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1 of 8
18
8
?
𝑎𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝐶𝐷𝐸𝐹
1
2
× 𝐵𝐶 × ℎ =
1
2
× 𝐸𝐹 + 𝐶𝐷 × ℎ
𝐵𝐶 = 𝐸𝐹 + 𝐶𝐷
𝐶𝐷 = 𝐵𝐶 − 𝐸𝐹
𝐶𝐷 = 18 − 8
= 10 𝑐𝑚
?
𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝐴𝑂𝐵 =
360°
𝑛
=
360°
8
= 45°
𝑥 = 3(45°)
= 135°
1
2
3
𝑛 = 8
𝑥 = 𝑂𝐵 → 𝑂𝐸
∆𝐷𝐴𝐵 ≡ ∆𝐸𝐵𝐶
120°
120°
120°
60°
120°
60°
120°
2𝑎
3𝑎 = 60°
𝑎 = 20°
r ↓ 20%
new radius, 𝑟1 = 0.8r
original radius = r
original area, A = π𝑟2
𝑛𝑒𝑤 𝑎𝑟𝑒𝑎, 𝐴1 = π 0.8𝑟 2
↓ ∆A% =
𝐴 − 𝐴1
𝐴
× 100%
= 1 −
𝐴1
𝐴
× 100%
= 1 −
𝜋 0.8𝑟 2
𝜋𝑟2
× 100%
= 36%
𝑥
𝑥 = 𝑏
34° 180° − 134°
2𝑥 = 34 + 180 − 134 − 15
2𝑥 + 15 = 34 + (180 − 134)
𝑥 =
65
2
= 32.5 °
𝑂 = (0, 0)
𝐴 = (2𝑝, 𝑝)
𝐵 = (2𝑝 + 1, 𝑝 − 3)
𝑂𝐴 = 𝑂𝐵
2𝑝 2 + 𝑝 2 = 2𝑝 + 1 2 + 𝑝 − 3 2
= 1 + 4𝑝 + 9 − 6𝑝
4𝑝2 + 𝑝2 = 4𝑝2 + 1 + 4𝑝 + 𝑝2 + 9 − 6𝑝
2𝑝 = 10
𝑝 = 5
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑄𝑅𝑆𝑇 = 30
∆𝑃𝑄𝑇~∆𝑃𝑅𝑆
1
2
𝑈𝑋 𝑄𝑇 + 𝑅𝑆 = 30
𝑈
𝑈𝑋 = 4 𝑐𝑚
1
2
𝑈𝑋 6 + 9 = 30
P
T
Q
P
S
R
6
U
9
X
𝑄𝑇
𝑅𝑆
=
𝑃𝑈
𝑃𝑋
6
9
=
𝑃𝑈
𝑃𝑈 + 𝑈𝑋
𝑃𝑈 = 8 𝑐𝑚
𝑃𝑋 = 𝑃𝑈 + 𝑈𝑋
= 8 + 4
= 12 𝑐𝑚

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JUEC 初中 (2016 Past Year)

  • 1. 18 8 ? 𝑎𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝐶𝐷𝐸𝐹 1 2 × 𝐵𝐶 × ℎ = 1 2 × 𝐸𝐹 + 𝐶𝐷 × ℎ 𝐵𝐶 = 𝐸𝐹 + 𝐶𝐷 𝐶𝐷 = 𝐵𝐶 − 𝐸𝐹 𝐶𝐷 = 18 − 8 = 10 𝑐𝑚
  • 2. ? 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝐴𝑂𝐵 = 360° 𝑛 = 360° 8 = 45° 𝑥 = 3(45°) = 135° 1 2 3 𝑛 = 8 𝑥 = 𝑂𝐵 → 𝑂𝐸
  • 4. r ↓ 20% new radius, 𝑟1 = 0.8r original radius = r original area, A = π𝑟2 𝑛𝑒𝑤 𝑎𝑟𝑒𝑎, 𝐴1 = π 0.8𝑟 2 ↓ ∆A% = 𝐴 − 𝐴1 𝐴 × 100% = 1 − 𝐴1 𝐴 × 100% = 1 − 𝜋 0.8𝑟 2 𝜋𝑟2 × 100% = 36%
  • 6. 34° 180° − 134° 2𝑥 = 34 + 180 − 134 − 15 2𝑥 + 15 = 34 + (180 − 134) 𝑥 = 65 2 = 32.5 °
  • 7. 𝑂 = (0, 0) 𝐴 = (2𝑝, 𝑝) 𝐵 = (2𝑝 + 1, 𝑝 − 3) 𝑂𝐴 = 𝑂𝐵 2𝑝 2 + 𝑝 2 = 2𝑝 + 1 2 + 𝑝 − 3 2 = 1 + 4𝑝 + 9 − 6𝑝 4𝑝2 + 𝑝2 = 4𝑝2 + 1 + 4𝑝 + 𝑝2 + 9 − 6𝑝 2𝑝 = 10 𝑝 = 5
  • 8. 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑄𝑅𝑆𝑇 = 30 ∆𝑃𝑄𝑇~∆𝑃𝑅𝑆 1 2 𝑈𝑋 𝑄𝑇 + 𝑅𝑆 = 30 𝑈 𝑈𝑋 = 4 𝑐𝑚 1 2 𝑈𝑋 6 + 9 = 30 P T Q P S R 6 U 9 X 𝑄𝑇 𝑅𝑆 = 𝑃𝑈 𝑃𝑋 6 9 = 𝑃𝑈 𝑃𝑈 + 𝑈𝑋 𝑃𝑈 = 8 𝑐𝑚 𝑃𝑋 = 𝑃𝑈 + 𝑈𝑋 = 8 + 4 = 12 𝑐𝑚