Chapter 2.10, Problem 28P

The measured dimensions of a shape or a solid are given to the nearest whole unit. Find the greatest percent error of each shape or solid.

The perimeter of a rectangle with length $127\text{ft}$ and width $211\text{ft}$.

To determine

To find : the greatest percent error perimeter of rectangle.

Answer to Problem 28P

0.3%

Explanation of Solution

Given information: The length and width of rectangle are 127 ft and 211ft when measured to the nearest ft

Calculation:

Perimeter of the rectangle with length and width are 127 ft and 211ft when measured to the nearest ft is

  $\begin{array}{l}P=2\left(l+w\right)\\ P=2\left(127+211\right)=676ft\end{array}$

The measurement was made to the nearest ft , so the greatest possible error is half of 1ft, that is 0.5 ft.

Therefore, minimum measurement of length and width of the rectangle are (127-0.5) ft by (211-0.5) ft, that is 126.5 ft by 210.5 ft respectively.

And maximum measurement of the dimensions of the rectangle are ((127+0.5) ft by (211+0.5) ft, that is 127.5 ft by 211.5 ft respectively.

Therefore, minimum possible perimter of the rectangle is

  $P_{\min }=2\left(126.5+210.5\right)=674ft$

And maximum possible perimeter of the rectangle is

  $P_{\max }=2\left(127.5+211.5\right)=678ft$

Now, absolute value of difference of perimeter P and $P_{\min }$ is 2 ft. and absolue value of difference of perimeter P and $P_{\max }$ is 2 ft .

Therefore, greatest percentage error in perimeter of rectangle = $\frac{2}{676}\times 100=0.3\%$