问题1.高度为14 cm的右圆柱的曲面面积为88 cm 2 。找到圆柱体底部的直径。
解决方案:
Given:
i) Height of cylinder(h)=14 cm
ii)Curved Surface Area of cylinder=88 cm2
Curved Surface Area = 2πrh
88 = 2*(22/7)*r*14
Hence, r=
= 1 cm
As diameter=2*radius,
The diameter of the base of cylinder =2*1=2 cm
问题2。要求制造一个距离金属板高1 m,底直径140 cm的密闭圆柱罐。同一张纸需要多少平方米?
解决方案:
Given: i) Height of cylindrical tank (h)=1 m
ii) Base diameter of cylinder (d)=140 cm
iii) Radius of cylinder (r)=d/2=140/2=70cm=0.7 m
As metal sheet requirement for closed cylindrical tank is asked,
Area of metal sheet required =Total surface area of the cylindrical tank
= 2πr(r+h)
= 2*(22/7)*0.7*(0.7+1)
= 7.48 m2
问题3.金属管长77厘米。横截面的内径为4厘米,外径为4.4厘米。求出其(i)内部曲面面积,(ii)外曲面面积,(iii)总表面积。
解决方案:
We know that pipe has a hollow cylindrical structure
1) Height of Cylindrical metal pipe (h) =77 cm
2) The inner diameter of metal pipe (d)=4cm,
Hence, inner radius(r)=d/2=2 cm
3)The outer diameter of metal pipe (D)=4.4cm,
Hence, outer radius(R)=D/2=2.2 cm
i) Inner curved surface area=2πrh =2*(22/7)*2*77
=968 cm2
ii) Outer curved surface area=2πrh =2*(22/7)*2.2*77
=1064.8 cm2
iii) Total Surface Area=Inner curved surface area+Outer curved surface area+area of two bases
As pipe is hollow, area of two bases is a ring area given by=2(πR2-πr2) = 2π(R2-r2)
= 2* (22/7)*((2.2)2-(2)2)
= 5.28 cm2
Hence, Total surface area=968+1064.8+5.28=2038.08 cm2
问题4.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到一个操场上。在m 2中找到操场的面积。
解决方案:
Roller is cylindrical in shape.
Diameter of a roller(d)=84 cm,
Hence, base radius of roller (r)=d/2=42cm =0.42 m
Height of cylinder(h)=Length of roller=120cm=1.2 m
Curved surface area of cylinder=2πrh =2*(22/7)*0.42*1.2 =3.168 m2
Revolutions done by roller=500
Hence, Area of playground= Curved surface area of roller*revolutions done by roller
= 3.168*500
=1584 m2
问题5.圆柱形的圆柱直径为50厘米,高度为3.5 m。找出以每m 2 12.50₹的速率喷涂支柱弯曲表面的成本。
解决方案:
Diameter of cylindrical pillar=50 cm
Hence radius (r)=25cm=0.25 m
Height of cylindrical pillar (h)=3.5 m
Curved surface area of cylindrical pillar=2πrh=2*(22/7)*0.25*3.5
=5.50 m2
Rate of painting=₹12.50 per m2
Hence, cost of painting the curved surface of the pillar=5.50*12.50=₹68.75
问题6右圆柱的弯曲表面积为4.4平方米。如果圆柱体底部的半径为0.7 m,请找到其高度。
解决方案:
Given: i) Radius of the base of cylinder(h)=0.7 m
ii)Curved Surface Area of cylinder=4.4 m2
Curved Surface Area =2πrh
4.4 =2*(22/7)*0.7*h
Hence, h=
= 1 m
问题7.圆形井的内径为3.5 m。它是10 m深。找
(i)其内曲面面积,
(ii)将该曲面抹灰的成本为每m 2 ₹40。
解决方案:
Well is cylindrical in nature.
Its inner diameter=3.5m, hence inner radius (r)=3.5/2=1.75 m
Its height (h)=10 m
i) Its inner curved surface area (A)=2πrh = 2*(22/7)*1.75*10
= 110 m2
ii) The cost of Plastering this inner curved surface = rate of plastering*A
= 40*110
= ₹4400
问题8.在热水加热系统中,有一个长28 m,直径5 cm的圆柱管。找到系统中的总辐射面。
解决方案:
Length of cylindrical pipe (h)= 28m = 2800 cm
Diameter of cylindrical pipe=5 cm
Hence, radius of cylindrical pipe=5/2=2.5 cm
Cylindrical pipe radiates from curved surface only. Hence, we have to calculating radiating surface we have to measure curved surface area of that pipe.
Total radiating surface in the system = Curved surface area of pipe
= 2πrh
= 2*(22/7)*2.5*2800
= 44000 cm2
= 4.4 m2
问题9.查找
(i)密闭的圆柱形汽油储罐的侧向或弯曲表面积,该储罐的直径为4.2 m,高度为4.5 m。
(ii)如果在制造储罐中浪费了实际使用的钢的1/12,则实际使用了多少钢。
解决方案:
Diameter of cylindrical tank=4.2 m
Hence, radius of cylindrical tank (r)=4.2/2=2.1 m
Height of cylindrical tank (h)=4.5 m
i) Curved surface area of tank=2πrh = 2*(22/7)*2.1*4.5
= 59.4 m2
ii) As it is closed cylindrical tank,
Area of steel required to make = Total surface area of cylindrical tank
= 2πr(r+h)
= 2*(22/7)*2.1*(2.1+4.5)
= 87.12 m2 (1)
Let actually used steel =x m2. (1/2) of actually used steel was wasted.
Hence, tank is made up of 1-(1/12) = 11/12 part of steel. (2)
Hence, from (1) and (2), we get,
(11/12)x=87.12
i.e. x== 95.04 m2
Hence, 95.04 m2 steel was actually used to make tank.
问题10.灯罩的圆柱形框架要用装饰布覆盖。框架的底直径为20厘米,高度为30厘米。需留出2.5厘米的边距,以便将其折叠在框架的顶部和底部。找出需要多少布来遮盖灯罩。
解决方案:
Base diameter of frame=20 cm
Hence, base radius of frame=20/2=10 cm
Height of frame (including margins provided for folding frame over top and bottom of frame) = 30+2.5+2.5=35 cm
Cloth required for covering the lampshade= Curved surface area of lampshade
= 2πrh
= 2*(22/7)*10*35
= 2200 cm2
问题11. Vidyalaya的学生被要求参加比赛,以纸板制作和装饰带有底座的圆柱状笔筒。每个笔杆的半径为3厘米,高度为10.5厘米。 Vidyalaya将为竞争对手提供纸板。如果有35名参赛者,那么比赛需要购买多少纸板?
解决方案:
Penholder is in shape of a cylinder with a base. That is with curved surface with one base is to be created using cardboard.
Radius of penholder=3 cm
Height of penholder=10.5 cm
Area of cardboard required to make one penholder (A)
= Curved surface area of penholder + area of one base
= 2πrh+πr2
= 2*(22/7)*3*10.5+(22/7)*(3)2
= 198+28.28
= 226.28 cm2
Total area of Cardboard required to be bought for the competition
= Number of competitors* cardboard required for one penholder
= 35*226.28
= 7919.99 ≈ 7920 cm2