Hello, Monocerotis!

A picture that measures 10 cm × 8 cm is to be surrounded by a mat.

The width of the mat is to be the same on all sides of the picture.

The area of the mat is to equal the area of the picture.

What is the width of the mat, to the nearest tenth of a centimeter?

I can't figure out how to begin working this one out. Start by making a sketch . . .

Code:

: x : - 8 - : x :
- * - - - - - - - * - -
x | | :
- | * - - - * | :
: | | | | :
: | | | | :
10 | |10 | | 2x+10
: | | | | :
: | | 8 | | :
- | * - - - * | :
x | | :
- * - - - - - - - * - -
: - - 2x+8 - - :

The dimensions of the picture is: .$\displaystyle 8\times10$

. . Its area is: .$\displaystyle 8\cdot10 \:=\:80\text{ cm}^2$

The dimensions of the frame is: .$\displaystyle (2x+8) \times (2x+10)$

. . Its area is: .$\displaystyle (2x+8)(2x+10) $

$\displaystyle \text{Area of mat} \;=\;\text{Area of frame} - \text{Area of picture}$

n . . . . . . .$\displaystyle = \;(2x+8)(2x+10) - 80$

n . . . . . . .$\displaystyle = \;4x ^2 + 36x$

We are told: .$\displaystyle \text{Area of mat} \:=\:\text{Area of picture}$

. . . . . . . . . . .$\displaystyle 4x^2 + 36x \;\;=\quad 80$

We have: .$\displaystyle 4x^2 + 36x - 80 \:=\:0 \quad\Rightarrow\quad x^2 + 9x - 20 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{-9\pm\sqrt{161}}{2} $

Therefore: .$\displaystyle x \:=\:\frac{-9 +\sqrt{161}}{2} \;\approx\;1.8\text{ cm}$