第 10 类 RD Sharma 解决方案 - 第 16 章表面积和体积 - 练习 16.1 |设置 2
问题 19. 必须熔化多少个直径为 1.75 厘米、厚度为 2 毫米的硬币才能形成一个 11 厘米 x 10 厘米 x 7 厘米的长方体?
解决方案:
Diameter of the coin = 1.75 cm
Radius = 1.75/2 = 0.875 cm
Thickness or height = 2 mm = 0.2 cm
Volume of the cylinder = πr2h
= π 0.8752 × 0.2
Volume of the cuboid = 11 × 10 × 7 cm3
Let the number of coins needed to be melted be n.
Volume of cuboid = n × Volume of cylinder
11 × 10 × 7 = π 0.8752 × 0.2 x n
11 × 10 × 7 = 22/7 x 0.8752 × 0.2 x n
n = 1600
Therefore, the number of coins required are 1600
问题 20. 固体金属球的表面积是 616 cm 2 。它被熔化并重铸成一个高 28 厘米的圆锥体。求这样形成的圆锥底的直径。
解决方案:
Radius of sphere = r cm
Surface area of the solid metallic sphere = 616 cm3
Surface area of the sphere = 4πr2
4πr2 = 616
r2 = 49
r = 7
Radius of the solid metallic sphere = 7 cm
Let radius of cone be x cm
Volume of the cone = 1/3 πx2h
= 1/3 πx2 (28) ….. (i)
Volume of the sphere = 4/3 πr3
= 4/3 π73 ………. (ii)
(i) and (ii) are equal
1/3 πx2 (28) = 4/3 π73
x2 (28) = 4 x 73
x2 = 49
r =7
Diameter of the cone = 7 x 2 = 14 cm
Therefore, the diameter of the base of the cone is 14 cm
问题 21.一个圆柱形桶,高 32 厘米,底部半径 18 厘米,装满沙子。这个桶倒在地上,就形成了一个圆锥形的沙堆。如果圆锥堆的高度是 24 厘米,求堆的半径和斜高。
解决方案:
Height of the cylindrical bucket = 32 cm
Radius of the cylindrical bucket = 18 cm
Volume of cylinder = π × r2 × h
Volume of cone = 1/3 π × r2 × h
Volume of the conical heap = Volume of the cylindrical bucket
1/3 π × r2 × 24 = π × 182 × 32
r2 = 182 x 4
r = 18 x 2 = 36 cm
Let slant height of conical heap be l cm
l = √(h2 + r2)
l = √(242 + 362) = √1872
l = 43.26 cm
Therefore, the radius = 36cm slant height of the conical heap = 43.26 cm
问题 22. 将一个半径为 5.6 厘米的实心金属球熔化,制成半径为 2.8 厘米、高 3.2 厘米的实心锥体。找出形成的这种圆锥的数量。
解决方案:
Let the number of cones be n
Radius of metallic sphere = 5.6 cm
Radius of the cone = 2.8 cm
Height of the cone = 3.2 cm
Volume of a sphere = 4/3 π × r3
= 4/3 π × 5.63
Volume of cone = 1/3 π × r2 × h
= 1/3 π × 2.82 × 3.2
Volume of sphere = n * Volume of each cone
Number of cones (n) = Volume of the sphere/ Volume of the cone
n = 4/3 π × 5.63/(1/3 π × 2.82 × 3.2)
n = (4 x 5.63)/(2.82 × 3.2)
n = 28
Therefore, 28 such cones can be formed.
问题 23. 将尺寸为 53 cm x 40 cm x 15 cm 的实心铁立方体熔化并重铸成圆柱形管。管道的外径和内径分别为 8 cm 和 7 cm。求管子的长度。
解决方案:
Volume of cuboid = (53 x 40 x 15) cm3
Internal radius of the pipe = 7/2 cm = r
External radius of the pipe = 8/2 = 4 cm = R
Let h be length of pipe
Volume of iron in the pipe = (External Volume) – (Internal Volume)
= πR2h – πr2h
= πh(R2– r2)
= πh(R – r) (R + r)
= π(4 – 7/2) (4 + 7/2) x h
= π(1/2) (15/2) x h
Volume of iron in the pipe = volume of iron in cuboid
π(1/2) (15/2) x h = 53 x 40 x 15
h = (53 x 40 x 15 x 7/22 x 2/15 x 2) cm
h = 2698.18 cm
Therefore, the length of the pipe is 2698.2 cm.
