第22章测定III(右圆柱的表面积和体积)–练习22.1 |套装1
问题11:找出在井的内表面抹灰的成本为Rs。 9.50 m 2 。如果深度为21 m,并且其顶部直径为6 m。
解决方案:
The details given about well are –
Height of well = 21 m
Diameter of well = 6 m
So radius of well = 6/2 = 3 m
Curved surface area of cylinder = 2 * (22 /7) * r * h
= 2 * (22/7) * 3 * 21
= 396 m2
Cost of plastering the inner surface of a well = 396 * 9.50
= Rs 3762
问题12:顶部开口的圆柱形容器的直径为20厘米,高度为14厘米。找到锡的成本-将其以每100平方厘米50帕的速度镀在里面。
解决方案:
The details given about cylinder are –
Diameter of cylinder = 20 cm
So, radius = 20/2 = 10 cm
Height of cylinder = 14 cm
Total surface area of cylinder = 2 * (22/7) * r * h + (22/7) * r2
= 2 * (22/7) * 10 * 14 + (22/7) * 14 * 14
= 880 + 2200/7
= ((880 * 7) + 2200)/7
= (6160 + 2200)/7
= 8360/7 cm2
It is given that code per 100 cm2 = 50 paise
So, cost per 1 cm2 = Rs 0.005
Cost of tin-plating the area inside the vessel = (8360/7) * 0.005
= Rs 5.97
问题13:圆形井的内径为3.5 m。它是10 m深。查找在其内部曲面上抹灰的成本为4平方米。
解决方案:
The details given about well are –
Inner diameter of circular well = 3.5 m
So radius = 3.5/2 m
Height of well = 10 m
Curved surface area of well = 2 * (22/7) * r * h
= 2 * (22/7) * (3.5/2) * 10
= 110 m2
Cost of painting 1 m2 = Rs. 4
Cost of painting 110 m2 = 4 * 110
= Rs 440
问题14:滚筒的直径为84厘米,长度为120厘米。需要500次完整的旋转才能移动到操场上。游乐场面积是多少?
解决方案:
The details given about roller are –
Diameter of roller = 84 cm
So radius = 84/2 = 42 cm
Length of roller = 120 cm
Curved surface area of roller = 2 * (22/7) * r * h
= 2 * (22/7) * 42 * 120
= 31680 cm2
Area of playground = Number of revolution * Curved surface area of roller
= 500 * 31680
= 15840000 cm2
= 1584 m2 (1 m = 100 cm)
问题15:要清洁国会大厦的二十一个圆柱支柱。如果每个支柱的直径为0.50m高,高度为4m,以每平方米2卢比的价格清洗它们的成本为50卢比?
解决方案:
The details given about pillar are –
Diameter of pillar = 0.50 m
So radius = 0.50/2 = 0.25 m
Height of roller = 4 m
Curved surface area of pillar = 2 * (22/7) * r * h
= 2 * (22/7) * 0.25 * 4
= 44/7 m2
Curved surface area of pillar = (21 * 44)/7
= 132 m2
Cost of cleaning the pillars = 2.50 * 132
= Rs 330
问题16:如果两侧为4620平方厘米,底环面积为115.5平方厘米,高度为7厘米,则从两侧开口的空心圆柱的总表面积。找到圆柱体的厚度。
解决方案:
The details given about cylinder are –
Total surface area of hollow cylinder = 4620 cm2
Area of base ring = 115.5 cm2
Height of cylinder = 7 cm
Let inner radius = r and outer radius = R
Area of hollow cylinder = 2 * (22/7) * (R2 – r2) + 2 * (22/7) * R * h + 2 * (22/7) * r * h
= 2 * (22/7) * (R + r) (R – r) + 2 * (22/7) * r * h (R + r)
= 2 * (22/7) * (R + r) * (h + R – r)
Area of base = (22/7) * R * R – (22/7) * r * r
= (22/7) (R + r) * (R – r)
Total Surface area/Area of base = 4620/115.5
(2 * (22/7) (R + r) * (h + R – r))/((22/7) (R + r) (R – r)) = 4620/115.5
2 * (h + R – r)/(R – r) = 4620/115.5
R – r = thickness of cylinder
Let thickness of cylinder = t
2 * (h + t)/t = 4620/115.5
2 * h + 2 * t = 40 * t
2h = 40t – 2t
2 * 7 = 38t
14/38 = t
t = 7/19 cm
问题17:底座的半径和实心圆柱体的高度之和为37厘米。如果实心圆柱体的总表面积为1628 cm 2 ,请找到其底部的圆周。
解决方案:
The details about the cylinder are –
Sum of radius of base and height of cylinder = 37 cm
Total surface area pf cylinder = 1628 cm2
Total surface area of cylinder = 2 * (22/7) * r * (h + r)
1628 = 2 * (22/7) * r * 37
(1628 * 7)/(2 * 22 * 37) = r
11396/1628 = r
r = 7 m
Circumference of base = 2 * (22/7) * r
= 44 m
问题18:求出圆柱体的总表面积与曲面表面积之间的比率,假定圆柱体的高度和半径分别为7.5 cm和3.5 cm。
解决方案:
The details given about the cylinder are –
Height of cylinder = 7.5 cm
Radius of cylinder = 3.5 cm
Total surface area of cylinder/Curved surface area of cylinder = (2 * (22/7) * r * (h + r))/(2 * (22/7) * r * h)
= (h + r)/h
= (7.5 + 3.5)/7.5
= 11/7.5
= 22/15
问题19:无盖的圆柱形容器必须在其两面都镀锡。如果底座的半径为70 cm,高度为1.4 m,则按每1000 cm 2 3.50卢比的比率计算镀锡的成本。
解决方案:
The details given about cylinder are –
Radius of base = 70 cm
Height of cylinder = 1.4 m = 140 cm (1 m = 100 cm)
Total surface area of vessel = Area of outer side of base + Area of inner and outer curved surface
= 2 ((22/7) * r * r + 2 * (22/7) * r * h)
= 2 * (22/7) * r * (r + 2 * h)
= 2 * (22/7) * 70 * (70 + 280)
= 2 * (22/7) * 70 * 350
= 154000 cm2
Cost of tin coating at the rate of Rs 3.50 per 1000 cm2 = (3.50 * 154000)/1000
= Rs 539