The area of the outermost square tile is [tex]\boxed{{x^2} - 6x + 9}[/tex].Option D is correct.
Further Explanation:
The formula of area of square can be expressed as follows,
[tex]\boxed{Area = \left( s \right) \times \left( s\right)}[/tex]
The formula of perimeter of the rectangle can be expressed as follows,
[tex]\boxed{{\text{Perimeter}} = 4 \times s}[/tex]
Here, s represents the side of square.
Given:
The options are as follows,
A. [tex]{x^2} + 6x - 6[/tex]
B. [tex]{x^2} - 9x + 6[/tex]
C. [tex]{x^2} + 6x - 9[/tex]
D. [tex]{x^2} - 6x + 9[/tex]
Explanation:
The side of the outermost square tile is [tex]x-3.[/tex]
The area of the square can be obtained as follows,
[tex]\begin{aligned}{\text{Area}} &= \left( {{\text{side}}} \right) \times \left( {{\text{side}}} \right) \\&=\left( {x - 3} \right)\left( {x - 3} \right)\\&= {\left( {x - 3} \right)^2} \\&= {x^2} - 2 \times \left( x \right) \times \left( 3 \right) + {\left( 9 \right)^2}\\&= {x^2} - 6x + 9\\\end{aligned}[/tex]
The area of the outermost square tile is [tex]\boxed{{x^2} - 6x + 9}[/tex]. Option D is correct.
Option A is not correct.
Option B is not correct.
Option C is not correct.
Option D is correct.
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Answer details:
Grade: Middle School
Subject: Mathematics
Chapter: Mensuration
Keywords: rectangle, length, square, tile, polynomial expression, area, outermost tile, square tile, side of square, x-3, width, breadth, area of rectangle, dimension, measure of sides, dimensions of rectangle, expression, square.