triangle with base \( b \) and height \( h \):
\[ A_{triangle} = \frac{1}{2} b h \]
For a triangle with all three sides known (\( a, b, c \)), Heron’s Formula applies:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \quad \text{where} \quad s = \frac{a+b+c}{2} \]
Circle: For a circle with radius \( r \):
\[ A_{circle} = \pi r^2 \qquad C = 2\pi r \]
Parallelogram: For a parallelogram with base \( b \) and height \( h \):
\[ A_{parallelogram} = b \times h \]
Trapezium: For a trapezium with parallel sides \( a \) and \( b \), and height \( h \):
\[ A_{trapezium} = \frac{1}{2}(a + b) \times h \]
Rhombus: For a rhombus with diagonals \( d_1 \) and \( d_2 \):
\[ A_{rhombus} = \frac{1}{2} d_1 \times d_2 \]
Measurement Formulas for 3D Objects
Three-dimensional shapes have length, breadth, and height. Their key measurements are volume, total surface area (TSA), and lateral surface area (LSA).
Cube: A cube with side \( a \):
\[ V_{cube} = a^3 \qquad TSA_{cube} = 6a^2 \qquad LSA_{cube} = 4a^2 \]
Cuboid: A cuboid with length \( l \), breadth \( b \), height \( h \):
\[ V_{cuboid} = l \times b \times h \qquad TSA_{cuboid} = 2(lb + bh + hl) \]
Cylinder: A cylinder with radius \( r \) and height \( h \):
\[ V_{cylinder} = \pi r^2 h \qquad TSA_{cylinder} = 2\pi r(r + h) \qquad LSA_{cylinder} = 2\pi r h \]
Cone: A cone with radius \( r \), height \( h \), and slant height \( l = \sqrt{r^2 + h^2} \):
\[ V_{cone} = \frac{1}{3}\pi r^2 h \qquad TSA_{cone} = \pi r(r + l) \qquad LSA_{cone} = \pi r l \]
Sphere: A sphere with radius \( r \):
\[ V_{sphere} = \frac{4}{3}\pi r^3 \qquad TSA_{sphere} = 4\pi r^2 \]
Hemisphere: A hemisphere with radius \( r \):
\[ V_{hemisphere} = \frac{2}{3}\pi r^3 \qquad TSA_{hemisphere} = 3\pi r^2 \qquad LSA_{hemisphere} = 2\pi r^2 \]
Complete Measurement Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Area of Rectangle | \( A = l \times b \) | l = length, b = breadth | m² | Class 6, Ch 10 |
| Perimeter of Rectangle | \( P = 2(l + b) \) | l = length, b = breadth | m | Class 6, Ch 10 |
| Area of Square | \( A = a^2 \) | a = side | m² | Class 6, Ch 10 |
| Area of Triangle | \( A = \frac{1}{2}bh \) | b = base, h = height | m² | Class 7, Ch 11 |
| Heron’s Formula | \( A = \sqrt{s(s-a)(s-b)(s-c)} \) | s = semi-perimeter, a,b,c = sides | m² | Class 9, Ch 12 |
| Area of Circle | \( A = \pi r^2 \) | r = radius | m² | Class 7, Ch 11 |
| Circumference of Circle | \( C = 2\pi r \) | r = radius | m | Class 7, Ch 11 |
| Area of Trapezium | \( A = \frac{1}{2}(a+b)h \) | a,b = parallel sides, h = height | m² | Class 8, Ch 11 |
| Area of Rhombus | \( A = \frac{1}{2}d_1 d_2 \) | d1, d2 = diagonals | m² | Class 8, Ch 11 |
| Volume of Cube | \( V = a^3 \) | a = side | m³ | Class 8, Ch 11 |
| TSA of Cuboid | \( TSA = 2(lb + bh + hl) \) | l, b, h = dimensions | m² | Class 9, Ch 13 |
| Volume of Cuboid | \( V = lbh \) | l, b, h = dimensions | m³ | Class 8, Ch 11 |
| LSA of Cylinder | \( LSA = 2\pi rh \) | r = radius, h = height | m² | Class 9, Ch 13 |
| TSA of Cylinder | \( TSA = 2\pi r(r+h) \) | r = radius, h = height | m² | Class 9, Ch 13 |
| Volume of Cylinder | \( V = \pi r^2 h \) | r = radius, h = height | m³ | Class 9, Ch 13 |
| LSA of Cone | \( LSA = \pi r l \) | r = radius, l = slant height | m² | Class 9, Ch 13 |
| TSA of Cone | \( TSA = \pi r(r + l) \) | r = radius, l = slant height | m² | Class 9, Ch 13 |
