Question

1. the length of a rectangle is 4 times its width. the area of the rectangle is 144 square inches. find the dimensions of the rectangle. let l = a = lw show set up, then solve. ( )( ) = 0 factor solve answer =

Explanation:

Step1: Define variables

Let $W$ be the width of the rectangle. Then the length $L = 4W$ since the length is 4 times the width. The area formula is $A = LW$, and $A=144$. Substitute $L = 4W$ into the area - formula: $144=(4W)W$.

Step2: Simplify the equation

$(4W)W = 4W^{2}$, so the equation becomes $4W^{2}=144$. Divide both sides of the equation by 4: $\frac{4W^{2}}{4}=\frac{144}{4}$, which simplifies to $W^{2}=36$.

Step3: Solve for $W$

Take the square - root of both sides. Since we are dealing with a dimension (a non - negative quantity), $W=\sqrt{36}=6$ (we ignore $W = - 6$ because width cannot be negative).

Step4: Solve for $L$

Since $L = 4W$, substitute $W = 6$ into this equation. So $L=4\times6 = 24$.

Answer:

The width of the rectangle is 6 inches and the length is 24 inches.