Square of Difference: Using quadrilaterals

Given the rectangle ABCD

AB=X the ratio between AB and BC is equal to$\sqrt{\frac{x}{2}}$

We mark the length of the diagonal $A$ with $m$

Check the correct argument:

Let's find side BC

Based on what we're given:

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

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

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

Let's divide by square root x:

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

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

Let's reduce the numerator and denominator by square root x:

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

We'll use the Pythagorean theorem to calculate the area of triangle ABC:

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

Let's substitute what we're given:

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

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

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

Given the rectangle ABCD

AB=Y AD=X

Triangular area DEC equal to S

Expresses the square of the difference of the sides of the rectangle

band means of X, Y and S

Since we are given the length and width, we will substitute them according to the formula:

$(AD-AB)^2=(x-y)^2$

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

The height is equal to side AD, meaning both are equal to X

Let's calculate the area of triangle DEC:

$S=\frac{xy}{2}$

$x=\frac{2S}{y}$

$y=\frac{2S}{x}$

$xy=2S$

Let's substitute the given data into the formula above:

$(\frac{2S}{y})^2-2\times2S+(\frac{25}{x})^2$

$\frac{4S^2}{y^2}-4S+\frac{4S^2}{x^2}$

$4S=(\frac{S^2}{y}+\frac{S^2}{x}-1)$

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

Given a trapezoid whose 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 equal to 5. We will mark the large base with X

Express the area of the trapezoid using X

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

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

and the large side with X

Check the correct argument:

$s=(x-3)^2-9$