Question

If one side of a square is increased by 3 m and other side is reduced by 2 m, a rectangle is formed whose area is 4 m2 more than the area of the original square. Find the side of the original square.

Answer 1

Answer:

3m 2m 4m2 3×2=6m 4×2=8m 6+8=14

Answer 2

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

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.

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