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Chapter 10: Problem 58

Sign Dimensions. The base of a triangular sign is 4 in. longer than twice the height. The area of the sign is 255 in \(^{2} .\) Find the dimensions of the sign.

Short Answer

Expert verified
Height = 15 inches, Base = 34 inches.

Step by step solution

01

- Define Variables

Let the height of the triangular sign be denoted as h inches. The base of the triangular sign is given to be 4 inches longer than twice the height, so we can express the base as 2h + 4 inches.
02

- Write the Area Formula for a Triangle

Recall the formula for the area of a triangle: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]. Given the area of the sign is 255 square inches, substitute the known values into the formula: \[ 255 = \frac{1}{2} \times (2h + 4) \times h \].
03

- Simplify the Equation

Simplify the area formula equation: \[ 255 = \frac{1}{2} \times (2h^2 + 4h) \] Multiply both sides by 2 to clear the fraction: \[ 510 = 2h^2 + 4h \]
04

- Rearrange the Equation

Rearrange the equation into standard quadratic form: \[ 2h^2 + 4h - 510 = 0 \]
05

- Solve the Quadratic Equation

Use the quadratic formula \[ h = \frac{-b \,\pm\, \sqrt{b^2 - 4ac}}{2a} \] where a = 2, b = 4, and c = -510. Substitute these values into the formula: \[ h = \frac{-4 \,\pm\, \sqrt{4^2 - 4 \times 2 \times (-510)}}{2 \times 2} \] Simplify under the square root: \[ h = \frac{-4 \,\pm\, \sqrt{16 + 4080}}{4} \] \[ h = \frac{-4 \,\pm\, 64}{4} \].This gives two solutions: \[ h = \frac{60}{4} = 15 \] and \[ h = \frac{-68}{4} = -17 \] Discard the negative value since height cannot be negative. Hence, h = 15 inches.
06

- Find the Base

Substitute h = 15 back into the expression for the base: \[ \text{base} = 2h + 4 = 2(15) + 4 = 34 \] inches.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

area of a triangle
Understanding the formula for the area of a triangle is crucial in solving geometry problems. The area of a triangle can be calculated using the formula:
  • Area = \( \frac{1}{2} \times \text{base} \times \text{height} \)
In this exercise, we are given that the area of the triangular sign is 255 square inches. This means that:
  • 255 = \( \frac{1}{2} \times \text{base} \times \text{height} \)
By substituting the expressions for the base and the height from our defined variables, we can simplify and solve this equation. Understanding and correctly applying this formula is the first step to solving problems involving the area of a triangle.
variable definition in algebra
In algebra, defining variables is a critical step in solving equations. Variables represent unknown values that we need to find. In this problem, we define:
  • \(h\) as the height of the triangular sign
  • \(2h + 4\) as the base of the triangular sign
By defining these variables clearly, they help us write equations that are easier to manage. For example, knowing that the base is 4 inches longer than twice the height allows us to substitute \(2h + 4\) in place of the base in our formula.
Clear variable definitions simplify complex problems, making it easier to understand and solve equations.
solving equations
Solving equations, especially quadratic equations, is a fundamental skill in algebra. Here's a quick overview of how to approach it:1. **Simplify the Equation:** Once we substitute our variables into the area formula, we simplify the equation:
  • 255 = \( \frac{1}{2} \times (2h + 4) \times h \)
  • 510 = 2h虏 + 4h
2. **Rearrange into Standard Form:** We convert it into the standard quadratic equation form:
  • 2h虏 + 4h - 510 = 0
3. **Use the Quadratic Formula:** To solve for \(h\), we use the quadratic formula:
  • \(h = \frac{-b \,\pm\, \sqrt{b虏 - 4ac}}{2a}\) where \(a = 2\), \(b = 4\), and \(c = -510\)
4. **Solve for the Values:** Calculating, we get two potential solutions:
  • \(h = 15\)
  • \(h = -17\) (discarded as height cannot be negative)
5. **Find the Base:** Substitute \(h = 15\) back to find the base:
  • \(2h + 4 = 34\) inches
By following these steps, we solve the quadratic equation and find the correct dimensions of the triangular sign.

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Most popular questions from this chapter

Oxford Builders has an extension cord on their generator that permits them to work, with electricity, anywhere in a circular area of \(3850 \mathrm{ft}^{2}\). Find the dimensions of the largest square room they could work on without having to relocate the generator to reach each corner of the floor plan.

Review simplifying rational expressions (Sections\(7.1-7.5)\) Perform the indicated operation and, if possible, simplify. $$ \frac{x^{2}-1}{x^{2}-4} \div \frac{x^{2}-x-2}{x^{2}+x-2}[7.2] $$

Let \(f(x)=x^{2} .\) Find each of the following. $$f(\sqrt{6}-\sqrt{3})$$

Multiply and simplify. Assume that no radicands were formed by raising negative numbers to even powers. $$ \sqrt{10} \sqrt{14} $$

Review simplifying rational expressions (Sections\(7.1-7.5)\) Perform the indicated operation and, if possible, simplify. $$ \frac{a^{-1}+b^{-1}}{a b}[7.5] $$
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