Question

If the area of a square reduces by 51%, the percentage reduction in its perimeter will be: ​

Answer 1

Step-by-step explanation:

let the side of the square be 'x'

So the area of the square

$= {x}^{2}$

and the perimeter = 4x

Reduction in area = 51 % of x^2

$= \frac{51 {x}^{2} }{100}$

$so \: reduced \: area \: of \: the \: square$

$= {x}^{2} - \frac{51 {x}^{2} }{100}$

$= \frac{49 {x}^{2} }{100}$

$so \: reduced \: side \: of \: the \: square$

$= \sqrt{ \frac{49 {x}^{2} }{100} }$

$= \frac{7x}{10}$

$now \: reduced \: perimeter \: of \: the \: square$

$= 4 \times \frac{7x}{10}$

$= \frac{14x}{5}$

$reduction \: in \: perimeter \: = 4x - \frac{14x}{5}$

$= \frac{20x - 14x}{5}$

$= \frac{8x}{5}$

$= \frac{ \frac{8x}{20} }{4x} \times 100\%$

$= \frac{8x}{20} \times \frac{1}{4x} \times 100\%$

$= \frac{100}{10} \%$

$= 10\%$