Question

The length of a rectangle is 4 cm more than its breadth. If the perimeter of the rectangle is 40 , find it’s length and breadth. What is the area of a square which has the same perimeter as this rectangle?

Answer 1

$\huge\purple{ \underline{ \boxed{\mathbb{\red{QUESTION : }}}}}$

The length of a rectangle is 4 cm more than its breadth. If the perimeter of the rectangle is 40 cm, find it’s length and breadth. What is the area of a square which has the same perimeter as this rectangle?

$\huge\purple{ \underline{ \boxed{\mathbb{\red{ANSWER : }}}}}$

Length of rectangle = 12 cm

Breadth of rectangle = 8 cm

Area of square = 100 ${ \sf cm}^{ \: 2}$

$\huge\purple{ \underline{ \boxed{\mathbb{\red{EXPLANATION : }}}}}$

$\green{ \large \underline{ \mathbb{\underline{GIVEN : }}}}$

The length of a rectangle is 4 cm more than its breadth.

The perimeter of the rectangle is 40 cm.

Perimeter of square = Perimeter of rectangle.

$\green{ \large \underline{ \mathbb{\underline{TO \: FIND : }}}}$

The length and breadth of the rectangle .

Area of square .

$\green{ \large \underline{ \mathbb{\underline{FORMULAS \: USED: }}}}$

$\purple{ \boxed{ \textsf{Perimeter of rectangle = 2 \: ( L + B)}}}$

$\purple{ \boxed{ \textsf{Perimeter of square = 4 \times side}}}$

$\purple{ \boxed{ \textsf{Area of square = side \times side}}}$

$\green{ \large \underline{ \mathbb{\underline{SOLUTION: }}}}$

$\red{ \underline{\underline{ \textbf{Case 1 : }}}}$

Let breadth of rectangle = x cm

Then length of rectangle = x + 4 cm

$\red{ \underline{\underline{ \textbf{According to question : }}}}$

$\textsf{Perimeter of rectangle = 2 \: ( L + B)}$

$\displaystyle \longrightarrow \bf40 \: cm = 2 \: (x + 4 + x) \\ \\ \displaystyle \longrightarrow \bf40 \: cm = 2 \: (2x + 4) \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf40 \:cm = 4x + 8 \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf4x + 8 = 40 \: cm \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf4x = 40 - 8 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf4x = 32 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf x = \frac{32}{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf x = 8 \: cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{ \boxed{ \begin{array}{l} \textsf{Breadth of rectangle = x = 8 cm}\\ \\ \textsf{Length of rectangle = x + 4 = 8 + 4 = 12 \: cm }\end{array}}}$

$\red{ \underline{\underline{ \textbf{Case 2 : }}}}$

Let side of square as x .

$\red{ \underline{\underline{ \textbf{According to question : }}}}$

$\textsf{Perimeter of square = Perimeter of rectangle} \\ \\ \longrightarrow \sf{4 \times side = 40 \: cm} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \sf{4 \times x = 40} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \sf{x = \frac{40}{4} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \sf{x = 10 \: cm} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\textsf{Side of square = 10 cm}$

$\textsf{Area of square = side \times side} \\ \\ \textsf{Area of square = 10 \times 10} \: \: \: \: \\ \\ \textsf{Area of square = 100 \: { \sf cm}^{2} }$

$\purple{ \boxed{ \begin{array}{l} \textsf{Area of square = 100 \: \sf{cm}^{2} }\end{array}}}$

Answer 2

$\huge\purple{ \underline{ \boxed{\mathbb{\red{QUESTION : }}}}}$

The length of a rectangle is 4 cm more than its breadth. If the perimeter of the rectangle is 40 cm, find it’s length and breadth. What is the area of a square which has the same perimeter as this rectangle?

$\huge\purple{ \underline{ \boxed{\mathbb{\red{ANSWER : }}}}}$

Length of rectangle = 12 cm

Breadth of rectangle = 8 cm

Area of square = 100 ${ \sf cm}^{ \: 2}$

$\huge\purple{ \underline{ \boxed{\mathbb{\red{EXPLANATION : }}}}}$

$\green{ \large \underline{ \mathbb{\underline{GIVEN : }}}}$

The length of a rectangle is 4 cm more than its breadth.

The perimeter of the rectangle is 40 cm.

Perimeter of square = Perimeter of rectangle.

$\green{ \large \underline{ \mathbb{\underline{TO \: FIND : }}}}$

The length and breadth of the rectangle .

Area of square .

$\green{ \large \underline{ \mathbb{\underline{FORMULAS \: USED: }}}}$

$\purple{ \boxed{ \textsf{Perimeter of rectangle = 2 \: ( L + B)}}}$

$\purple{ \boxed{ \textsf{Perimeter of square = 4 \times side}}}$

$\purple{ \boxed{ \textsf{Area of square = side \times side}}}$

$\green{ \large \underline{ \mathbb{\underline{SOLUTION: }}}}$

$\red{ \underline{\underline{ \textbf{Case 1 : }}}}$

Let breadth of rectangle = x cm

Then length of rectangle = x + 4 cm

$\red{ \underline{\underline{ \textbf{According to question : }}}}$

$\textsf{Perimeter of rectangle = 2 \: ( L + B)}$

$\displaystyle \longrightarrow \bf40 \: cm = 2 \: (x + 4 + x) \\ \\ \displaystyle \longrightarrow \bf40 \: cm = 2 \: (2x + 4) \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf40 \:cm = 4x + 8 \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf4x + 8 = 40 \: cm \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf4x = 40 - 8 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf4x = 32 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf x = \frac{32}{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \longrightarrow \bf x = 8 \: cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{ \boxed{ \begin{array}{l} \textsf{Breadth of rectangle = x = 8 cm}\\ \\ \textsf{Length of rectangle = x + 4 = 8 + 4 = 12 \: cm }\end{array}}}$

$\red{ \underline{\underline{ \textbf{Case 2 : }}}}$

Let side of square as x .

$\red{ \underline{\underline{ \textbf{According to question : }}}}$

$\textsf{Perimeter of square = Perimeter of rectangle} \\ \\ \longrightarrow \sf{4 \times side = 40 \: cm} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \sf{4 \times x = 40} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \sf{x = \frac{40}{4} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \sf{x = 10 \: cm} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\textsf{Side of square = 10 cm}$

$\textsf{Area of square = side \times side} \\ \\ \textsf{Area of square = 10 \times 10} \: \: \: \: \\ \\ \textsf{Area of square = 100 \: { \sf cm}^{2} }$

$\purple{ \boxed{ \begin{array}{l} \textsf{Area of square = 100 \: \sf{cm}^{2} }\end{array}}}$