Rectangle Area with Variables: Solving for -a+3x and -5+4x Dimensions

Look at the rectangle in the figure below. What is its area?

What do a and x need to be for the rectangle to exist?

To solve this problem, we'll follow these steps:

• Step 1: Calculate the area of the rectangle using the area formula for a rectangle.

• Step 2: Identify the conditions required for a valid rectangle by ensuring positive dimensions.

• Step 3: Analyze the provided choices to identify the correct answer.

Now, let's work through each step:

Step 1: The width of the rectangle is given as $-a + 3x$, and the height is $-5 + 4x$. The area of a rectangle is calculated by multiplying these two dimensions:

$\text{Area} = (-a + 3x)(-5 + 4x)$

Step 2: We'll expand the expression for the area:

$= (-a + 3x)(-5 + 4x) = (-a)(-5) + (-a)(4x) + (3x)(-5) + (3x)(4x)$

Step 3: Simplifying each term, we get:

$= 5a - 4ax - 15x + 12x^2$

Step 4: Reorganize the terms:

$= 12x^2 - 15x - 4ax + 5a$

Next, let's determine the conditions for the rectangle to exist, which means both dimensions must be positive:

• Width: $-a + 3x > 0 \implies 3x > a$

• Height: $-5 + 4x > 0 \implies 4x > 5 \implies x > \frac{5}{4} = 1\frac{1}{4}$

Therefore, the conditions for the rectangle to exist are $3x > a$ and $x > 1\frac{1}{4}$.

By evaluating the provided choices, we can see the correct choice is:

Area: $12x^2-15x-4ax+5a$

Conditions: $x > 1\frac{1}{4}$ and $3x > a$.

Thus, the correct choice is option 4. Confirming with the given correct answer, our solution matches perfectly.