Square of sum: Using multiple rules

Given the rectangle ABCD

AB=X

The ratio between AB and BC is $\sqrt{\frac{x}{2}}$

We mark the length of the diagonal A the rectangle in m

Check the correct argument:

Given that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

Given that AB equals X

We will substitute accordingly in the formula:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's focus on triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

Let's substitute the known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

We'll add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$

$m^2+1=(x+1)^2$

$(a+3b)^2-(3b-a)^2=\text{?}$

$12ab$

$(x+3)^2+(x-3)^2=\text{?}$

$2x^2+18$

Find $a ,b$ such that:

$(a+b)(a-b)=(a+b)^2$

$a=-b$ o

$0=b$

$(x+y)^2-(x-y)^2+(x-y)(x+y)=\text{?}$

$x^2+4xy-y^2$

Find a X given the following equation:

$(x+3)^2+(2x-3)^2=5x(x-\frac{3}{5})$

$6$

Given a circle whose center O. From the center of the circle go out 2 radii that cut the circle at the points A and B.

Given AO⊥OB.

The side AB is equal to and+2.

Express band and the area of the circle.

$\frac{\pi}{2}[y^2+4y+4]$

$(x+3)^2=(x-3)^2$

$x=0$