| Volume of Cone | \( V = \frac{1}{3}\pi r^2 h \) | r = radius, h = height | m³ | Class 9, Ch 13 |
| TSA of Sphere | \( TSA = 4\pi r^2 \) | r = radius | m² | Class 9, Ch 13 |
| Volume of Sphere | \( V = \frac{4}{3}\pi r^3 \) | r = radius | m³ | Class 9, Ch 13 |
| TSA of Hemisphere | \( TSA = 3\pi r^2 \) | r = radius | m² | Class 9, Ch 13 |
| Volume of Hemisphere | \( V = \frac{2}{3}\pi r^3 \) | r = radius | m³ | Class 9, Ch 13 |
| Slant Height of Cone | \( l = \sqrt{r^2 + h^2} \) | r = radius, h = height | m | Class 9, Ch 13 |
Measurement Formulas — Solved Examples
Example 1 (Class 8-9 Level): Area and Perimeter of a Composite Shape
Problem: A rectangular park measures 40 m in length and 25 m in breadth. A circular fountain of radius 3.5 m is built at its centre. Find the area of the park excluding the fountain. (Use \( \pi = \frac{22}{7} \))
Given: Length of rectangle \( l = 40 \) m, breadth \( b = 25 \) m, radius of circle \( r = 3.5 \) m
Step 1: Calculate the area of the rectangular park.
\( A_{rectangle} = l \times b = 40 \times 25 = 1000 \) m²
Step 2: Calculate the area of the circular fountain.
\( A_{circle} = \pi r^2 = \dfrac{22}{7} \times (3.5)^2 = \dfrac{22}{7} \times 12.25 = 38.5 \) m²
Step 3: Subtract the fountain area from the park area.
\( A_{remaining} = 1000 – 38.5 = 961.5 \) m²
Answer
The area of the park excluding the fountain is 961.5 m².
Example 2 (Class 10-11 Level): Surface Area and Volume of a Combined Solid
Problem: A toy is made by placing a cone of radius 3 cm and height 4 cm on top of a hemisphere of the same radius. Find the total surface area and total volume of the toy. (Use \( \pi = 3.14 \))
Given: Radius \( r = 3 \) cm, height of cone \( h = 4 \) cm
Step 1: Find the slant height of the cone.
\( l = \sqrt{r^2 + h^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) cm
Step 2: Calculate the curved surface area (CSA) of the cone.
\( CSA_{cone} = \pi r l = 3.14 \times 3 \times 5 = 47.1 \) cm²
Step 3: Calculate the curved surface area of the hemisphere.
\( CSA_{hemisphere} = 2\pi r^2 = 2 \times 3.14 \times 9 = 56.52 \) cm²
Step 4: Add both CSAs for total surface area. (The flat circular base of the hemisphere is the only external base.)
\( TSA_{toy} = CSA_{cone} + CSA_{hemisphere} = 47.1 + 56.52 = 103.62 \) cm²
Step 5: Calculate the volume of the cone.
\( V_{cone} = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times 3.14 \times 9 \times 4 = 37.68 \) cm³
Step 6: Calculate the volume of the hemisphere.
\( V_{hemisphere} = \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times 3.14 \times 27 = 56.52 \) cm³
Step 7: Add both volumes.
\( V_{total} = 37.68 + 56.52 = 94.2 \) cm³
Answer
Total Surface Area of the toy = 103.62 cm². Total Volume of the toy = 94.2 cm³.
Example 3 (JEE/NEET Level): Optimisation Using Measurement Formulas
Problem: A cylindrical can is to be made from a fixed amount of sheet metal such that its total surface area is 150π cm². If the radius of the cylinder is 5 cm, find the height. Then determine the volume of the cylinder.
Given: \( TSA = 150\pi \) cm², \( r = 5 \) cm
Step 1: Write the TSA formula for a cylinder.
\( TSA = 2\pi r(r + h) \)
Step 2: Substitute the known values.
\( 150\pi = 2\pi \times 5 \times (5 + h) \)
Step 3: Simplify by dividing both sides by \( 2\pi \).
\( 75 = 5(5 + h) \)
Step 4: Solve for \( h \).
