0580/41 Oct/Nov 2024, Question 10: Equations That Become Quadratics
Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 14 marks
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The question, in short
Rectangle with sides 2x+3 and x+1 has area 190 cm²; sector OAB radius r, angle 50°, shaded segment area 30 cm². Form equations and find x, perimeter, r and segment perimeter.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (ai) · 4 marks
Form an equation from the rectangle area 190 cm² and solve it to show x = 8.5.
- 1(2x + 3)(x + 1) = 190
- 22x² + 5x + 3 = 190
- 32x² + 5x − 187 = 0
- 4(2x − 17)(x + 11) = 0
- 5x = 8.5 or x = −11 (length cannot be negative)
- 6x = 8.5
AnswerLeading to x = 8.5 with no errors
Part (aii) · 2 marks
Using x = 8.5, calculate the perimeter of the rectangle.
- 1Area: (2x + 3)(x + 1) = 190, so 2x² + 5x − 187 = 0
- 2Solve: x = (−5 + √1521)/4 = (−5 + 39)/4 = 8.5
- 3Perimeter = 2(2x + 3) + 2(x + 1) = 6x + 8
- 46 × 8.5 + 8 = 59 cm
Answer59
Part (bi) · 4 marks
Use the shaded segment area 30 cm² and angle 50° to show r = 23.7 cm to 1 decimal place.
- 1Sector area = 50/360 × π × r²
- 2Triangle area = 1/2 × r² × sin 50°
- 350/360 × π × r² − 1/2 × r² × sin 50° = 30
- 4r² × 0.05331 = 30, so r² = 562.7…
- 5r = 23.72… = 23.7 cm (1 d.p.)
Answer23.709 to 23.72…
Part (bii) · 4 marks
Calculate the perimeter of the shaded segment, including chord AB and arc AB.
- 1Chord AB = 2 × 23.7 × sin 25° = 20.03… cm
- 2Arc AB = 50/360 × 2 × π × 23.7 = 20.68… cm
- 3Perimeter = 20.03… + 20.68… = 40.71… ≈ 40.7 cm
Answer40.7 or 40.8 or 40.71 to 40.75…
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Also searched as: 0580/41/O/N/24 · 0580_w24_qp_41 · 0580 w24 41 · question 10.
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