Math Problem Solving: Step-by-Step Solutions for Algebra & Geometry
Work through a diverse set of word problems covering circles, rectangles, squares, right triangles, and profit calculations. Try setting up the algebraic equations yourself first, then open each solution to compare your work with the detailed steps.
Useful Formulas for These Problems
- Circumference of a circle: \(C = 2\pi r\)
- Rectangle perimeter: \(P = 2L + 2W\)
- Rectangle area: \(A = L \times W\)
- Square area: \(A = s^2\)
- Profit: \(P = \text{Revenue} - \text{Cost}\)
- Pythagorean theorem: \(h^2 = a^2 + b^2\)
- Quadratic formula: \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Practice Problems and Solutions
Problem 1: Circumference of a circle
Question: The circumference of a circle is \(72\pi\). Find the radius of this circle.
Solution:
Use the circumference formula:
\[ C = 2\pi r \]Substitute \(C = 72\pi\):
\[ 72\pi = 2\pi r \]Divide both sides by \(2\pi\):
\[ r = \frac{72\pi}{2\pi} = 36 \]Problem 2: Rectangular garden dimensions
Question: The length of a rectangular garden is 2 feet longer than 3 times its width. If the perimeter is 100 feet, find the width and length.
Solution:
Let \(L\) and \(W\) be the length and width. From the description:
\[ L = 3W + 2 \]The perimeter is \(P = 2L + 2W\). Substituting \(P = 100\) and the expression for \(L\):
\[ 100 = 2(3W + 2) + 2W \] \[ 100 = 6W + 4 + 2W \] \[ 100 = 8W + 4 \] \[ 96 = 8W \implies W = 12 \ \text{feet} \]Find the length:
\[ L = 3(12) + 2 = 38 \ \text{feet} \]Check: \(2(12) + 2(38) = 24 + 76 = 100\). ✔
Problem 3: Rectangular field area
Question: A rectangular field has a length 10 feet more than its width. If the area is 264 square feet, what are the dimensions?
Solution:
Let \(x\) be the length and \(y\) the width:
\[ x = y + 10 \]The area is \(A = xy = 264\). Substitute \(x\):
\[ (y + 10)y = 264 \] \[ y^2 + 10y - 264 = 0 \]Factor (two numbers multiplying to \(-264\) and adding to \(10\)):
\[ (y - 12)(y + 22) = 0 \]Reject the negative root. Thus \(y = 12\) ft, and:
\[ x = 12 + 10 = 22 \ \text{feet} \]Check: \(12 \times 22 = 264\). ✔
Problem 4: Profit from units sold
Question: A company's revenue is \(x^2 + 100x\) dollars and costs are \(240x + 500\) dollars. How many units must be sold to make a profit of \(10{,}000\) dollars?
Solution:
Profit is revenue minus cost:
\[ P = (x^2 + 100x) - (240x + 500) = x^2 - 140x - 500 \]Set \(P = 10{,}000\):
\[ 10000 = x^2 - 140x - 500 \] \[ x^2 - 140x - 10500 = 0 \]Apply the quadratic formula:
\[ x = \frac{140 \pm \sqrt{(-140)^2 - 4(1)(-10500)}}{2} = \frac{140 \pm \sqrt{61600}}{2} \] \[ x \approx \frac{140 \pm 248.19}{2} \]Take the positive root:
\[ x \approx 194.1 \]The company must sell approximately 194 units.
Problem 5: Squares with related areas
Question: A square has a side 5 cm shorter than a second square. The area of the larger square is four times the area of the smaller square. Find the side of each square.
Solution:
Let \(x\) be the smaller side and \(y\) the larger side:
\[ x = y - 5 \] \[ y^2 = 4x^2 \]Substitute \(x = y - 5\):
\[ y^2 = 4(y - 5)^2 \]Take the positive square root of both sides:
\[ y = 2(y - 5) \] \[ y = 2y - 10 \implies y = 10 \ \text{cm} \]Find the smaller side:
\[ x = 10 - 5 = 5 \ \text{cm} \]Problem 6: Two numbers with given sum and product
Question: Find two numbers whose sum is 26 and product is 165.
Solution:
Let the numbers be \(a\) and \(b\):
\[ a + b = 26 \implies b = 26 - a \] \[ ab = 165 \]Substitute \(b\):
\[ a(26 - a) = 165 \] \[ a^2 - 26a + 165 = 0 \]Factor:
\[ (a - 11)(a - 15) = 0 \]The two numbers are 11 and 15.
Problem 7: Rectangle with given area and perimeter
Question: The area of a rectangle is 15 and the perimeter is 16. Find its dimensions.
Solution:
Let \(x\) and \(y\) be the sides:
\[ xy = 15 \] \[ 2x + 2y = 16 \implies x + y = 8 \]From the perimeter: \(x = 8 - y\). Substitute into the area equation:
\[ (8 - y)y = 15 \] \[ y^2 - 8y + 15 = 0 \] \[ (y - 3)(y - 5) = 0 \]The rectangle is 3 by 5.
Problem 8: Two numbers with a linear relation
Question: The sum of two numbers is 20. The larger number is four less than twice the smaller. Find the numbers.
Solution:
Let \(x\) be the larger number and \(y\) the smaller:
\[ x + y = 20 \] \[ x = 2y - 4 \]Substitute \(x\):
\[ (2y - 4) + y = 20 \] \[ 3y - 4 = 20 \implies 3y = 24 \implies y = 8 \]Find \(x\):
\[ x = 2(8) - 4 = 12 \]The numbers are 8 and 12.
Problem 9: Right triangle hypotenuse
Question: The hypotenuse of a right triangle is 2 cm more than the longer side. The shorter side is 7 cm less than the longer side. Find the hypotenuse.
Solution:
Let \(h\) be the hypotenuse, \(a\) the longer leg, and \(b\) the shorter leg:
\[ h = a + 2 \] \[ b = a - 7 \]Apply the Pythagorean theorem:
\[ (a + 2)^2 = a^2 + (a - 7)^2 \] \[ a^2 + 4a + 4 = a^2 + a^2 - 14a + 49 \] \[ 0 = a^2 - 18a + 45 \] \[ (a - 15)(a - 3) = 0 \]If \(a = 3\), then \(b = 3 - 7 = -4\), which is impossible. So \(a = 15\) and:
\[ h = 15 + 2 = 17 \ \text{cm} \]Check: \(15^2 + 8^2 = 225 + 64 = 289 = 17^2\). ✔
Key Takeaways
- Translate the verbal description into algebraic equations before solving.
- Many word problems reduce to a quadratic equation — factor when possible, otherwise use the quadratic formula.
- Reject any solution that makes a physical quantity (length, area, units) negative.
- Always check your answer against the original conditions.