Problem 77 - Prealgebra

Question

The base of a triangle is \(\frac{4}{3}\) feet and the height is \(\frac{2}{3}\) feet. Find the area.

Step-by-Step Solution

Answer

The area of the triangle is \( \frac{4}{9} \) square feet.

Step 1: Recall the Area Formula for a Triangle

To find the area of a triangle, use the formula: \( A = \frac{1}{2} \times \text{base} \times \text{height} \). This formula calculates the area by taking half of the product of the base and the height.

Step 2: Plug in the Given Values

Substitute the given base and height values into the formula. The base is \( \frac{4}{3} \) feet and the height is \( \frac{2}{3} \) feet. Therefore, \( A = \frac{1}{2} \times \frac{4}{3} \times \frac{2}{3} \).

Step 3: Multiply the Fractions

Calculate the product of the base and height. First, multiply the numerators: \( 4 \times 2 = 8 \), and then the denominators: \( 3 \times 3 = 9 \). So the product is \( \frac{8}{9} \).

Step 4: Calculate Half of the Product

Now, calculate half of the product \( \frac{8}{9} \). This is done by multiplying \( \frac{1}{2} \times \frac{8}{9} \), which results in \( \frac{8}{18} \).

Step 5: Simplify the Fraction

Simplify \( \frac{8}{18} \) by dividing both the numerator and the denominator by their greatest common divisor, which is \( 2 \). \( \frac{8}{18} = \frac{4}{9} \).

Step 6: Conclude the Area Calculation

The final area of the triangle is \( \frac{4}{9} \) square feet.

Key Concepts

triangle area formula

The area of a triangle can be found using the formula \( A = \frac{1}{2} \times \text{base} \times \text{height} \). This formula exists because the shape of a triangle can be thought of as half of a rectangle. In a rectangle, the area is calculated by multiplying its length by its width. Similarly, for a triangle, by taking half of the base times the height, we account for the triangle's shape, which is essentially half of a rectangle.

• Base: This is the length of the bottom side of the triangle.
• Height: This is the perpendicular distance from the base to the opposite vertex.

Understanding this basic formula is crucial, as it's used often in geometry to solve for the area of various triangles, whether they are right, equilateral, or isosceles.

fraction multiplication

Multiplying fractions is a straightforward process, but it requires attention to detail: simply multiply the numerators together and then multiply the denominators together. In this problem, the numerators 4 and 2 multiply to give 8, and the denominators 3 and 3 multiply to give 9, resulting in the fraction \( \frac{8}{9} \).

• Step 1: Multiply the numerators (top numbers) together.
• Step 2: Multiply the denominators (bottom numbers) together.
• Result: The new fraction is \( \frac{8}{9} \).

This basic skill of fraction multiplication is indispensable not only in finding areas but also in many other areas of algebra and geometry.

simplifying fractions

Simplifying a fraction involves reducing it to its simplest form, where the numerator and denominator have no common factors other than 1. To simplify \( \frac{8}{18} \), notice that both the numerator and the denominator share a common factor of 2. We divide both by 2, resulting in \( \frac{4}{9} \).

• Find the greatest common divisor (GCD): In this case, it's 2.
• Divide both numbers: 8 divided by 2 is 4, and 18 divided by 2 is 9.
• Simplified fraction: \( \frac{4}{9} \).

Simplifying fractions helps make calculations easier and solutions more elegant. It's an essential skill in arithmetic that supports advanced mathematical concepts.

geometry basics

Geometry is the study of shapes, sizes, and the properties of space. It's an essential branch of mathematics that deals with lines, angles, surfaces, and solids. Understanding how to calculate the area of simple shapes like triangles is foundational in geometry.

• Triangles: Figures with three sides and three angles.
• Importance of Area: Knowing how to calculate area helps in understanding real-world space and in scientific calculations.
• Application: Used in architecture, engineering, and various fields to optimize space and design.

By grasping these basic geometry concepts, students can build a strong foundation for more advanced topics in mathematics and related fields.