Square of Difference: Using quadrilaterals

To express the area of the trapezoid in terms of $X$, we follow these steps:

• Step 1: Identify given measurements of the trapezoid.
  • Large base $= X$.
  • Small base $= X - 5$.
  • Height $= 2X - 5$.
• Step 2: Use the trapezoid area formula: $\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}$
• Step 3: Substitute the values: $\text{Area} = \frac{1}{2} \times (X + (X - 5)) \times (2X - 5)$
• Step 4: Simplify the expression: $\text{Area} = \frac{1}{2} \times (2X - 5) \times (2X - 5)$ $\text{Area} = \frac{1}{2} \times (2X - 5)^2$ $\text{Area} = \frac{1}{2} \times (4X^2 - 20X + 25)$
• Step 5: Conclude with the expression for the area: $\text{Area} = \frac{1}{2}[4X^2 - 20X + 25]$

Therefore, the expression for the area of the trapezoid in terms of $X$ is $\frac{1}{2}[4X^2 - 20X + 25]$.

To solve this problem, we need to express the area of the rectangle in terms of $X$, accounting for the relationship between its sides.

Step 1: Calculate the area of the rectangle.
The area $s$ of a rectangle is obtained by multiplying the length by the width:

$s = X \times (X - 6)$

Expanding the expression gives us:

$s = X^2 - 6X$

We want to express this in a form that matches the given answer choices. Recognizing the square of a binomial will help us reformulate:

$s = X^2 - 6X$ can be related to a square of a binomial by adjusting it:

Step 2: Recast into a recognizable square:
We want to find a relation to $(x-3)^2$, so bear in mind:

$(X - 3)^2 = X^2 - 6X + 9$

Therefore, rearranging in the form of $(X - 3)^2$, we derive:

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

Therefore, the area of the rectangle, expressed in a way to match the correct choices, is $s = (x - 3)^2 - 9$, which corresponds to choice 3 from the given options.

The correct choice is: Choice 3: $s=(x-3)^2-9$.

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:

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

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

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:

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

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

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