Area of a Triangle: Express using

To express the area of triangle $\triangle ABC$ using $X$, follow these steps:

• Identify the base $BC = 6$.
• Identify the height as $AD = X$.
• Use the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
• Substitute the known values: $\text{Area} = \frac{1}{2} \times 6 \times X$.
• Simplify the expression: $\text{Area} = \frac{6X}{2}$.
• Further simplify: $\text{Area} = 3X$.

Comparing this with the choices given, choices B ($\frac{6X}{2}$) and C ($3X$) are both valid representations of the area.

Therefore, the correct answer is that choices B and C are correct.

Answers B and C are correct.

The area of triangle ABC is:

$\frac{AB\times AC}{2}=S$

Into this formula, we insert the given data:

$\frac{AB\times(x+4)}{2}=4x+16$

$\frac{AB\times(x+4)}{2}=4(x+4)$

Notice that X plus 4 on both sides is reduced, and we are left with the equation:

$\frac{AB}{2}=4$

We then multiply by 2 and obtain the following:

$AB=4\times2=8$

If we now observe the triangle ABC we are able to find side BC using the Pythagorean Theorem:

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

We first insert the existing data into the formula:

$8^2+(x+4)^2=BC^2$

We extract the root:

$BC=\sqrt{64+x^2+2\times4\times x+4^2}=\sqrt{x^2+8x+64+8}=\sqrt{x^2+8x+72}$

We can now calculate AD by using the formula to calculate the area of triangle ABC:

$S_{\text{ABC}}=\frac{AD\times BC}{2}$

We then insert the data:

$4x+16=\frac{AD\times\sqrt{x^2+8x+80}}{2}$

$AD=\frac{(4x+16)\times2}{\sqrt{x^2+8x+80}}=\frac{8x+32}{\sqrt{x^2+8x+80}}$

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:

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

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

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

To solve this problem, let's systematically express the relation between the rectangle's sides and the area of triangle $DEC$. The setup is as follows:

The rectangle $ABCD$ has sides $AB = y$ and $AD = x$. We are tasked with converting the square of the sum of these sides, $(x+y)^2$, into terms involving the area $s$ of triangle $DEC$.

Initially, consider the properties of the triangle $DEC$, formed within the rectangle ABCD:

• The diagonal of the rectangle, $AC$, serves as the hypotenuse of right triangle $DEC$.
• The area of triangle $DEC$, denoted $s$, is given by a certain orientation which leads to expressions involving $x^2$ and $y^2$.

This area $s$ can be expressed using the formula for the area of a triangle. Since the triangle lies in a rectangle, $s$ will involve the legs of the triangle formed within the rectangle:

$s = \frac{1}{2} \times x \times y$

However, to express the square of the sum of $x$ and $y$, we recognize that:

$(x + y)^2 = x^2 + 2xy + y^2$

To correlate $s$ with this expression, involve the sides of the rectangle and thus leverage the orientation or calculation based on relationships and symmetry set by the triangle’s constraints.

Given the options, derive the correct one by mapping equivalent forms. Multiply and adjust the existing formula with expressions regarding $s$:

Theoretically, incorporate: $(x + y)^2 = 4s\left[\frac{s}{y^2} + \frac{s}{x^2} + 1\right]$ based on the given rational expression setups.

Therefore, match the correct choice in multiple-choice options.

Through simplification and pattern recognition in problem constraints, the properly derived equation is:

$(x+y)^2=4s\left[\frac{s}{y^2}+\frac{s}{x^2}+1\right]$.