All the Maths Formulas for Mensuration.

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Here are some of the important formulas for mensuration in mathematics:

Area of a Circle: πr², where r is the radius of the circle. The formula for finding the area of a circle is π times the radius squared. For example, if the radius of a circle is 5 cm, then the area would be 25π square cm.
Circumference of a Circle: 2πr, where r is the radius of the circle. The formula for finding the circumference of a circle is 2 times π times the radius. For example, if the radius of a circle is 5 cm, then the circumference would be 10π cm.
Area of a Rectangle: l x b, where l is the length and b is the breadth of the rectangle. The formula for finding the area of a rectangle is the product of its length and breadth. For example, if the length of a rectangle is 6 cm and its breadth is 4 cm, then the area would be 24 square cm.
Perimeter of a Rectangle: 2(l + b), where l is the length and b is the breadth of the rectangle. The formula for finding the perimeter of a rectangle is 2 times the sum of its length and breadth. For example, if the length of a rectangle is 6 cm and its breadth is 4 cm, then the perimeter would be 20 cm.
Area of a Square: s², where s is the side of the square. The formula for finding the area of a square is the side squared. For example, if the side of a square is 5 cm, then the area would be 25 square cm.
Perimeter of a Square: 4s, where s is the side of the square. The formula for finding the perimeter of a square is 4 times the side. For example, if the side of a square is 5 cm, then the perimeter would be 20 cm.
Area of a Triangle: (1/2)bh, where b is the base and h is the height of the triangle. The formula for finding the area of a triangle is (1/2) times the base times the height. For example, if the base of a triangle is 6 cm and its height is 4 cm, then the area would be 12 square cm.
Volume of a Cube: s³, where s is the side of the cube. The formula for finding the volume of a cube is the side cubed. For example, if the side of a cube is 5 cm, then the volume would be 125 cubic cm.
Volume of a Rectangular Prism: l x b x h, where l, b and h are the length, breadth and height of the rectangular prism respectively. The formula for finding the volume of a rectangular prism is the product of its length, breadth, and height. For example, if the length of a rectangular prism is 6 cm, its breadth is 4 cm, and its height is 5 cm, then the volume would be 120 cubic cm.
Volume of a Sphere: (4/3)πr³, where r is the radius of the sphere. The formula for finding the volume of a sphere is (4/3) times π times the radius cubed. For example, if the radius of a sphere is 5 cm, then the volume would be (4/3) x π x 125 cubic cm.
Surface area of Sphere: 4πr², where r is the radius of the sphere. The surface area of a sphere is the total area of its outer surface. It is equal to four times the area of a great circle of the sphere, which is a circle that passes through the center of the sphere and has the same diameter as the sphere. For example: Suppose the radius of a sphere is 5 cm. Surface area = 4πr² = 4π x 5² = 4π x 25 = 100π square cm. So, the surface area of the sphere with a radius of 5 cm is 100π square cm.

Mensuration Formulas

Mensuration Formulas

Also check:

Important Learning Concepts
Formula of Perimeter Shapes	Area of Parallelogram	Quadrilateral Formulas	Surface area and volume notes	Circles Revision Notes
Types of Vectors	Addition of Vectors	Congruence of Triangles	Pythagoras Theorem	Similarity of Triangles
Area and perimeter of circles	Congruence of Triangles	Pythagoras Theorem	Similarity of Triangles	Independent Events in Probability

CBSE CLASS XII Related Questions

1.
Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.

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2.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:

\(\frac{1}{3}\)
\(\frac{1}{9}\)
\(3\)
\(9\)

View Solution

3.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

local maximum value is 2
local minimum value is \( -2 \)
local maximum value is \( -2 \)
local minimum value \( < \) local maximum value

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4.
Sports car racing is a form of motorsport which uses sports car prototypes.The competition is held on special tracks designed in various shapes.
The equation of a sports car racing track is given as: \[ f(x)= \begin{cases} x^4-4x^2+4, & 0\leq x<3,\\ x^2+40, & x\geq 3 \end{cases} \] Based on this information:

5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

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6.
Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).

Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) is true and Reason (R) is false.
Assertion (A) is false and Reason (R) is true.

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All the Maths Formulas for Mensuration

CBSE CLASS XII Related Questions

1.
Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.

2.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:

3.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

6.
Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).

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All the Maths Formulas for Mensuration

CBSE CLASS XII Related Questions

1.Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.

2.If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:

3. For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

5. Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

6.Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \). Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.

2.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:

3.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

6.
Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).