Jim built two square-shaped walls which he named them as A & B. If the side of wall A is 2 times the side of wall B and the difference of their areas is $12 {ft}^{2}$ , then the side(in feet) of wall A is .

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Solution

Let the side of wall B be $x$ ft.
Accordingly, the side of wall A would be $2 x$ ft.
We know that area of square = ${side}^{2}$
Given that:
The difference of their areas $= 12 {ft}^{2}$
$\Rightarrow (2 x)^{2} - x^{2} = 12$
$\Rightarrow 4 x^{2} - x^{2} = 12$
$\Rightarrow 3 x^{2} = 12$
$\Rightarrow x^{2} = 4$
$\Rightarrow x = 2$ ft.
$∴$ The side of wall A is $2 x = 4$ ft.

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Jim built two square-shaped walls which he named them as A & B. If the side of wall A is 2 times the side of wall B and the difference of their areas is 12 ft2, then the side(in feet) of wall A is .

Let the side of wall B be x ft. Accordingly, the side of wall A would be 2x ft. We know that area of square = side2 Given that: The difference of their areas =12 ft2 ⇒(2x)2−x2=12 ⇒4x2−x2=12 ⇒3x2=12 ⇒x2=4 ⇒x=2 ft. ∴ The side of wall A is 2x=4 ft.

Jim built two square-shaped walls which he named them as A & B. If the side of wall A is 2 times the side of wall B and the difference of their areas is $12 {ft}^{2}$ , then the side(in feet) of wall A is .