Question

Find the length and breadth of a room, if its area is 120 m² and perimeter 44m.​

Answer 1

$\maltese\LARGE\textsf{\underline{ FiGuRe :-}}$

$\large\textsf{ }$

$\large\textsf{Length = 12m}$

$\boxed{\begin{array}{ c c c c } \; \; \; \; & \; \; \; \; & \; \; \; \; & \; \; \; \; \\\\ \; \; \; \; & \; \; \; \; & \; \; \; \; & \; \; \; \; \\\\ \; \; \; \; & \; \; \; \; & \; \; \; \; & \; \; \; \;\end{array}}$ Breadth = 10m

$\large\textsf{ }$

$\maltese\LARGE\textsf{\underline{ SoLuTioN :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{Assume the length as ' x '}$

$\qquad\tt{:}\longrightarrow\large\textsf{Assume the breadth as ' y ' }$

$\large\textsf{ }$

↦ We know that :-

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{purple}{ {\large\textsf{Area}}_{\large\textsf{( \; Rectangle \; )}} = \large\sf{length \times breadth} }}\\\\\\\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{purple}{ {\large\textsf{Perimeter}}_{\large\textsf{( \; Rectangle \; )}} = \large\textsf{2 ( length + breadth )} }}$

$\large\textsf{ }$

↦ According to the 1st Condition :-

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf{xy = 120 ----- ( i )}}$

$\large\textsf{ }$

↦ According to 2nd Condition :-

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{2 ( x + y ) = 44}$

$\qquad\tt{:}\longrightarrow\large\textsf{x + y = \cfrac{44}{2} }$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf{x + y = 22 ----- ( ii )}}$

$\large\textsf{ }$

↦ Multiple eq. ( ii ) by x :-

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{x ( x + y ) = 22x}$

$\qquad\tt{:}\longrightarrow\large\textsf{x² + xy = 22x}$

$\qquad\tt{:}\longrightarrow\large\textsf{x² + 120 = 22x ----- [ From ( i ) ]}$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf{x² - 22x + 120 = 0}}$

$\large\textsf{ }$

↦ Solving the following equation :-

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{x² - 22x + 120 = 0}$

$\qquad\tt{:}\longrightarrow\large\textsf{x² - 10x - 12x + 120 = 0}$

$\qquad\tt{:}\longrightarrow\large\textsf{x ( x - 10 ) - 12 ( x - 10 ) = 0}$

$\qquad\tt{:}\longrightarrow\large\textsf{( x - 10 ) ( x - 12 ) = 0}$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf{x = 10 or x = 12}}$

$\large\textsf{ }$

↦ Substituting the value of x in ( ii ) :-

↦ When x = 10

↦ y = 22 - 10

↦ y = 12

$\large\textsf{ }$

↦ When x = 12

↦ y = 22 - 12

↦ y = 10

$\large\textsf{ }$

↦ So now two possibilities arise , but we know that most of the times length is greater than breadth :-

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf\textcolor{orange}{Length = 12m}$

$\qquad\tt{:}\longrightarrow\large\textsf\textcolor{orange}{Breadth = 10m}$

$\large\textsf{ }$

Answer 2

Answer:

refer the above attachment...

length and breadth of a room = 10,12