中三數學三題【求完整解答及說明】

5.
VOB 為一等腰三角形，VO = OB = 11 cm

圓錐的體積
= (1/3) × π × (11)² × 11 cm³ (畢氏定理)
= 1331π/3 cm³
≈ 1394cm³ (四位有效數字)

VB² = OB² +VO²
VB² = 11² + 11² cm²
VB = 11√2 cm

曲面面積
= (1/2) × (11√2) × (2 × π × 11) cm²
= 121(√2)π cm²
≈ 537.6cm² (四位有效數字)


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6.
設底半徑為 r cm。

π × r² = 4π
r = 2

斜高² = 2² + 2.5² m² (畢氏定理)
斜高² = 2² + (5/2)² m²
斜高 = (√41)/2 m

總面積
= (1/2) × [(√41)/2] × (2 × π × 2) + 4π cm²
= (4 + √41)π cm²
= 32.68 cm² (四位有效數字)


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7.
設底半徑為 r m，並設高為 h m。

2 × π × r = 2.8π
r = 1.4

h² + (1.4 m)² = (1.8m)² (畢氏定理)
h = (0.8√2) m

這堆米的重量
= (1/3) × 2.8π × (0.8√2) × 125 kg
= 280(√2)π/3 kg
= 414.7 kg (至四位有效數字)


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8.
設圓柱形杯內酒的高度為 h cm。

π × 3² × h = (1/3) × π × 4² × 7
h = 112/27

圓柱形杯內酒的高度 = 112/27cm

2014-11-20 22:40:46 補充：
修正：
7.
設底半徑為 r m，並設高為 h m。

2 × π × r = 2.8π
r = 1.4

h² + (1.4 m)² = (1.8m)² (畢氏定理)
h = (0.8√2) m

這堆米的重量
= (1/3) × π × 1.4² × (0.8√2) × 250 kg
= 392(√2)π/3 kg
= 580.5 kg (至四位有效數字)

2014-11-20 22:42:55 補充：
8.
答案 112/27 cm = 4.148 cm (四位有效數字)

5. In triangle VOB, ∠VBO=∠BVO=45°, so h=VO=OB=11
By the Pythagoras Theorem, VB^2=VO^2+OB^2
s=VB= √(11^2+11^2)=11√2
(a) Volume of cone=π r^2 h/3=π*11^2*11/3=1331π/3=1393.8199 (cm^3)
(b) Curved surface area of cone=π r s=π*11*11√2=(121√2)π=537.5888 (cm^2)

6. Given h=2.5, base radius of tent r= √(4π/π)=√4=2, slant of tent s= √(2.5^2+2^2)=(√41)/2
Curved surface area of tent=π r s=π*2*(√41)/2=(√41)π
Total surface area of tent=(√41)π + 4π =(4+√41)π =32.6824 (m^2)

7. Given s=1.8, base radius of tent r= 2.8π/(2π)=1.4 height of tent h= √(1.8^2-1.4^2)=√1.28,
Volume of rice=π r^2 h/3=π*1.4^2*(√1.28)/3=[1.96(√1.28)/3]π=2.3221
Mass of rice=volume*250=2.3221*250=580.5367 (kg)

8. For wine in cone-shape cup: r=4, h=7; for wine in cylinder-shape cup: radius R=3, let height=H
Volume of wine in cone-shape cup=π r^2 h/3=π*4^2*7/3=112π/3
Volume of wine in cylinder-shape cup=π R^2 H=π*3^2*H=9Hπ
Both volume of wine are equal, so
9Hπ=112π/3
H=112/27=4.1481 (cm)