Problem 88 - Elementary Algebra

Question

For exercises 87-88, use the five steps and a polynomial equation to find the base \(b\) and height \(h\) of the triangle. The formula for the area \(A\) of a triangle is \(A=\frac{1}{2} b h\). The height of a triangle is \(4 \mathrm{ft}\) more than the length of its base. The area of the triangle is \(70 \mathrm{ft}^{2}\).

Step-by-Step Solution

Answer

Base = 10 ft, Height = 14 ft

Step 1: Set up equations

Use the formula for the area of a triangle, given as: \[ A = \frac{1}{2} b h \] We know the height (h) is 4 ft more than the length of its base (b), so we can write: \[ h = b + 4 \] Also, given the area (A) is 70 ft².

Step 2: Substitute the known values

Replace A with 70 and h with b+4 in the formula: \[ 70 = \frac{1}{2} b (b + 4) \]

Step 3: Simplify the equation

Multiply both sides by 2 to get rid of the fraction: \[ 140 = b (b + 4) \] Simplify to: \[ 140 = b^2 + 4b \]

Step 4: Form a polynomial equation

Rearrange the equation to standard polynomial form: \[ b^2 + 4b - 140 = 0 \]

Step 5: Solve the quadratic equation

Solve the quadratic equation using the quadratic formula: \[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \] Here, A = 1, B = 4, and C = -140. Substituting these values into the equation: \[ b = \frac{-4 \pm \sqrt{4^2 - 4(1)(-140)}}{2(1)} \] Simplify: \[ b = \frac{-4 \pm \sqrt{16 + 560}}{2} \] \[ b = \frac{-4 \pm \sqrt{576}}{2} \] Since \( \sqrt{576} = 24 \), we get: \[ b = \frac{-4 \pm 24}{2} \] So, \[ b = \frac{20}{2} = 10 \] or \[ b = \frac{-28}{2} = -14 \] (which is not a valid base length)

Step 6: Calculate height

Using the valid value of b, which is 10, and substituting in \( h = b + 4 \): \[ h = 10 + 4 = 14 \]

Key Concepts

Quadratic Formula

The Quadratic formula is a powerful tool used in algebra to solve quadratic equations. These are equations of the form: \[ ax^2 + bx + c = 0 \] The formula to find the roots (values of x) is: \[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \] Here's how it works:

• A = the coefficient of the x² term.
• B = the coefficient of the x term.
• C = the constant term.

To apply the quadratic formula, simply plug in the values of A, B, and C. Then, use the discriminant, \( B^2 - 4AC \), to determine if the solutions are real or imaginary. If the discriminant is positive, you get two real solutions. If it's zero, you get one real solution. If it's negative, there are no real solutions, only imaginary ones.

Area of a Triangle

The area of a triangle can be calculated using the basic area formula for triangles: \[ A = \frac{1}{2} bh \] where:

• A = area of the triangle.
• b = base of the triangle.
• h = height of the triangle.

In this specific problem, we are given the area (70 ft²) and the relation between the base and height: the height is 4 ft more than the base. By substituting these values into the area formula, we set up the equation: \[ 70 = \frac{1}{2} b (b + 4) \] Multiplying both sides by 2 to simplify, we get: \[ 140 = b(b + 4) \]

Polynomial Equation

A polynomial equation is an equation made up of variables and coefficients, involving terms with non-negative integer exponents. In this case, we transformed our area equation into a standard polynomial form. From:\[ 140 = b(b + 4) \] we rearranged it to:\[ b^2 + 4b - 140 = 0 \] This is a quadratic polynomial equation because the highest exponent of the variable b is 2. Solving such an equation typically involves either factoring, completing the square, or using the quadratic formula.

Geometry

Geometry helps us understand the shapes and spaces around us. In this problem, we deal with the geometry of a triangle. Using the given information that the height of the triangle is 4 feet more than the base, we linked geometry with algebra to set up our polynomial equation. Recognizing these geometric relationships is critical for solving real-world problems. By understanding the properties of geometric figures like triangles, we can create algebraic equations to find unknown measurements.

Algebraic Substitution

Algebraic substitution is the process of replacing a variable with its equivalent value. Here, we knew that the height (h) is 4 feet more than the base (b), allowing us to write: \[ h = b + 4 \] We substituted this into our area formula: \[ 70 = \frac{1}{2} b (b + 4) \] This substitution simplified our equation into one variable, making it easier to solve. It's a fundamental technique in algebra, essential for solving equations where variables are expressed in terms of others.