Problem 86 - Intermediate Algebra

Question

Find the area and perimeter of the trapezoid. (Hint: The area of a trapezoid is the product of half the height \(6 \sqrt{3}\) meters and the sum of the bases \(2 \sqrt{63}\) and \(7 \sqrt{7}\) meters.)

Step-by-Step Solution

Answer

Area is \(39\sqrt{21}\) square meters. Perimeter needs additional side lengths.

Step 1: Identify the Given Values

We are given the height of the trapezoid as \( h = 6 \sqrt{3} \) meters, base1 (\( b_1 \)) as \( 2 \sqrt{63} \) meters, and base2 (\( b_2 \)) as \( 7 \sqrt{7} \) meters.

Step 2: Simplify the Expressions for the Bases

We simplify the given bases:- \( 2 \sqrt{63} = 2 \cdot 3 \cdot \sqrt{7} = 6\sqrt{7} \)- \( 7 \sqrt{7} \) is already simplified.

Step 3: Calculate the Sum of the Bases

Add the simplified bases together:\(b_1 + b_2 = 6\sqrt{7} + 7\sqrt{7} = 13\sqrt{7}\)

Step 4: Compute the Area of the Trapezoid

The formula for the area \( A \) of a trapezoid is:\[ A = \frac{1}{2} \times (b_1 + b_2) \times h \]Substitute the known values:\(A = \frac{1}{2} \times 13\sqrt{7} \times 6\sqrt{3}\)Simplify:\(A = 39 \sqrt{21} \text{ square meters}\)

Step 5: Consider Perimeter of the Trapezoid

The perimeter involves knowing all side lengths. However, since only the two bases are given with their exact measures, additional information about the non-parallel sides is required to compute the exact perimeter, which is not provided in the problem statement.

Key Concepts

Understanding the Perimeter of a Trapezoid

The perimeter of a trapezoid is the sum of the lengths of all its sides. To find the perimeter, you need the two bases and the lengths of the non-parallel sides, known as the legs. In the context of this problem, only the two bases are given: base1 (\( b_1 \) = \( 6\sqrt{7} \text{ meters} \)) and base2 (\( b_2 \) = \( 7\sqrt{7} \text{ meters} \)). However, information on the legs is missing, making it impossible to find the complete perimeter.To completely determine the perimeter, you would need:

• The lengths of each base
• The lengths of the legs

Once these values are known, simply add them all together for the total perimeter.

Exploring Geometry Formulas

Geometry is full of handy formulas that help us understand different shapes. For a trapezoid, the most important formulas relate to its area and perimeter.The formula for the area of a trapezoid is:\[ A = \frac{1}{2} \, \times \, (b_1 + b_2) \, \times \, h \]This formula allows you to find the area using the average length of the bases multiplied by the height. It is essential for swiftly calculating the surface that a trapezoid covers.Another important formula is for the perimeter:\[ P = b_1 + b_2 + \text{leg}_1 + \text{leg}_2 \]These fundamental geometry formulas are powerful tools that make calculations easier, saving time and effort.

Trapezoid Properties Simplified

A trapezoid, known as a trapezium in some countries, is a quadrilateral with one set of parallel sides, called bases. The other two sides are non-parallel and referred to as the legs. Here's a quick rundown of key properties:

• Has one pair of parallel sides (bases)
• The height is the perpendicular distance between the bases
• The non-parallel sides are called legs
• Diverse types of trapezoids include isosceles (non-parallel sides are equal) or right trapezoids (one angle is 90 degrees)

Understanding these properties is crucial when working with a trapezoid's area or perimeter, as they dictate how you apply geometry formulas.

Simplifying Radicals in Trapezoid Problems

Radicals are numbers under a square root sign, and simplifying them is a common task in geometry problems involving trapezoids. In our problem, we encountered expressions like \( 6\sqrt{7} \) and \( 7\sqrt{7} \).When simplifying radicals, follow these steps:

• Factor the number under the radical to its prime factors
• Take out pairs of factors from under the square root
• Multiply remaining factors within the radical

For example, simplifying \( 2\sqrt{63} \) involves breaking it down to \( 2 \times 3 \times \sqrt{7} = 6\sqrt{7} \). This skill is useful for making expressions easier to work with and thus facilitates smoother calculations in geometry problems.