Question

If the sides of a square are increased by x% find the percentage increase in its area

Answer 1

Let us have a square of side $\displaystyle\sf {a}$ unit so that it's area will be $\displaystyle\sf {A=a^2}$ unit².

If the sides are increased by $\displaystyle\sf{x\%,}$ length of new side becomes,

$\displaystyle\sf{\longrightarrow a'=a+\dfrac {ax}{100}}$

$\displaystyle\sf{\longrightarrow a'=a\left (1+\dfrac {x}{100}\right)}$

And then area will be,

$\displaystyle\sf{\longrightarrow A'=(a')^2}$

$\displaystyle\sf{\longrightarrow A'=a^2\left (1+\dfrac {x}{100}\right)^2}$

$\displaystyle\sf{\longrightarrow A'=a^2\left (1+\dfrac {2x}{100}+\dfrac {x^2}{10000}\right)}$

Increase in area is,

$\displaystyle\sf{\longrightarrow \Delta A=A'-A}$

$\displaystyle\sf{\longrightarrow\Delta A=a^2\left (1+\dfrac {2x}{100}+\dfrac {x^2}{10000}\right)-a^2}$

$\displaystyle\sf{\longrightarrow\Delta A=a^2\left (\dfrac {2x}{100}+\dfrac {x^2}{10000}\right)}$

$\displaystyle\sf{\longrightarrow\Delta A=A\left (\dfrac {2x}{100}+\dfrac {x^2}{10000}\right)}$

$\displaystyle\sf{\longrightarrow\dfrac {\Delta A}{A}=\dfrac {2x}{100}+\dfrac {x^2}{10000}}$

Then percentage increase in area,

$\displaystyle\sf{\longrightarrow\delta A=\dfrac {\Delta A}{A}\times 100}$

$\displaystyle\sf{\longrightarrow\delta A=\left (\dfrac {2x}{100}+\dfrac {x^2}{10000}\right)100}$

$\displaystyle\sf{\longrightarrow\underline {\underline {\delta A=\left (2x+\dfrac {x^2}{100}\right)\%}}}$