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primary 6 | Maths
| Ratio
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Date Posted: 2 years ago
Views: 233
①Area of quarter circle BCX + area of triangle ACX
= shaded area BYC + unshaded area CXY + shaded area AYX + unshaded area CXY
= shaded area BYC + shaded area AYX + 2 unshaded area CXY
= 514cm² + 2 unshaded area CXY
② When calculated,
Area of quarter circle BCX + area of triangle ACX
= ¼ × πr² + ½ × base × height
= ¼ × 22/7 × 28cm × 28cm + ½ × 28cm × 28cm
= 616cm² + 392cm²
= 1008cm²
Comparing this to part ①,
1008cm² = 514cm² + 2 unshaded area CXY
2 unshaded area CXY = 1008cm² - 514cm²
2 unshaded area CXY = 494cm²
1 unshaded area CXY = 494cm² ÷ 2
= 247cm²
= shaded area BYC + unshaded area CXY + shaded area AYX + unshaded area CXY
= shaded area BYC + shaded area AYX + 2 unshaded area CXY
= 514cm² + 2 unshaded area CXY
② When calculated,
Area of quarter circle BCX + area of triangle ACX
= ¼ × πr² + ½ × base × height
= ¼ × 22/7 × 28cm × 28cm + ½ × 28cm × 28cm
= 616cm² + 392cm²
= 1008cm²
Comparing this to part ①,
1008cm² = 514cm² + 2 unshaded area CXY
2 unshaded area CXY = 1008cm² - 514cm²
2 unshaded area CXY = 494cm²
1 unshaded area CXY = 494cm² ÷ 2
= 247cm²
Ans: 247cm²
½ × 28² = 392
¼ × (22/7) × 28² = 616
shaded area
u = (392+616-514)÷2
= 247cm² (Ans)
½ × 28² = 392
¼ × (22/7) × 28² = 616
shaded area
u = (392+616-514)÷2
= 247cm² (Ans)
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