Square of sum: Using quadrilaterals

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:

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

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

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

Given the trapezoid where the height is equal to the sum of the two bases.

It is known that the difference between the large base and the small base is 5

We will mark the small base with X

Express the area of the trapezoid using X

$\frac{1}{2}\lbrack4x^2+20x+25\rbrack$

Given two squares, one side of the squares is larger by 2 than the other. The area of the large square is larger than the perimeter of the small square by 20

Find the length of the small square

4

Given a rectangle whose side is greater by 6 than the other side. We mark the area of the rectangle with S

What is the correct argument?

9+S equal to the smaller side plus 3 squared (the two squared).

Given a square of side length X

We will mark the area of the square by S and the perimeter of the square by P

Check the correct statement

$P+S+4=(x+2)^2$

Shown below is the rectangle ABCD.

AB = y

AD = x

Express the square of the sum of the sides of the rectangle using the area of the triangle DEC.

$(x+y)^2=4s\lbrack\frac{s}{y^2}+\frac{s}{x^2}+1\rbrack$