Formula Rearrangement

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Answer Key & Explanations

Formula Rearrangement · Grade 9 · Worksheet 2

1. Solve for r: V = (4/3)πr³ Answer: r = ∛(3V/(4π)) Solution: Start with V = (4/3)πr³ Multiply both sides by 3 to eliminate the fraction: 3V = 4πr³ Divide both sides by 4π to isolate r³: 3V/(4π) = r³ Take the cube root of both sides to solve for r: r = ∛(3V/(4π)) The answer is r = ∛(3V/(4π)).
   Full step-by-step solution

   Step 1: Start with V = (4/3)πr³ Step 2: Multiply both sides by 3 to eliminate the fraction: 3V = 4πr³ Step 3: Divide both sides by 4π to isolate r³: 3V/(4π) = r³ Step 4: Take the cube root of both sides to solve for r: r = ∛(3V/(4π)) The answer is r = ∛(3V/(4π)).

2. Liam is designing a rectangular garden with an area of 120 square meters. The length of the garden is 4 meters more than its width. Write an equation in terms of the width w that represents this situation, then solve for w to find the garden's dimensions. Answer: w = 10 meters, length = 14 meters Solution: Let \( w \) = width of the garden in meters. The length is 4 meters more than the width, so: length \( l = w + 4 \). Area of a rectangle = length × width.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Define variables** Let \( w \) = width of the garden in meters. The length is 4 meters more than the width, so: length \( l = w + 4 \). --- **Step 2: Write the area equation** Area of a rectangle = length × width. Given area = 120 m², so: \[ w \times (w + 4) = 120 \] --- **Step 3: Expand and rearrange** \[ w^2 + 4w = 120 \] Subtract 120 from both sides: \[ w^2 + 4w - 120 = 0 \] --- **Step 4: Solve the quadratic equation** We can factor: Look for two numbers whose product is -120 and whose sum is 4. Those numbers are 10 and -12. So: \[ (w + 12)(w - 10) = 0 \] --- **Step 5: Find possible values of w** \[ w + 12 = 0 \quad \text{or} \quad w - 10 = 0 \] \[ w = -12 \quad \text{or} \quad w = 10 \] Since width can't be negative, \( w = 10 \). --- **Step 6: Find length** \[ l = w + 4 = 10 + 4 = 14 \] --- **Final Answer:** Width \( w = 10 \) meters, length \( = 14 \) meters.

3. Liam is designing a rectangular garden with a perimeter of 40 meters. He wants the length to be 4 meters more than twice the width. Write an equation for the perimeter in terms of the width, then solve to find the dimensions of Liam's garden. Answer: width = 6 meters, length = 16 meters Solution: In geometry problems involving rectangles, the perimeter formula P = 2(l + w) is fundamental.
   Full step-by-step solution

   In geometry problems involving rectangles, the perimeter formula P = 2(l + w) is fundamental. When one dimension is defined relative to the other (like 'length is 3 more than twice the width'), you can express both dimensions in terms of a single variable. This creates a solvable equation that reveals both measurements. This approach of substituting relationships into standard formulas is useful in many real-world design and construction scenarios.

4. Emma is looking at a rectangular prism that has a volume of 315 cubic centimeters. The length of the prism is 9 cm, and the width is 5 cm. If she rearranges the formula for the volume of a rectangular prism to solve for the height, what is the height of the prism? Answer: 7 Solution: Write the formula for the volume of a rectangular prism: V = l * w * h, where V is volume, l is length, w is width, and h is height. Substitute the given values: 315 = 9 * 5 * h.
   Full step-by-step solution

   Step 1: Write the formula for the volume of a rectangular prism: V = l * w * h, where V is volume, l is length, w is width, and h is height. Step 2: Substitute the given values: 315 = 9 * 5 * h. Step 3: Multiply the length and width: 9 * 5 = 45. So the equation becomes 315 = 45 * h. Step 4: Solve for h by dividing both sides of the equation by 45: h = 315 / 45. Step 5: Perform the division: 315 / 45 = 7. The answer is 7.

