Question

the length of the sides of a squre was decreased by 20% By how many percent did the area decrease?​

Answer 1

Answer:

$let \: the \: side \: be \: x \\ area = {x}^{2} \\ side \: decreased \: by \: 20\% = x - x \times 20\% \\ = x - \frac{x}{5} \\ = \frac{5x - x}{5} = \frac{4x}{5} \\ area \: of \: side \: decreased \: by \: 20\% = ({ \frac{4x}{5} })^{2} = \frac{16 {x}^{2} }{25} \\ area \: decreased = {x}^{2} - \frac{16 {x}^{2} }{25} = \frac{25 {x}^{2} - 16 {x}^{2} }{25} = \frac{ {9x}^{2} }{25} \\ persent = \frac{ {9x}^{2} }{25} \div x \times 100 \\ = \frac{9 {x}^{2} }{25} \times \frac{1}{x} \times 100 = 36\%$

Answer 2

Answer:

let the side be = x

therefore area = x²

now 20% decrease in side = x - 20%of'x'

= x - 20x/100

= x - x/5

= 4x/5

therefore, area = 16x²/25

now , difference = x²-16x²/25

= (25x²-16x²)/25

= 9x²/25

therefore, decrease in percentage =

[(difference)/(initial area)] × 100

= [(9x²/25)/(x²)] × 100

= (9/25) × 100

= 9 × 4

= 36%

Hope it helped :)