Question

The perimeter of a rectangle is 100m. If its length is increased 10 % and the breadth remains the same, the perimeter is increased by 6%. Find the length and breadth of rectangle​

Answer 1

Answer:

$\large{\underline{\boxed{\sf\green{Length \: \& \: breadth = 30m \: \& \: 20m}}}}$

$\rule{200}{1}$

Let length & breadth be x & y m respectively.

We know that,

$\dag \: \:\bf\blue{Perimeter\: of \: rectangle = 2(l + b)}$

${\underline{\underline{\bold\red{ATQ:-}}}}$

$\hookrightarrow\sf 2(x + y) = 100 \\\\\hookrightarrow\sf (x + y) = 100/2 \\\\\hookrightarrow\sf (x + y) = 50 \\\\\hookrightarrow\sf x = 50 - y \dots \dots \dots (eq.1)$

$\rule{100}{2}$

$\rightarrow\sf 2 \bigg[(x + x \times 10\%) + y \bigg] = 100 + 100 \times 6\% \\\\\rightarrow\sf 2 \bigg[x + \dfrac{x}{10} + y \bigg] = 100 + 6 \\\\\rightarrow\sf \cancel{2}\bigg[\dfrac{10x + x + 10Y}{\cancel{10}} \bigg] = 106 \\\\\rightarrow\sf \dfrac{11x + 10Y}{5} = 106 \\\\\rightarrow\sf 11x + 10Y = 106 * 5 \\\\\rightarrow\sf 11x + 10Y = 530 \\\\\scriptsize\sf\red{\: \: \: [From \: (eq.1) \: :-} ]\\\\\rightarrow\sf 11(50 -y) + 10y = 530 \\\\\rightarrow\sf 550 - 11y + 10y = 530 \\\\\rightarrow\sf - y = 530 - 550 \\\\\rightarrow\large{\boxed{\sf\green{ y = 20 \: m}}}$

${\underline{\dag{\textsc{Putting \: value \: of \: y \: in \: (eq.1) :- }}}}$

$\rightarrow\sf x = 50 - 20 \\\\\rightarrow\large{\boxed{\sf\pink{x = 30 \: m }}}$

$\therefore\sf{Length \: and \: breadth} = \textbf {30m \: \& \: 20m} \: \sf{ respectively}$

Answer 2

AnswEr :

Let the Length be l and Breadth be b of the Rectangle.

$\underline{\bigstar\:\textsf{Perimeter of Rectangle :}}$

$\dashrightarrow\:\:\tt Perimeter =2(Length+ Breadth)\\\\\\\dashrightarrow\:\:\tt 100 = 2(l + b)\\\\\\\dashrightarrow\:\:\tt \dfrac{100}{2}=(l+b)\\\\\\\dashrightarrow\:\:\tt 50=(l+b)\\\\\\\dashrightarrow\:\:\tt b=(50-l)\qquad \dfrac{\quad}{}\:eq.(1)$

$\rule{200}{1}$

$\underline{\bigstar\:\textsf{According to the Question Now :}}$

$:\implies\texttt{Perimeter Increases by 6\% = 2(Length Increases by 10\% + Breadth)}\\\\\\:\implies\tt Perimeter \times (100+6)\% = 2 \times (l (100+10)\% + b)\\\\{\scriptsize\qquad\bf{\dag}\:\:\textsf{Putting value of b from eq.(1)}}\\\\:\implies\tt Perimeter \times 106\% =2 \times (l(110\%)+(50-l))\\\\\\:\implies\tt100 \times \dfrac{106}{100} = 2 \times \bigg(\dfrac{110l}{100} + 50 - l \bigg)\\\\\\:\implies\tt106 = 2 \times \bigg( \dfrac{110l + 5000 - 100l}{100} \bigg)\\\\\\:\implies\tt \dfrac{106}{2} = \dfrac{10l + 5000}{100}\\\\\\:\implies\tt 53 \times 100 = 10l + 5000\\\\\\:\implies\tt5300 - 5000 = 10l\\\\\\:\implies\tt300 = 10l\\\\\\:\implies\tt \dfrac{300}{10} = l\\\\\\:\implies\underline{\boxed{\red{\tt Length = 30\:m}}}$

$\rule{150}{2}$

$\underline{\bigstar\:\textsf{Putting the value of l in eq.(1) :}}$

$:\implies\tt b =(50-l)\\\\\\:\implies\tt b =(50-30)\\\\\\:\implies\underline{\boxed{\red{\tt Breadth = 20\:m}}}$

⠀

$\therefore\:\underline{\textsf{Length \& Breadth is \textbf{30 m \& 20 m} respectively.}}$