Problem 140 - Elementary Algebra

Question

An architect is designing a hotel lobby. She wants to have a triangular window looking out to an atrium, with the width of the window 6 feet more than the height. Due to energy restrictions, the area of the window must be 140 square feet. Solve the equation \(\frac{1}{2} h^{2}+3 h=140\) for \(h,\) the height of the window.

Step-by-Step Solution

Answer

The height of the window is 14 feet.

Step 1: Write down the given equation

The provided equation for the height of the window is: \[ \frac{1}{2} h^{2} + 3h = 140 \]

Step 2: Eliminate the fraction

Multiply the entire equation by 2 to eliminate the fraction: \[ h^{2} + 6h = 280 \]

Step 3: Rearrange into standard quadratic form

Rearrange the equation to the form \( ax^2 + bx + c = 0 \): \[ h^{2} + 6h - 280 = 0 \]

Step 4: Use the quadratic formula

The quadratic formula is given by \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 6 \), and \( c = -280 \). Substitute these values into the formula: \[ h = \frac{-6 \pm \sqrt{6^2 - 4(1)(-280)}}{2(1)} \] Simplify under the square root: \[ h = \frac{-6 \pm \sqrt{36 + 1120}}{2} \] \[ h = \frac{-6 \pm \sqrt{1156}}{2} \]

Step 5: Calculate the solutions

Calculate the values under the square root and then solve for both possible values of \( h \): \[ h = \frac{-6 \pm 34}{2} \] This gives us two solutions: \[ h = \frac{28}{2} = 14 \] \[ h = \frac{-40}{2} = -20 \] Since height cannot be negative, we discard \( h = -20 \).

Step 6: Verify the solution

The height of the window is thus \( h = 14 \). To verify, plug \( h = 14 \) back into the original equation: \[ \frac{1}{2}(14^2) + 3(14) = 140 \] Simplify: \[ \frac{1}{2}(196) + 42 = 140 \] \[ 98 + 42 = 140 \], which holds true.

Key Concepts

solving quadratic equations

Solving quadratic equations can seem tricky, but it's easier when broken down step by step.
Here’s a quick overview of the process used to solve the equation \( \frac{1}{2} h^{2}+3 h=140 \).

• Step 1: Identify the equation. First, take note of the given equation, which will typically be in the form of \( ax^2 + bx + c = 0 \), but might need some adjustments.
• Step 2: Eliminate fractions. If there are fractions, as seen here with \( \frac{1}{2} h^{2} + 3h \), multiply through by the denominator to simplify.


• Step 3: Rearrange into standard form. Ensure the equation is in the form \( ax^2 + bx + c = 0 \). In this case, it becomes \( h^{2} + 6h - 280 = 0 \).


• Step 4: Apply the quadratic formula. Use \[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] by substituting the values of a, b, and c.
  For example, with \( a = 1, b = 6, \) and \( c = -280 \), solve for h.


• Step 5: Simplify and solve. Calculate the values under the square root and solve for both possible solutions, then determine which is viable. Here, \( h = 14 \) is the valid height.


• Step 6: Verify. Always plug your solution back into the original equation to ensure it holds true. This confirms your answer is correct.

Understanding each step will help you solve any quadratic equation confidently and accurately.

area of triangles

Calculating the area of a triangle is fundamental, especially in architectural design.
The standard formula for the area of a triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

• Base and height relationship: In many problems, including our example, you might relate the base and height. For our triangular window, the base is 6 feet more than the height.


• Given area formula: The given equation \[ \frac{1}{2} h^2 + 3h = 140 \] is derived by substituting the base (height + 6) into the standard area formula.


• Solving for height: From the derived equation, solving for h helps you determine the dimensions of the triangle.

Applying these concepts ensures the accuracy of dimensions and compliance with design constraints such as energy efficiency.
Always cross-check values after calculation to avoid errors.

architectural design problems

Architectural design often involves geometry and algebra to solve practical problems like window dimensions.
Analyzing the area and height relationship can provide efficient design solutions.

• Practical applications: Consider relevant constraints, for instance, the energy restrictions limiting the window area to 140 square feet in this problem.


• Formulate equations: Translate the problem conditions into mathematical equations as seen here with \[ \frac{1}{2} h^2 + 3h = 140 \]. This represents the energy-efficient area requirement in relation to the window's dimensions.


• Problem-solving: Use algebraic techniques like solving quadratic equations to find viable solutions. Here, it means determining the exact height and subsequently the width of the window.

This methodical approach ensures the design not only meets aesthetic standards but also adheres to practical guidelines.
Always check calculations against the original problem to verify your solutions meet all given criteria.