\( 15 = 5 + h \implies h = 10 \) cm
Step 5: Calculate the volume.
\( V = \pi r^2 h = \pi \times 25 \times 10 = 250\pi \approx 785.4 \) cm³
Answer
Height of the cylinder = 10 cm. Volume = 250π cm³ ≈ 785.4 cm³.
CBSE Exam Tips 2025-26
- Memorise the formula sheet: We recommend writing all measurement formulas on a single sheet and revising it daily in the week before exams. CBSE Class 9 and 10 papers consistently carry 8-12 marks on mensuration.
- Identify the shape first: Before writing any formula, clearly identify whether the problem involves a 2D or 3D shape. Many students lose marks by applying the TSA formula instead of the LSA formula, or vice versa.
- Always find slant height for cones: In CBSE 2025-26 papers, cone problems almost always require you to first calculate slant height using \( l = \sqrt{r^2 + h^2} \). Make this your automatic first step.
- Use \( \pi = \frac{22}{7} \) unless specified: CBSE questions specify which value of \( \pi \) to use. If not specified, use \( \frac{22}{7} \) for cleaner calculations.
- Draw the figure: Our experts suggest always sketching the shape and labelling all given dimensions. This prevents errors in identifying which dimension is height and which is slant height.
- Check units at every step: CBSE deducts marks for missing or incorrect units. Write m² for area and m³ for volume in every step, not just the final answer.
Common Mistakes to Avoid
- Confusing TSA and LSA: The Total Surface Area includes all faces (top, bottom, and curved/lateral surfaces). The Lateral Surface Area includes only the curved or side surfaces. For a cylinder, \( TSA = 2\pi r(r+h) \) but \( LSA = 2\pi rh \). Students frequently forget to add the circular bases.
- Using diameter instead of radius: All circle and sphere formulas use the radius \( r \). If the problem gives the diameter \( d \), always convert first: \( r = d/2 \). This is one of the most common errors in CBSE and JEE problems.
- Forgetting the factor of \( \frac{1}{3} \) in cone and pyramid volumes: The volume of a cone is \( \frac{1}{3}\pi r^2 h \), not \( \pi r^2 h \). The factor \( \frac{1}{3} \) is frequently dropped under exam pressure.
- Applying Heron’s formula without computing semi-perimeter: Students sometimes substitute the full perimeter instead of the semi-perimeter \( s = (a+b+c)/2 \) in Heron’s formula. Always compute \( s \) as the first step.
- Incorrect slant height for cones: The slant height \( l \) is NOT the same as the vertical height \( h \). Always calculate \( l = \sqrt{r^2 + h^2} \) separately before using the cone’s LSA or TSA formula.
JEE/NEET Application of Measurement Formulas
In our experience, JEE aspirants encounter measurement formulas most frequently in coordinate geometry, physics (calculating cross-sectional areas and volumes in fluid mechanics and thermodynamics), and in three-dimensional geometry problems.
Pattern 1 — Combined Solids: JEE Main regularly presents problems involving two or more solids joined together. A common example is a cylinder topped with a hemisphere or cone. Students must add volumes and subtract or add surface areas carefully, depending on which surfaces are exposed.
Pattern 2 — Rate of Change of Volume: In JEE Advanced calculus problems, the volume formula of a sphere \( V = \frac{4}{3}\pi r^3 \) is differentiated with respect to time. This gives the rate at which volume changes as the radius changes: \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \). Recognising that \( 4\pi r^2 \) is the surface area of the sphere is a key insight tested in JEE Advanced.
Pattern 3 — NEET Physics Applications: NEET tests measurement formulas indirectly. The cross-sectional area of a blood vessel (modelled as a cylinder) appears in the flow-rate formula \( Q = Av \), where \( A = \pi r^2 \). Similarly, the surface area of a cell (modelled as a sphere) determines diffusion rates. Students who know their measurement formulas solve these problems significantly faster.
We recommend practising at least five JEE-level mensuration problems per week during your preparation. Focus especially on combined solids and optimisation problems, as these carry the highest marks.
FAQs on Measurement Formulas
Explore more related formula articles on ncertbooks.net to strengthen your preparation. Visit our Physics Formulas hub for a complete collection of NCERT-aligned formulas. You may also find our articles on the Flow Rate Formula, the Spring Constant Formula, and the Angular Displacement Formula useful for your JEE and NEET preparation. For official NCERT textbook references, visit ncert.nic.in.