What Is the formula of finding breadth of Rectangle

Content Curator

The formula for finding the breadth (b) of a rectangle is b = Area / Length. Where Length (l) is the longer side of the rectangle, and Area (A) is the product of the length and breadth.

Note: The breadth and width of a rectangle are used interchangeably, and they refer to the same dimension.

Related Links

Length and Breadth of Rectangle

Length and Breadth of Rectangle

Let us Understand the Formula through Examples

Example 1: A rectangle has an area of 120 square meters and a length of 20 meters. What is the breadth of the rectangle?

Solution: Breadth = 120 ÷ 20 = 6 meters

Example 2: A rectangle has an area of 144 square inches and a length of 12 inches. What is the breadth of the rectangle?

Solution: Breadth = 144 ÷ 12 = 12 inches

Example 3: A rectangle has an area of 150 square centimeters and a length of 30 centimeters. What is the breadth of the rectangle?

Solution: Breadth = 150 ÷ 30 = 5 centimeters

Important Points to Remember while Finding Breadth of a Rectangle

The breadth of a rectangle is one of the four important measurements required to fully describe the size and shape of a rectangle.
The breadth of a rectangle is equal to the width of the rectangle if it is oriented with its sides parallel to the x- and y-axes.
The breadth and length of a rectangle can be interchanged to create a different rectangle with the same area.
In a square, the breadth and length are equal, which makes finding the breadth of a square easier than a rectangle.

Also check:

Related Topics
Quadrilateral Angle Sum Property	Area of Parallelogram	Perimeter of a Parallelogram
Area of Parallelogram and Triangles	Difference between Square and Rectangle	Geometry Formula

CBSE CLASS XII Related Questions

1.
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

View Solution

2.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

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, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.

View Solution

3.
In a school, the probability of holding a debate competition is \( \frac{1}{3} \) and that of a quiz competition is \( \frac{2}{3} \). In the two participating teams, A has 4 girls and 6 boys and B has 7 girls and 3 boys. If a debate competition is held, the students are selected from team A and for the quiz competition they are selected from team B. If only two students are to be chosen from the teams, then find the probability that one will be a girl and the other a boy.

View Solution

4.
The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:

\([-1, 1]\)
\([4, 6]\)
\([-7, -3]\)
\([2, 3]\)

View Solution

5.
A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.

View Solution

6.
For a square matrix \(A\), \[ (3A)^{-1}= \]

\( 3A^{-1} \)
\( 9A^{-1} \)
\( \frac{1}{3} A^{-1} \)
\( \frac{1}{9} A^{-1} \)

View Solution

What Is the formula of finding breadth of Rectangle?

Let us Understand the Formula through Examples

CBSE CLASS XII Related Questions

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

2.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

4.
The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:

5.
A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.

6.
For a square matrix \(A\), \[ (3A)^{-1}= \]

Comments

SUBSCRIBE TO OUR NEWS LETTER

What Is the formula of finding breadth of Rectangle?

Let us Understand the Formula through Examples

CBSE CLASS XII Related Questions

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

2.For two vectors \(\vec{a}\) and \(\vec{b}\): Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

4.The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:

5. A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.

6. For a square matrix \(A\), \[ (3A)^{-1}= \]

Comments

SUBSCRIBE TO OUR NEWS LETTER

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

2.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

4.
The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:

5.
A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.

6.
For a square matrix \(A\), \[ (3A)^{-1}= \]