Final answer:
The width of the painting is 23.5025 inches, and the length, being 5.54 inches taller, is 29.0425 inches.
Step-by-step explanation:
We are tasked with finding the dimensions of John's famous rectangular painting given its perimeter is 105.09 inches and that its length is 5.54 inches taller than its width. Let's tackle this problem step by step.
a) Length in terms of width:
Let x represent the width of the painting. Then, the length would be x + 5.54 inches.
b) Expression for the perimeter:
The perimeter P of a rectangle is given by P = 2(length + width). Therefore, the expression for the perimeter using x is P = 2(x + (x + 5.54)).
c) Equation based on the given perimeter:
To set up the equation, we'll use the given perimeter P = 105.09 inches and substitute into the equation from step b), yielding 105.09 = 2(x + (x + 5.54)).
d) Solve the equation:
Simplifying the equation gives: 105.09 = 2(2x + 5.54), thus 105.09 = 4x + 11.08. Subtracting 11.08 from both sides results in: 94.01 = 4x, and dividing by 4 gives x = 23.5025 inches.
e) Dimensions of the painting:
The width of the painting is 23.5025 inches and the length, being 5.54 inches greater, is 29.0425 inches.