Question

ivanna wants to paint a rectangular region of a wall with one layer of paint. she has enough paint to cover a region that is 16 m². before painting, she wants to put a strip of tape around the border of the region. she is considering three regions with widths of 1 m, 4 m, and 8 m. answer the questions below to find which of these regions would require the least tape. (a) fill in the table to find the length and the perimeter for each region. | | width | length | perimeter | |---|---|---|---| | region #1 | 1 m | □ m | □ m | | region #2 | 4 m | □ m | □ m | | region #3 | 8 m | □ m | □ m | (b) which of these regions would require the least tape? ○ region #1 ○ region #2 ○ region #3

Explanation:

Step1: Find length of Region #1

Since area $A = \text{width} \times \text{length}$, length $l = \frac{A}{w}$. For $A=16\ \text{m}^2$, $w=1\ \text{m}$:
$l = \frac{16}{1} = 16\ \text{m}$

Step2: Find perimeter of Region #1

Perimeter $P = 2(w + l)$. Substitute $w=1$, $l=16$:
$P = 2(1 + 16) = 2\times17 = 34\ \text{m}$

Step3: Find length of Region #2

Use $l = \frac{A}{w}$. For $w=4\ \text{m}$:
$l = \frac{16}{4} = 4\ \text{m}$

Step4: Find perimeter of Region #2

Substitute $w=4$, $l=4$ into perimeter formula:
$P = 2(4 + 4) = 2\times8 = 16\ \text{m}$

Step5: Find length of Region #3

Use $l = \frac{A}{w}$. For $w=8\ \text{m}$:
$l = \frac{16}{8} = 2\ \text{m}$

Step6: Find perimeter of Region #3

Substitute $w=8$, $l=2$ into perimeter formula:
$P = 2(8 + 2) = 2\times10 = 20\ \text{m}$

Step7: Compare perimeters

Compare $34\ \text{m}$, $16\ \text{m}$, $20\ \text{m}$ to find the smallest.

Answer:

(a) Completed Table:

	Width	Length	Perimeter
Region #2	4 m	4 m	16 m
Region #3	8 m	2 m	20 m

(b)

Region #2