问题 24. 空心球壳的内表面和外表面的直径分别为 6 厘米和 10 厘米。如果将其熔化并重铸成直径为 14 厘米的实心圆柱体,求圆柱体的高度。
解决方案:
Internal diameter of hollow spherical shell = 6 cm
Internal radius of hollow spherical shell = 6/2 = 3 cm = r
External diameter of hollow spherical shell = 10 cm
External diameter of hollow spherical shell = 10/2 = 5 cm = R
Diameter of the cylinder = 14 cm
Radius of cylinder = 14/2 = 7 cm
Let the height of cylinder be taken as h cm
Volume of cylinder = Volume of spherical shell
π × r2 × h = 4/3 π × (R3 – r3)
π × 72 × h = 4/3 π × (53 – 33)
h = 4/3 x 2
h = 8/3 cm
Therefore, the height of the cylinder = 8/3 cm
问题 25. 一个内径和外径分别为 4 cm 和 8 cm 的空心球被熔化成一个底径为 8 cm 的圆锥体。计算圆锥的高度?
解决方案:
Internal diameter of hollow sphere = 4 cm
Internal radius of hollow sphere = 2 cm
External diameter of hollow sphere = 8 cm
External radius of hollow sphere = 4 cm
Volume of the hollow sphere = 4/3 π × (43 – 23) … (i)
Diameter of the cone = 8 cm
Radius of the cone = 4 cm
Let the height of the cone be x cm
Volume of the cone =1/3 π × 42 × (x) ….. (ii)
(i) and (ii) are equal
4/3 π × (43 – 23) = 1/3 π × 42 × h
4 x (64 – 8) = 16 x h
h = 14
Therefore, the height of the cone = 14 cm
问题 26 。一个内半径和外半径分别为 2 cm 和 4 cm 的空心球被熔化成一个底半径为 4 cm 的圆锥体。求圆锥的高度和斜高。
解决方案:
The internal radius of hollow sphere = 2 cm
The external radius of hollow sphere = 4 cm
Volume of the hollow sphere = 4/3 π × (43 – 23) … (i)
The base radius of the cone = 4 cm
Let the height of the cone be x cm
Volume of the cone = 1/3 π × 42 × h ….. (ii)
4/3 π × (43 – 23) = 1/3 π × 42 × h
4 x (64 – 8) = 16 x h
h = 14
Let Slant height of the cone be l
l = √(h2 + r2)
l = √(142 + 42) = √212
l = 14.56 cm
Therefore, the height = 14 cm and slant height = 14.56 cm
问题 27. 一个半径为 3 厘米的球被熔化并重铸成三个球。其中两个球的半径分别为 1.5 厘米和 2 厘米。求第三个球的直径。
解决方案:
Radius of the spherical ball = 3 cm
Volume of the sphere = 4/3 πr3
Volume = 4/3 π33
Volume of first ball = 4/3 π 1.53
Volume of second ball = 4/3 π23
Let the radius of the third ball = r cm
Volume of third ball = 4/3 πr3
Volume of the spherical ball is equal to sum of volumes of three balls
4/3 π33 = 4/3 π 1.53+ 4/3 π23 + 4/3 πr3
(3)3 = (2)3 + (1.5)3 + r3
r3 = 33– 23– 1.53 cm3
r3 = 15.6 cm3
r = (15.625)1/3 cm
r = 2.5 cm
Diameter = 2 x radius = 2 x 2.5 cm
= 5.0 cm.
Therefore, the diameter of the third ball = 5 cm
问题 28. 一条 2 m 宽的路径环绕着一个直径为 40 m 的圆形池塘。需要多少立方米的砾石才能将小路挖到 20 厘米的深度?
解决方案:
Diameter of the circular pond = 40 m
Radius of the pond = 40/2 = 20 m = r
Thickness (width of the path) = 2 m
Height = 20 cm = 0.2 m
Thickness (t) = R – r
2 = R – 20
R = 22 m
Volume of the hollow cylinder = π (R2– r2) × h
= π (222– 202) × 0.2
= 52.8 m3
Therefore, the volume of the hollow cylinder = 52.8 m3
问题 29. 挖出一口 16 m 深、直径 3.5 m 的井,将其中的泥土均匀地铺开,形成一个 27.5 m × 7 m 的平台。找到平台的高度?
解决方案:
Radius(r) of the cylinder = 3.5/2 m = 1.75 m
Depth of the well or height of the cylinder (h) = 16 m
Volume of the cylinder= πr2h
= π × 1.752 × 16
The length of the platform (l) = 27.5 m
Breadth of the platform (b) =7 m
Let the height of the platform be x m
Volume of the cuboid = l×b×h
= 27.5×7×(x)
Volume of cylinder = Volume of cuboid
π × 1.75 × 1.75 × 16 = 27.5 × 7 × x
x = 0.8 m = 80 cm
Therefore, the height of the platform = 80 cm.
问题 30. 一口直径为 2 m 的井被挖到 14 m 深。从中取出的土均匀地铺在四周,形成一个40厘米高的堤坝。求路堤的宽度?
