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Section A, Problem 1: The area of the rectangle is given by length × width. So, x(x + 5) = 300. This simplifies to x² + 5x - 300 = 0. Factoring gives (x + 20)(x - 15) = 0, so x = 15 or x = -20. Since a length cannot be negative, x = 15.
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Section A, Problem 2: The perimeter of a rectangle is 2(length + width). With x = 15, the sides are 15 cm and 20 cm. Perimeter = 2(15 + 20) = 70 cm.
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Section A, Problem 3: The area of the triangle is (1/2) × base × height. So, (1/2) × w × (w + 1) = 6. This simplifies to w(w + 1) = 12, or w² + w - 12 = 0. Factoring gives (w + 4)(w - 3) = 0, so w = 3 or w = -4. Since a length cannot be negative, w = 3.
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Section A, Problem 4: Using the Pythagorean theorem, the hypotenuse h is √(w² + (w+1)²). With w = 3, this is √(3² + 4²) = √(9 + 16) = √25 = 5.00 cm.
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Section A, Problem 5: The surface area of a cylinder is 2πr² + 2πrh. Given diameter d, radius r = d/2 and height h = d. So, SA = 2π(d/2)² + 2π(d/2)d = 2π(d²/4) + 2π(d²/2) = (πd²/2) + πd² = (3πd²)/2. Setting this equal to 150π gives (3πd²)/2 = 150π. Dividing both sides by π: (3d²)/2 = 150. Multiplying by 2: 3d² = 300. Dividing by 3: d² = 100. Taking the square root: d = 10 (since diameter must be positive).
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Section A, Problem 6: The volume of a cylinder is V = πr²h. With d = 10, r = 5 and h = 10. So, V = π(5)²(10) = π(25)(10) = 250π cm³.
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Section B, Problem 1: By the Pythagorean theorem: x² + (x + 2)² = (2x - 2)². Expanding: x² + x² + 4x + 4 = 4x² - 8x + 4. Simplifying: 2x² + 4x + 4 = 4x² - 8x + 4. Bringing all terms to one side: 0 = 2x² - 12x. Factoring: 0 = 2x(x - 6). So, x = 0 or x = 6. Since side lengths must be positive, x = 6. The sides are 6, 8, and 10.
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Section B, Problem 2: Let the number be x. The equation is 14/x - 5 = x. Multiplying by x: 14 - 5x = x². Rearranging: x² + 5x - 14 = 0. Factoring: (x + 7)(x - 2) = 0. So, x = -7 or x = 2. Since the number is positive, x = 2.
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Section B, Problem 3: Let the number be x. The equation is x + 1/x = 2.9. Multiplying by x: x² + 1 = 2.9x. Rearranging: x² - 2.9x + 1 = 0. Multiplying by 10 to eliminate decimal: 10x² - 29x + 10 = 0. Using quadratic formula: x = [29 ± √(841 - 400)] / 20 = [29 ± √441] / 20 = [29 ± 21] / 20. So, x = 50/20 = 2.5 or x = 8/20 = 0.4.
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Section B, Problem 4: Lucy's number is x, David's is x + 4. The difference of their squares is (x + 4)² - x² = 88. Expanding: x² + 8x + 16 - x² = 88. Simplifying: 8x + 16 = 88. Solving: 8x = 72, so x = 9. Lucy thought of 9, David thought of 13.
Parent Tip: Review the logic above to help your child master the concept of quadratic equations worksheet with answers.