5. Solve for r: A = πr² + 2πrh Answer: r = (-2πh ± √(4π²h² + 4πA)) / (2π) Solution: Start with A = πr² + 2πrh. Rearrange to standard quadratic form: πr² + 2πrh - A = 0. Identify coefficients: a = π, b = 2πh, c = -A.
   Full step-by-step solution

   Step 1: Start with A = πr² + 2πrh. Step 2: Rearrange to standard quadratic form: πr² + 2πrh - A = 0. Step 3: Identify coefficients: a = π, b = 2πh, c = -A. Step 4: Apply the quadratic formula: r = (-b ± √(b² - 4ac)) / (2a). Step 5: Substitute: r = (-2πh ± √((2πh)² - 4π(-A))) / (2π). Step 6: Simplify inside the square root: (2πh)² = 4π²h², and -4π(-A) = +4πA. Step 7: So r = (-2πh ± √(4π²h² + 4πA)) / (2π). Step 8: This is the final expression for r.

6. Aisha is designing a triangular garden with sides of length 8 meters, 15 meters, and 17 meters. She needs to calculate the area to determine how much soil to buy. Using Heron's formula, what is the area of Aisha's garden in square meters? Answer: 60 Solution: Calculate the semi-perimeter (s). s = (a + b + c) / 2 = (8 + 15 + 17) / 2 = 40 / 2 = 20 meters. Apply Heron's formula: Area = sqrt(s(s-a)(s-b)(s-c)).
   Full step-by-step solution

   Step 1: Calculate the semi-perimeter (s). s = (a + b + c) / 2 = (8 + 15 + 17) / 2 = 40 / 2 = 20 meters. Step 2: Apply Heron's formula: Area = sqrt(s(s-a)(s-b)(s-c)). Step 3: Substitute the values: Area = sqrt(20 * (20-8) * (20-15) * (20-17)) = sqrt(20 * 12 * 5 * 3). Step 4: Calculate inside the square root: 20 * 12 = 240; 240 * 5 = 1200; 1200 * 3 = 3600. Step 5: Take the square root: sqrt(3600) = 60. The area of the garden is 60 square meters.

7. Liam is designing a rectangular garden with an area of 120 square meters. The length of the garden is 4 meters more than its width. Write an equation in terms of width w that represents this situation, then solve for the dimensions of Liam's garden. Answer: width = 10 meters, length = 14 meters Solution: This type of problem involves setting up and solving quadratic equations from geometric contexts.
   Full step-by-step solution

   This type of problem involves setting up and solving quadratic equations from geometric contexts. The key is to translate the word problem into mathematical expressions - defining variables for unknown quantities, writing relationships between them, and substituting into known formulas like area. Solving the resulting quadratic equation gives you the dimensions that satisfy all the given conditions.

8. Emma is designing a rectangular garden. The length of the garden is 5 meters more than twice its width. If the area of the garden is 75 square meters, write an equation in terms of the width w that represents this situation, then solve for the width of Emma's garden. Answer: 5 Solution: Let w = width of the garden in meters. The length is 5 meters more than twice the width, so length = 2w + 5. The area of a rectangle is A = length × width.
   Full step-by-step solution

   Step 1: Let w = width of the garden in meters. Step 2: The length is 5 meters more than twice the width, so length = 2w + 5. Step 3: The area of a rectangle is A = length × width. Substitute the given area and the expression for length: 75 = (2w + 5) × w. Step 4: Expand the right side: 75 = 2w² + 5w. Step 5: Rearrange into standard quadratic form: 2w² + 5w - 75 = 0. Step 6: Factor the quadratic. Find two numbers that multiply to (2 × -75 = -150) and add to 5. These numbers are 15 and -10. Step 7: Rewrite the middle term: 2w² + 15w - 10w - 75 = 0. Step 8: Factor by grouping: w(2w + 15) - 5(2w + 15) = 0, so (2w + 15)(w - 5) = 0. Step 9: Set each factor to zero: 2w + 15 = 0 gives w = -7.5, and w - 5 = 0 gives w = 5. Step 10: Since width cannot be negative, the width is 5 meters. The width of Emma's garden is 5 meters.