The perimeter of the printed part of the page is 29 inches.
Explanation
Length of the sheet of paper is 11 inches and width is [tex]8\frac{1}{2}[/tex] or [tex]\frac{17}{2}[/tex] inches.
Width of the margin is [tex]1\frac{1}{4}[/tex] or [tex]\frac{5}{4}[/tex] inches.
As the margin is on all sides of the paper, that means both across the length and width of the paper we need to leave [tex]\frac{5}{4}[/tex] inches for that margin.
So, the length of the printed part [tex]=11-(2*\frac{5}{4})= 11- \frac{5}{2} =\frac{17}{2}[/tex] inches
and the width of the printed part [tex]= \frac{17}{2}-(2*\frac{5}{4})= \frac{17}{2}-\frac{5}{2}= \frac{12}{2}=6[/tex] inches.
Thus the perimeter of the printed part of the page will be: [tex]2(\frac{17}{2} +6)= 2(\frac{29}{2})= 29[/tex] inches.
Answer:
Step-by-step explanation:
Given:
Dimensions of the paper:
Length, l = 11 in
Width, b = 8 × 1/2 in
= 17/2 in
Printed paper dimensions:
Length, lp = length, l - margin
= 11 - 1 × 1/4
= 39/4 in
Width, bp = width, w - margin
= 17/2 - 1 × 1/4
= 29/4 in
Perimeter of a rectangle, P = 2 × (lp × bp)
= 2 × (39/4 + 29/4)
= 68/2 in
= 34 in
