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

Question

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?

-a+3x-a+3x-a+3x-5+4x

Video Solution

Step-by-Step Solution

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

Area=(a+3x)(5+4x) \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) = (-a + 3x)(-5 + 4x) = (-a)(-5) + (-a)(4x) + (3x)(-5) + (3x)(4x)

Step 3: Simplifying each term, we get:

=5a4ax15x+12x2 = 5a - 4ax - 15x + 12x^2

Step 4: Reorganize the terms:

=12x215x4ax+5a = 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: 12x215x4ax+5a 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.

Answer

Area:

12x215x4ax+5a 12x^2-15x-4ax+5a

Conditions:

x > 1\frac{1}{4}

3x>a