Math, asked by Neea, 10 months ago

if one side of a square is increased by 3m and the other side is reduced by 2m , the rectangle formed whose area is 4 m square more than the area of the original square. find the side of the original square​

Answers

Answered by Anonymous
13

Answer:

Let x be the side of the square.

Area of the square= x^2

Area of the rectangle

= length * breadth

= (x+3)(x-2)

Given that area if the rectangle is 4 m.sq more than area of the square

(x+3)(x-2)= (x^2) + 4

(x^2) + 3x - 2x -6 = (x^2) + 4

(x^2) + x -6 = (x^2) + 4

x -6 = 4

x= 4+6

x=10.

hope it helps you mate.

please thank and mark my answer as brainliest.

@ ANUSHA

Answered by nilesh102
16

 \mathfrak {\huge \underline \red{question} :  - }

if one side of a square is increased by 3m and the other side is reduced by 2m , the rectangle formed whose area is 4 m square more than the area of the original square. find the side of the original square.

\mathfrak{ \huge  \underline \red{solution : -  }} \\  \\  \underline\purple{according \: to \: question} \: \\  \\  \red{  \: one \: side \: of \: square \: increased \: by \: 3 \: m} \\  \red { \: and \: other \: side \: redused \: by \: 2 \: m.} \\  \\  \purple{ \: let \: side \: be \: x \:and \: assume \: }  \\ \:\:  \: \\  \\ length \:  = ( x + 3) \:  \:  and \: \: breadth = (x - 2)  \\  \\ Area \: of \: rectangle \:  =  length \:  \times breadth \\  \\ Area \: of \: rectangle \:  =  (x + 3) \times (x - 2)\:  \:  \:  \:  ..........(1)\\  \\  \red{the \:  rectangle  \: formed \:  whose \:  area \:  is  \: 4  \: m \: }  \\  \red{square  \: more \:  than  \: the  \: area  \: of  \: the  \: original } \\  \red{square.} \\  \\  \purple{so}  \\  \\  =  >Area \:of \: square \:   +   \:   4\\ \\  =  >  {(side)}^{2}   +  4 \\  \\  =  >  {(x)}^{2}    +  4 \:  \:  \:  \:  \: .............(2) \\  \\  \underline \red{by \: question \: form \: (1)  \:  \: and \:  \: (2).} \\  \\  =  > (x + 3)(x -2) =  {(x)}^{2}  +  4 \\  \\  =  >  {x}^{2}  - 2x + 3x - 6 =  {x}^{2}   +  4 \\  \\  =  >  {x}^{2}  + x - 6 =  {x}^{2}  +  4 \\  \\  =  > x - 6 = 4 \\  \\  =  > x = 4 + 6 \\  \\ =  >  x = 10 m \\  \\  \fcolorbox{red}{white}{hence \: original \: side \: of \: square \: is \: 10 \: m} \\  \\ \\  \\   \fcolorbox{red}{aqua}{i \: hope \: it \: helps \: you.}

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