Problem 19
Question
Find the area of the triangle with base \(b\) and height \(h .\) \(b=z, h=5 z\)
Step-by-Step Solution
Verified Answer
The area of the triangle is \(\frac{5}{2} z^2\).
1Step 1: Understand Triangle Area Formula
The area of a triangle can be calculated using the formula:\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]We will use the given values for base and height in this formula.
2Step 2: Substitute Values into Formula
Substitute the given values into the formula. Here, the base \(b\) is given as \(z\) and the height \(h\) is given as \(5z\). So the formula becomes:\[ \text{Area} = \frac{1}{2} \times z \times 5z \]
3Step 3: Simplify Expression
Now, simplify the expression: \[ \text{Area} = \frac{1}{2} \times 5z^2 \]\[ \text{Area} = \frac{5}{2} z^2 \]
4Step 4: Conclude Area Calculation
The area of the triangle is calculated as:\[ \frac{5}{2} z^2 \]
Key Concepts
Base and Height in TrianglesUnderstanding the Area FormulaWorking with Algebraic Expressions
Base and Height in Triangles
When talking about triangles, the terms 'base' and 'height' are essential. The base of a triangle is any one of its sides, usually chosen for convenience based on the problem context. It's where the height falls perpendicularly. While the base can be any side, the height, also known as the altitude, is always perpendicular to the base. This means the height forms a right angle with the base.
Imagine a right-triangle—its height meets the base at a 90-degree angle. In our exercise, the base is represented by the variable \( z \), while the height is five times longer, given as \( 5z \). Recognizing these elements will always help in calculating the triangle's area.
Imagine a right-triangle—its height meets the base at a 90-degree angle. In our exercise, the base is represented by the variable \( z \), while the height is five times longer, given as \( 5z \). Recognizing these elements will always help in calculating the triangle's area.
Understanding the Area Formula
The area of a triangle can be calculated with a simple yet powerful formula: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). This formula shows that the area of a triangle is half the product of its base and height, reflecting the triangle's half nature of a rectangle.
A triangle shares its base and height with any rectangle that might complement it to form a square or rectangle. Knowing this relation, especially in algebra, helps to use other methods or formulas, enhancing problem-solving flexibility. It's crucial to commit to memory how to apply these measurements effectively for various arithmetic tasks.
A triangle shares its base and height with any rectangle that might complement it to form a square or rectangle. Knowing this relation, especially in algebra, helps to use other methods or formulas, enhancing problem-solving flexibility. It's crucial to commit to memory how to apply these measurements effectively for various arithmetic tasks.
Working with Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and operations. They let us solve problems in an organized way. Here, our task was to find the area of a triangle with the variables \( z \) and \( 5z \) for its base and height, respectively. Substituting these into our formula—\( \text{Area} = \frac{1}{2} \times z \times 5z \)—leads to \( \text{Area} = \frac{1}{2} \times 5z^2 \).
When simplifying, you multiply the coefficients and apply the rules of exponents to get \( \frac{5}{2} z^2 \). By mastering the art of simplifying expressions, students unlock the door to more complex algebraic manipulations. Always remember: take your time to ensure each step is clear before moving to the next.
When simplifying, you multiply the coefficients and apply the rules of exponents to get \( \frac{5}{2} z^2 \). By mastering the art of simplifying expressions, students unlock the door to more complex algebraic manipulations. Always remember: take your time to ensure each step is clear before moving to the next.
Other exercises in this chapter
Problem 19
Use grouping to factor the polynomial. \(x^{3}+3 x^{2}+2 x+6\)
View solution Problem 19
Simplify. $$ \frac{4}{7} \div \frac{8}{7} $$
View solution Problem 20
Add the polynomials. $$\left(5 y^{2}+y^{3}\right)+\left(12 y^{2}-5 y^{3}\right)$$
View solution Problem 20
Exercises \(17-34:\) Evaluate the expression by hand. Check your result with a calculator. $$ (-2)^{4} $$
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