Area of a Rectangle: Using short multiplication formulas

First, let's recall the formula for calculating the area of a rectangle:

The area of a rectangle (which has two pairs of equal opposite sides and all angles are $90\degree$) with sides of length $a,\hspace{4pt} b$ units, is given by the formula:

$\boxed{ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=a\cdot b }$(square units)

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After recalling the formula for the area of a rectangle, let's solve the problem:

First, let's denote the area of the given rectangle as: $S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}$ and write (in mathematical notation) the given information:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2-13$

Let's continue and calculate the area of the rectangle given in the problem:

$$

Using the rectangle area formula mentioned earlier:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=a\cdot b\\ \downarrow\\ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=(x-4)(x+1)$

Let's continue and simplify the expression we got for the rectangle's area, using the distributive property:

$(c+d)(h+g)=ch+cg+dh+dg$

Therefore, we get that the area of the rectangle by

using the distributive property is:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=(x-4)(x+1) \\ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2+x-4x-4\\ \boxed{ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2-3x-4}$

Now let's recall the given information:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2-13$

Therefore, we can conclude that:

$x^2-3x-4=x^2-13 \\ \downarrow\\ -3x=4-13\\ -3x=-9\hspace{6pt}\text{/}(-3)\\ \boxed{x=3}$

We solved the resulting equation simply by combining like terms, isolating the expression with the unknown on one side and dividing both sides by the unknown's coefficient in the final step,

Note that this result satisfies the domain of definition for x, which was given as:

-1\text{<}x\text{<}4 and therefore it is the correct result

Therefore, the correct answer is answer C.

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:

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

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

We know that:

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

We also know that AB equals X.

First, we will substitute the given data into the formula accordingly:

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

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

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

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

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

Now let's look at triangle ABC and use the Pythagorean theorem:

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

We substitute in our known values:

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

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

Finally, we will add 1 to both sides:

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

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

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:

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

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

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