Question

a rectangular painting measures 11 inches by 17 inches and contains a frame of uniform width around the four edges. the perimeter of the rectangle formed by the painting and its frame is 72 inches. determine the width of the frame. the width of the frame is inch(es).

Explanation:

Step1: Express outer - rectangle dimensions

Let the width of the frame be $x$ inches. The length of the outer - rectangle is $(17 + 2x)$ inches and the width of the outer - rectangle is $(11+2x)$ inches.

Step2: Use the perimeter formula

The perimeter formula for a rectangle is $P = 2(l + w)$. Here, $P=72$, $l = 17 + 2x$, and $w = 11 + 2x$. So, $72=2((17 + 2x)+(11 + 2x))$.

Step3: Simplify the equation

First, simplify the expression inside the parentheses: $(17 + 2x)+(11 + 2x)=28 + 4x$. Then the equation becomes $72 = 2(28 + 4x)$. Distribute the 2 on the right - hand side: $72=56 + 8x$.

Step4: Solve for $x$

Subtract 56 from both sides of the equation: $72−56=8x$, which gives $16 = 8x$. Divide both sides by 8: $x = 2$.

Answer:

2