## Algebra 1

To solve for the perimeter of B, we would first find the dimensions of rectangle A. (1) Since the Area of A is 180 $ft^{2}$ and the width of A is 20 ft we can find length by dividing the area by width as A=L$\times$W and therefore 180 = L$\times$20 and so L=$\frac{180}{20}$ = 9ft. (2) Since the Area of B is 45 $ft^{2}$, we can let width=x ft and we can find length by dividing the area by width as A=L$\times$W and therefore 45 = L$\times$x and so L=$\frac{45}{x}$ ft (3) So using similar triangles we can say that $\frac{L_{Rectangle A}}{L_{Rectangle B}}$ = $\frac{W_{Rectangle A}}{W_{Rectangle B}}$ $\frac{9}{\frac{45}{x}}$ = $\frac{20}{x}$. --> using cross multiplication bring $\frac{45}{x}$ to the other side 9= $\frac{20}{x}$ $\times$ $\frac{45}{x}$. 9= $\frac{900}{x^{2}}$ x = $\sqrt\frac{900}{9}$ = $\sqrt{100}$ x=10, x=-10 (x=-10 is not possible because negative width is not possible) Width is 10 ft and so Length is $\frac{45}{10}$ = 4.5 ft. Therefore the perimeter of Rectangle B = 2(10)+2(4.5) = 29 ft.