Formula Rearrangement

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Answer Key & Explanations

Formula Rearrangement · Grade 9 · Worksheet 3

1. Solve for r: S = 2πr(r + h) Answer: r = (-2πh ± √(4π²h² + 8πS)) / (4π) Solution: Expand the right side: S = 2πr² + 2πrh Bring all terms to one side: 2πr² + 2πrh - S = 0 This is a quadratic in r: (2π)r² + (2πh)r - S = 0 Use the quadratic formula r = [-b ± √(b² - 4ac)] / (2a) with a = 2π, b = 2πh, c = -S Discriminant: b² - 4ac = (2πh)² - 4(2π)(-S) = 4π²h² + 8πS Substitute into…
   Full step-by-step solution

   Step 1: Expand the right side: S = 2πr² + 2πrh Step 2: Bring all terms to one side: 2πr² + 2πrh - S = 0 Step 3: This is a quadratic in r: (2π)r² + (2πh)r - S = 0 Step 4: Use the quadratic formula r = [-b ± √(b² - 4ac)] / (2a) with a = 2π, b = 2πh, c = -S Step 5: Discriminant: b² - 4ac = (2πh)² - 4(2π)(-S) = 4π²h² + 8πS Step 6: Substitute into formula: r = [-2πh ± √(4π²h² + 8πS)] / (4π) The answer is r = (-2πh ± √(4π²h² + 8πS)) / (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. Aisha is designing a triangular support structure for a bridge. The area of the triangular section is 84 square meters, and the base is 7 meters longer than the height. Write an equation in terms of the height h that represents this situation, then solve for the height of the triangular section. Answer: h = 8 Solution: The area of a triangle is given by the formula A = (1/2) * base * height. When the base is expressed in terms of the height, this formula becomes a quadratic equation.
   Full step-by-step solution

   The area of a triangle is given by the formula A = (1/2) * base * height. When the base is expressed in terms of the height, this formula becomes a quadratic equation. Solving such equations often involves using the quadratic formula or factoring, which requires setting the equation to zero and finding the positive root since physical dimensions cannot be negative.

4. A drone is flying in a parabolic path to deliver a package. Its height above ground is modeled by the equation h(t) = -5t² + 40t + 20, where h is height in meters and t is time in seconds. The drone needs to maintain a minimum height of 50 meters to avoid obstacles. During what time interval will the drone be at or above this minimum height? Answer: 2 ≤ t ≤ 6 Solution: When working with quadratic inequalities, we find where the parabola intersects the horizontal line representing our target value.
   Full step-by-step solution

   When working with quadratic inequalities, we find where the parabola intersects the horizontal line representing our target value. The solution will be the interval between these intersection points for downward-opening parabolas. This concept applies to various real-world scenarios like projectile motion and profit optimization.

5. Mere is designing a cylindrical water tank. The formula for the volume of a cylinder is V = πr²h, where V is the volume, r is the radius, and h is the height. If the tank must have a volume of 1152π cubic meters and a radius of 12 meters, what height must the tank be? Answer: 8 Solution: Start with the formula for volume of a cylinder: V = πr²h Substitute the known values: V = 1152π and r = 12 1152π = π(12²)h Simplify 12² = 144, so 1152π = π(144)h Divide both sides by π to get: 1152 = 144h Divide both sides by 144 to solve for h: h = 1152/144 Simplify: h = 8 The answer is 8.
   Full step-by-step solution

   Step 1: Start with the formula for volume of a cylinder: V = πr²h Step 2: Substitute the known values: V = 1152π and r = 12 Step 3: 1152π = π(12²)h Step 4: Simplify 12² = 144, so 1152π = π(144)h Step 5: Divide both sides by π to get: 1152 = 144h Step 6: Divide both sides by 144 to solve for h: h = 1152/144 Step 7: Simplify: h = 8 The answer is 8.

6. Solve for h: S = 2πr² + 2πrh Answer: h = (S - 2πr²) / (2πr) Solution: Start with S = 2πr² + 2πrh Subtract 2πr² from both sides: S - 2πr² = 2πrh Divide both sides by 2πr: (S - 2πr²) / (2πr) = h Rewrite: h = (S - 2πr²) / (2πr) The answer is h = (S - 2πr²) / (2πr).
   Full step-by-step solution

   Step 1: Start with S = 2πr² + 2πrh Step 2: Subtract 2πr² from both sides: S - 2πr² = 2πrh Step 3: Divide both sides by 2πr: (S - 2πr²) / (2πr) = h Step 4: Rewrite: h = (S - 2πr²) / (2πr) The answer is h = (S - 2πr²) / (2πr).

7. Aisha is designing a triangular sail for her sailboat. The sail has a base that is 2 meters less than twice its height. If the area of the sail is 30 square meters, write an equation in terms of the height h that represents this situation, then solve for the height of Aisha's sail. Answer: 6 Solution: The area of a triangle is A = (1/2) × base × height Let h = height of the sail The base is 2 meters less than twice the height, so base = 2h - 2 Substitute into the area formula: 30 = (1/2) × (2h - 2) × h Multiply both sides by 2: 60 = (2h - 2) × h Expand the right side: 60 = 2h² - 2h Divide all…
   Full step-by-step solution

   Step 1: The area of a triangle is A = (1/2) × base × height Step 2: Let h = height of the sail Step 3: The base is 2 meters less than twice the height, so base = 2h - 2 Step 4: Substitute into the area formula: 30 = (1/2) × (2h - 2) × h Step 5: Multiply both sides by 2: 60 = (2h - 2) × h Step 6: Expand the right side: 60 = 2h² - 2h Step 7: Divide all terms by 2: 30 = h² - h Step 8: Rearrange to standard quadratic form: h² - h - 30 = 0 Step 9: Factor the quadratic: (h - 6)(h + 5) = 0 Step 10: Solve for h: h = 6 or h = -5 Step 11: Since height cannot be negative, h = 6 The height of Aisha's sail is 6 meters.

8. Solve for r: S = 3πr² + 5πrh Answer: r = (-5πh ± sqrt(25π²h² + 12πS)) / (6π) Solution: Start with S = 3πr² + 5πrh Rearrange to standard quadratic form: 3πr² + 5πrh - S = 0 Identify coefficients: a = 3π, b = 5πh, c = -S Apply the quadratic formula: r = (-b ± sqrt(b² - 4ac)) / (2a) Substitute: r = (-5πh ± sqrt((5πh)² - 4(3π)(-S))) / (2(3π)) Simplify inside sqrt: r = (-5πh ±…
   Full step-by-step solution

   Step 1: Start with S = 3πr² + 5πrh Step 2: Rearrange to standard quadratic form: 3πr² + 5πrh - S = 0 Step 3: Identify coefficients: a = 3π, b = 5πh, c = -S Step 4: Apply the quadratic formula: r = (-b ± sqrt(b² - 4ac)) / (2a) Step 5: Substitute: r = (-5πh ± sqrt((5πh)² - 4(3π)(-S))) / (2(3π)) Step 6: Simplify inside sqrt: r = (-5πh ± sqrt(25π²h² + 12πS)) / (6π) The answer is r = (-5πh ± sqrt(25π²h² + 12πS)) / (6π).