Math Problem Statement

The diagram shows a rectangle with area 72 cm². The length is given as 3x + 5 and the width as 3x. Find the value of x and the dimensions of the rectangle.

Solution

The given diagram shows a rectangle with dimensions 3x+53x + 5 (length) and 3x3x (width). The area of the rectangle is 72 cm². We can set up an equation based on the formula for the area of a rectangle, which is:

Area=Length×Width\text{Area} = \text{Length} \times \text{Width}

Substituting the given values:

72=(3x+5)(3x)72 = (3x + 5)(3x)

Expanding the right-hand side:

72=9x2+15x72 = 9x^2 + 15x

Now, we have the quadratic equation:

9x2+15x72=09x^2 + 15x - 72 = 0

We can solve this quadratic equation using the quadratic formula. The quadratic formula is:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For the equation 9x2+15x72=09x^2 + 15x - 72 = 0, we have:

  • a=9a = 9,
  • b=15b = 15,
  • c=72c = -72.

Now, let's calculate the discriminant:

Δ=b24ac=1524(9)(72)\Delta = b^2 - 4ac = 15^2 - 4(9)(-72) Δ=225+2592=2817\Delta = 225 + 2592 = 2817

Now, applying the quadratic formula:

x=15±28172(9)=15±53.0918x = \frac{-15 \pm \sqrt{2817}}{2(9)} = \frac{-15 \pm 53.09}{18}

This gives two possible solutions for xx:

  1. x=15+53.0918=38.09182.12x = \frac{-15 + 53.09}{18} = \frac{38.09}{18} \approx 2.12
  2. x=1553.0918=68.09183.78x = \frac{-15 - 53.09}{18} = \frac{-68.09}{18} \approx -3.78

Since xx represents a dimension, we discard the negative value. Thus, x2.12x \approx 2.12.

Now, substituting x=2.12x = 2.12 back into the expressions for the dimensions:

  • Length: 3x+5=3(2.12)+5=6.36+5=11.36cm3x + 5 = 3(2.12) + 5 = 6.36 + 5 = 11.36 \, \text{cm}
  • Width: 3x=3(2.12)=6.36cm3x = 3(2.12) = 6.36 \, \text{cm}

Thus, the dimensions of the rectangle are approximately 11.36cm×6.36cm11.36 \, \text{cm} \times 6.36 \, \text{cm}.

Would you like further details on this? Here are some questions to expand your understanding:

  1. What would the area be if xx were a different value?
  2. Can you solve the quadratic equation using factoring instead of the quadratic formula?
  3. How would the dimensions change if the area were doubled?
  4. How do you find the perimeter of this rectangle?
  5. What are the possible values for the width and length if we allow negative dimensions?

Tip: When solving quadratic equations, always check for negative or non-real solutions, especially when dealing with physical dimensions like length and width.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Area of Rectangles

Formulas

Area of a rectangle = length × width
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Quadratic formula

Suitable Grade Level

Grades 9-10