Question

A Rectangular Field has area 45m and perimeter 28m.Its dimensions are

Answer 1

$\bf\purple{\underline{\boxed{Question}}}$

$\:$

$\small\text{Rectangular Field has area 45m and}\\{\small\text{ perimeter 28m . Its dimensions are}}$

$\:$

$\bf\purple{\underline{\boxed{Solution}}}$

$\:$

✏

$\small\text{As we know , The perimeter of}\\{\small\text{ a rectangle is 2(L+B)}}$

$\:$

$\small\text{perimeter of a rectangle is = 28 }$

$\:$

$\small\text{Acc. to Question}$

$\:$

$\small\text{2(L+B) = 28 }$

$\:$

$\small\tt{L+B\:=\:{\cancel\frac{28}{2}}}$

$\:$

$\small\tt{L+B\:=\:14}$

$\:$

$\small\tt{L\:=\:14\:-\:B\:\:\:\:......(1)}$

$\:$

$\small\text{As we know , The area of}$

$\small\tt{ a \:rectangle\: is\: (L\times B)}$

$\:$

$\small\text{perimeter of a rectangle is = 45 }$

$\:$

$\small\text{Acc. to Question}$

$\:$

$\small\tt{L\times B\: =\: 45 }$

$\:$

$\small\text{Substituting value of L }$

$\:$

$\small\tt{(14-B)\times B \:=\: 45 }$

$\:$

$\small\tt{14B-B^{2}\:=\: 45 }$

$\:$

$\small\tt{14B-B^{2}\:-45\:=\: 0 }$

$\:$

$\small\text{Taking (-) common }$

$\:$

$\small\tt{B^{2}\:-14B\:+\:45\:=\: 0 }$

$\:$

$\small\tt{B^{2}\:-\:14B\:+\:45}$

$\:$

$\small\tt{Let's\: factorise\: this\: equation}$

$\:$

$\small\tt{We\:will\:get,}$

$\:$

$\small\tt{B^{2}-9B\:-\:5B\:+\:45}$

$\:$

$\small\tt{B(B-9)-5(B-9)}$

$\:$

$\small\tt{(B-5)\:(B-9)=0}$

$\:$

$\small\tt{Therefore,}$

$\:$

$\small\tt\red{B={5}\:\:,\:B\:=\:{9}}$

$\:$

$\small\tt\red{Hence\:,\:Length\:of\: the\: rectangle\:is\:=\:9}$

$\:$

$\small\tt\red{And\:,\:Breadth\:of\: the\: rectangle\:is\:=\:5}$

$\:$

$\bf\purple{\underline{\boxed{Thanks}}}$