Area and Perimeter Practice Test 4

Area of the rectangle = xy
On reducing the breadth by A% and increasing the length by 20%
Length of the new rectangle = $\frac{120\mathit{x}}{100}\mathit{x}$ 1.2x
Breadth of the new rectangle = y – A% of y = $\mathit{y}\left(1-\frac{\mathit{A}}{100}\right)$
∴ Area of the new rectangle = $1.2\mathit{x}\times \mathit{y}\left(1-\frac{\mathit{A}}{100}\right)$
Now, xy = $1.2\mathit{x}\mathit{y}\left(1-\frac{\mathit{A}}{100}\right)$ ⇒ 1 = $1.2\frac{\left(100-\mathit{A}\right)}{100}$
⇒ 1.2 A = 120 – 100
∴ A = $\frac{20}{1.2}$ = $16\frac{2}{3}$ %