Area and Perimeter Practice Test 3

Area of the square = ${\mathit{x}}^2$ sq. m.
Side of the new square = x – 50% of x = $\frac{\mathit{x}}{2}$ m
∴ Area of the new square = ${\left(\frac{\mathit{x}}{2}\right)}^2$ = $\frac{{\mathit{x}}^2}{4}$ sq. m.
∴ Reduction in area of the square = ${\mathit{x}}^2-\frac{{\mathit{x}}^2}{4}$ = $\frac{3{\mathit{x}}^2}{4}$ sq. m.
∴ Percentage reduction = $\frac{3{\mathit{x}}^2∕4}{{\mathit{x}}^2}$ × 100 = 75 %