If the length of one side of a square is triple and the length of an adjacent side is increased by 10, the resulting rectangle has an area that is 6 times the area of the original square. Find the length of a side pf the original square.

Answer:10Step-by-step explanation:Given: If the length of one side of a square is triple and the length of an adjacent side is increased by [tex]10[/tex].To Find: If area is [tex]6[/tex] times the area of original square find length of a side of original square.Solution:Let the side of original square be [tex]=\text{x}[/tex]area of original square [tex]=\text{x}^2[/tex]when length of side is tripled,new length of one side of square [tex]=3\text{x}[/tex]length of other side is increased by 10 unitnew length of other side of square [tex]=\text{x}+10[/tex]new area of resulting rectangle [tex]=\text{length of one side}\times\text{length of other side}[/tex]                     [tex]=(\text{x}+10)\times(3\text{x})[/tex]                     [tex]3\text{x}^{2}+30[/tex]As area of resulting rectangle is 6 times the original square            [tex]3\text{x}^{2}+30=6\text{x}^2[/tex]            [tex]3\text{x}^2-30=0[/tex]            [tex]3\text{x}(\text{x}-10)=0[/tex]            [tex]\text{x}=10,0[/tex]as length cannot be zero            [tex]\text{x}=10[/tex]Hence the length of a side of original square is [tex]10[/tex]