Solve the problem.
-A rug is to fit in a room so that a border of even width is left on all four sides. If the room is 10 feet by 14 feet and the area of the rug is 21 square feet, how wide will the border be?
A) $5.5 \mathrm{ft}$
B) $3.5 \mathrm{ft}$
C) $4.5 \mathrm{ft}$
D) $6 \mathrm{ft}$

Question

Quizplus · Accepted Answer

The width of the border can be found by first calculating the dimensions of the rug and then the width of the border. Let $x$ be the width of the border. The area of the room is $10 \times 14 = 140 \, \text{sq ft}$, and the area of the rug is $21 \, \text{sq ft}$. The dimensions of the rug would then be $10 - 2x$ by $14 - 2x$, since the border is on all sides. The equation for the area of the rug is $(10 - 2x)(14 - 2x) = 21$. Simplifying this equation directly to find $x$ is complex, but knowing the total area and the rug's area helps us understand the problem better. The correct approach involves recognizing that the area outside the rug (which is the area of the border around the rug) is $140 - 21 = 119 \, \text{sq ft}$. However, the direct calculation involves setting up the equation based on the rug's area and solving for $x$, which represents the border's width. The mistake in the explanation is assuming a direct calculation from the given areas without correctly setting up or solving the equation. The correct method involves understanding that the dimensions of the rug and the room lead to a linear equation in $x$, which upon solving gives the width of the border. Given the error in the initial explanation, the focus should be on the relationship between the room's dimensions, the rug's area, and the border's width. Solving for $x$ correctly involves using the given dimensions and the area of the rug to find that the border width is $3.5 \, \text{ft}$, making option B correct.

Solve the Problem 5.5ft5.5 \mathrm{ft}5.5ft B) 3.5ft3.5 \mathrm{ft}3.5ft C) 4.5ft4.5 \mathrm{ft}4.5ft D) 6ft6 \mathrm{ft}6ft

Solve the Problem $5.5 \mathrm{ft}$ B) $3.5 \mathrm{ft}$ C) $4.5 \mathrm{ft}$ D) $6 \mathrm{ft}$