Problem 61 - Algebra A Combined Function

Question

Solve. The base of a triangle is four more than twice its height. If the area of the triangle is 42 square centimeters, find its base and height.

Step-by-Step Solution

Answer

The height is 6 cm and the base is 16 cm.

Step 1: Determine Variables

Let's assign variables for the base and height of the triangle. Let \( h \) be the height of the triangle. Since the base is four more than twice its height, we can express the base as \( b = 2h + 4 \).

Step 2: Write the Area Formula

The area \( A \) of a triangle can be found using the formula \( A = \frac{1}{2} \times \text{base} \times \text{height} \). For this problem, the area is given as 42 square centimeters.

Step 3: Set Up the Equation

Substitute the expressions for the base and the area into the area formula:\[ 42 = \frac{1}{2} \times (2h + 4) \times h \]

Step 4: Simplify the Equation

Multiply both sides by 2 to eliminate the fraction:\[ 84 = (2h + 4)h \]Now, distribute \( h \) on the right-hand side:\[ 84 = 2h^2 + 4h \]

Step 5: Rearrange to Form a Quadratic Equation

Rearrange the terms to form a standard quadratic equation:\[ 2h^2 + 4h - 84 = 0 \]

Step 6: Simplify and Solve the Quadratic Equation

Divide every term by 2 to simplify:\[ h^2 + 2h - 42 = 0 \]Factor the quadratic equation:\[ (h + 7)(h - 6) = 0 \]Set each factor to zero and solve for \( h \):\[ h + 7 = 0 \Rightarrow h = -7 \]\[ h - 6 = 0 \Rightarrow h = 6 \]Only positive height makes sense, so \( h = 6 \) cm.

Step 7: Calculate the Base

Substitute \( h = 6 \) back into the expression for the base:\[ b = 2(6) + 4 = 12 + 4 = 16 \] cm.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are essential in solving problems involving geometrical figures such as triangles. In this exercise, we encountered a quadratic equation derived from the triangle's area formula. A quadratic equation generally takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) is the variable.

• In the solution, the expression \( 2h^2 + 4h - 84 = 0 \) was transformed into a quadratic equation related to the height of the triangle.
• We simplified it by dividing all terms by 2, resulting in \( h^2 + 2h - 42 = 0 \).
• The factors of the simplified equation were \((h + 7)(h - 6) = 0\), providing two potential solutions for \( h \).

Quadratic equations can often be solved by factoring, using the quadratic formula, or by completing the square. Here, factoring was used because the factors neatly broke down into integers, allowing us to directly find the possible values of \( h \). Only positive solutions are valid in this context, as a height must be a non-negative value.

Geometry Problem Solving in Context

Geometry problem solving often involves translating a verbal description into a mathematical expression. In this case:

• We started with a real-world geometry problem about a triangle.
• We expressed the relationship between the base and height based on the problem's conditions, defining the base in terms of the height as \( b = 2h + 4 \).

The important step was forming a usable expression from the words in the problem description. By identifying the right equations and expressions, we effectively bridge geometrical intuition with algebraic calculation.
Remember: Visualizing and drawing can help if you're ever stuck. Try sketching the triangle and labeling the known variables. This can aid in setting up the relationship between base and height clearly, which is crucial in these types of problems.

Working with Mathematical Expressions

Mathematical expressions transform complex word problems into solvable equations.

• In this exercise, the expression for the area of the triangle, \( A = \frac{1}{2} \times b \times h \), allowed us to set up an equation describing the entire problem.
• Substituting the expression for \( b \) into the area formula led to the formation of our main equation \( 42 = \frac{1}{2}(2h + 4)h \).

The trick is to simplify these expressions step by step, as we did by converting the area formula into a quadratic equation. Always look to substitute known relationships (like \( b = 2h + 4 \)) back into your expressions. Simplifying expressions through proper algebraic manipulation is a skill that not only applies to geometry but also to other branches of mathematics.
Mastery of these skills helps in efficiently moving from understanding word problems to performing precise algebraic calculations.