解决方案:
Radius of the circular cylinder (r) = 2/2 m = 1 m
Height of the well (h) = 14 m
Volume of the solid circular cylinder = π r2h
= π × 12× 14 …. (i)
The height of the embankment = 40 cm = 0.4 m
Let the width of the embankment be (x) m.
External radius = (1 + x)m
Internal radius = 1 m
Volume of the embankment = π × r2 × h
= π × [(1 + x)2 – (1)2]× 0.4 ….. (ii)
(i) and (ii) are equal
π × 12 × 14 = π × [(1 + x)2 – (1)2] x 0.4
14/0.4 = 1 + x2 + 2x – 1
35 = x2 + 2x
x2 + 2x – 35 = 0
(x + 7) (x – 5) = 0
x = 5 or -7
Therefore, the width of the embankment = 5 m.
问题 31. 挖出一口内径 4 m 深 14 m 的井。从中取出的土均匀地散布在3m宽的周围,形成了一个路堤。找出路堤的高度?
解决方案:
Inner radius of the well = 4 m
Depth of the well = 14 m
Volume of the cylinder = π r2h
= π × 42 × 14 …. (i)
Width of the embankment = 3 m
Outer radius of the well = 3 + 4 m = 7 m
Volume of the hollow embankment = π (R2 – r2) × h
= π × (72 – 42) × h …… (ii)
(i) and (ii) are equal
π × 42 × 14 = π × (72 – 42) × h
h = 42 × 14 / (33)
h = 6.78 m
Therefore, the height of the embankment = 6.78 m.
问题 32. 一口直径为 3 m 的井被挖到 14 m 深。挖出的土在四周均匀铺开,宽度为4m,形成路堤。求路堤的高度。
解决方案:
Diameter of the well = 3 m
Radius of the well = 3/2 m = 1.5 m
Depth of the well (h) = 14 m
Width of the embankment (thickness) = 4 m
Radius of the outer surface of the embankment = (4 + 1.5) m = 5.5 m
Volume of the embankment = π(R2 – r2) × h
= π(5.52 – 1.52) × h ….. (i)
Volume of earth dug out = π × r2 × h
= π × (3/2)2 × 14 ….. (ii)
(i) and (ii) are equal
π(5.52 – 1.52) × h = π × (3/2)2 × 14
(5.5+1.5)(5.5-1.5) x h = 9 × 14/ 4
h = 9 × 14/ (4 × 28)
h = 9/8 m
= 1.125m
Therefore, the height of the embankment =1.125 m
问题 33. 求从边长为 9 厘米的立方体中可以切出的最大直圆锥的体积。
解决方案:
The side of the cube = 9 cm
Diameter of the cone = Side of cube
2r = 9
Radius of cone = 9/2 cm = 4.5 cm
Height of cone = side of cube
Height of cone (h) = 9 cm
Volume of the largest cone = 1/3 π × r2 × h
= 1/3 π × 4.52 × 9
= 190.93 cm3
Therefore, the volume of the largest cone to fit in the cube has a volume of 190.93 cm3
问题 34. 一个圆柱形桶,高 32 厘米,底部半径 18 厘米,里面装满沙子。这个桶倒在地上,就形成了一个圆锥形的沙堆。如果圆锥堆的高度是 24 厘米,求堆的半径和斜高。
解决方案:
Height of the cylindrical bucket = 32 cm
Radius of the cylindrical bucket = 18 cm
Height of conical heap = 24 cm
Volume of cylinder = π × r2 × h
Volume of cone = 1/3 π × r2 × h
Volume of the conical heap = Volume of the cylindrical bucket
1/3 π × r2 × 24 = π × 182 × 32
r2 = 182 X 4
r = 18 × 2 = 36 cm
Slant height (l) = √(h2 + r2)
l = √(242 + 362) = √576+1296
= √1872
= 43.26 cm
Therefore, the radius =36 cm and slant height = 43.26 cm
问题 35. 落在长 6 m、宽 4 m 的平坦矩形表面上的雨水被转移到内径为 20 cm 的圆柱形容器中。如果降雨量为 1 厘米,圆柱形容器中的水高度是多少?
解决方案:
Length of the rectangular surface = 6 m = 600 cm
Breadth of the rectangular surface = 4 m = 400 cm
Height of the rain = 1 cm
Volume of the rectangular surface = length * breadth * height
= 600×400×1 cm3
= 240000 cm3 …………….. (i)
Radius of the cylindrical vessel = 20 cm
Let the height of the cylindrical vessel be h cm
Volume of the cylindrical vessel = π × r2 × h
= π × 202 × h ……….. (ii)
(i) is equal to (ii)
240000 = π × 202 × h
h = 190.9 cm
Therefore, the height of the cylindrical vessel = 191 cm