1. ## algebra problem

Hi friends.
Please can you say me what is the correct way to solve this problem? Thank you

A rectangular area with a square at its center. If the long side of rectangle is more large than short side in 8 and the side of square is (1/10) of the short side of rectangle, the area A (area without square) is.....

if x: short side, then A=x(x+8)-(1/10)(x)

or x: long side, then A=(x-8)x-(1/10)(x-8)

2. ## Re: algebra problem

Hello, user!

A rectangular area with a square at its center.
The long side of rectangle is 8 more than its short side.
The side of square is (1/10) of the short side of rectangle.
Find the area A (area without square).

Let $x$ = short side of rectange.
Then $x+8$ = long side of rectange.
. . Area of rectangle: $x(x+8)$

Then $\tfrac{1}{10}x$ = side of the square.
. . Area of square: ${\color{blue}\left(\tfrac{x}{10}\right)^2}$

Therefore: $A \;=\;x(x+8) - \tfrac{x^2}{100}$

3. ## Re: algebra problem

Thank you Soroban. But if i take x:long side of rectangle ?

4. ## Re: algebra problem

Hello again, user!

Thank you, Soroban.
But if i take $x$ = long side of rectangle?

Let $x$ = long side of the rectangle.
Then $x-8$ = short side of the rectangle.
. . Area of rectangle: $x(x-8)$

Then $\frac{x-8}{10}$ = side of the square.
. . Area of square: $\left(\frac{x-8}{10}\right)^2$

Therefore: . $A \;=\;x(x-8) - \left(\frac{x-8}{10}\right)^2$

5. ## Re: algebra problem

thanks Soroban, when I obtain the both solutions, they are equals. Have a nice day.