问题1.使用以下方法找到右圆锥的体积:
(i)半径6厘米,高度7厘米
(ii)半径3.5厘米,高度12厘米
(iii)高度为21厘米,倾斜高度为28厘米
解决方案:
(i) Radius of Cone(r)=6cm
Height of Cone(h)=7cm
As we know that the Volume of a Right Circular Cone = 1/3 πr2h
By putting the values in formula we get,
= 1/3 x 3.14 x 62 x 7 = 264
Hence, Volume of a Right Circular Cone is 264 cm3
(ii) Radius of Cone(r)=3.5 cm
Height of Cone(h)=12cm
Volume of a Right Circular Cone = 1/3 πr2h
By putting the values in formula we get,
= 1/3 x 3.14 x 3.52 x 12 =154
Hence, Volume of a Right Circular Cone is 154 cm3
(iii) Height of Cone(h)=21 cm
Slant height of Cone(l) = 28 cm
As we know that, l2 = r2 + h2
282 = r2 + 212
r = 7√7
As we know that Volume of a Right Circular Cone = 1/3 πr2h
By putting the values in formula we get,
= 1/3 x 3.14 x (7√7)2 x 21 = 7546
Hence, Volume of a Right Circular Cone is 7546 cm3
问题2。用以下公式找到圆锥形容器的容量(以升为单位):
(i)半径7厘米,倾斜高度25厘米
(ii)高度12厘米,倾斜高度13厘米。
解决方案:
(i)圆锥半径(r)= 7厘米
圆锥体的倾斜高度(l)= 25厘米
我们知道l 2 = r 2 + h 2
252 = 72 + h 2
h = 24
我们知道正圆锥= = 1/3πR2小时的该卷
通过将值放在公式中,我们得到,
= 1/3 x 3.14 x(7) 2 x 24 = 1232
因此,右圆锥的体积为1232 cm 3或1.232升
(ii)锥体高度(h)= 12厘米
圆锥体的倾斜高度(l)= 13厘米
我们知道l 2 = r 2 + h 2
132 = r 2 + 122
r = 5
我们知道正圆锥的那个体积= 1/3πR2小时
通过将值放在公式中,我们得到,
= 1/3 x 3.14 x(5) 2 x 12 = 314.28
因此,右圆锥的体积为314.28 cm 3或0.314升。
问题3。两个圆锥的高度比例为1:3,底面的半径为比例3:1。找到它们的体积比。
解决方案:
Let us assume that the heights of the cones is h and 3h and radii of their bases is 3r and r. Then, their volumes are
Volume of 1st Cone (V1) = 1/3 π(3r)2h
Volume of 2nd Cone (V2) = 1/3 πr2(3h)
V1/V2 = 3/1
Hence, Ratio of two volumes is 3:1.
问题4.直角圆锥的半径和高度之比为5:12。如果其体积为314立方米,请找到倾斜高度和半径。 (使用π = 3.14)。
解决方案:
Let us assume that the ratio of radius and the height of a Right Circular Cone to be x.
Then the radius be 5x and height be 12x
As we know that l2 = r2 + h2
= (5x) 2 + (12x)2
= 25 x2 + 144 x2
l = 13x
Hence, Slant Height is 13 m.
Now it is given that the Volume of Cone = 314 m3
= 1/3πr2h = 314
= 1/3 x 3.14 x (25x2 ) x (12x) = 314
=x3=1
x = 1
Hence, Radius = 5x 1 = 5 m
Therefore, Slant height = 13m
问题5.直圆锥的半径和高度之比为5:12,其体积为2512立方厘米。找到圆锥体的倾斜高度和半径。 (使用π = 3.14)。
解决方案:
Let us assume that the ratio of radius and height of a right circular cone be y.
Radius of Cone(r) = 5y
Height of Cone (h) =12y
As we know that, l2 = r2 + h2
= (5y) 2 + (12y)2
= 25 y2 + 144 y2
= l = 13y
Volume of Cone, given 2512cm3
= 1/3πr2h = 2512
= 1/3 x 3.14 x (5y)2 x 12y = 2512
= y3 = (2512 x 3)/(3.14 x 25 x 12) = 8
= y = 2
Therefore,
Radius of Cone = 5y = 5×2 = 10cm
Slant Height (l) =13y = 13×2 = 26cm
问题6.两个圆锥体的体积比为4:5,其底面半径的比例为2:3。求出它们的垂直高度的比例。
解决方案:
Let us assume that the ratio of the Radius is x and Ratio of the Volume is y.
Radius of 1st Cone (r1) =2x
Radius of 2nd Cone (r2) =3x
Volume of 1st Cone (V1)= 4y
Volume of 2nd Cone (V2)= 5y
As we know that the formula for Volume of a Cone = 1/3πr2h
Let’s assume that h1 and h2 be the heights of respective cones.
V1/V2 = 4/5 = (1/3 x π x (r1)2 x h1) / (1/3 x π x (r2)2 x h2)
4/5 = 4h1/9h2 = 9/5
Hence, Heights are in the ratio of 9 : 5.
问题7.圆柱和圆锥的底边半径和高度相等。证明它们的体积比例为3:1。
解决方案:
Given that,
Cylinder and Cone are having equal radii of their bases and heights.
Let’s assume that Radius of the Cone = Radius of the Cylinder = r &
Height of the Cone = Height of the Cylinder = h
Volume of Cylinder / Volume of Cone = (πr2h) / (1/3 x π x r2 x h) = 3:1
Hence, The Ratio of their Volumes is 3:1.
问题8.如果将圆锥形底部的半径减半,并保持高度不变,则缩小后的圆锥体与原始圆锥体的体积之比是多少?
解决方案:
Let us assume that r be the radius and h be the height of the cone,
As we know that Volume of Cone = 1/3 πr2h
Put the values of radius and same height we get,
Volume = 1/3 π(r/2)2h = 1/3 x π x r2/2 x h
= 1/4 x (1/3 x π x r2 x h)
Ratio of two cone’s = 1/3 πr2h : 1/4 x (1/3 πr2h) = 1 : 1/4 = 4:1
Hence, the ratio between Cones are 1:4
问题9.小麦堆呈圆锥形,直径为9 m,高度为3.5 m。找到它的体积。仅覆盖堆放,需要多少帆布布?
解决方案:
Given that,
Diameter of conical heap of wheat = 9 m
Radius = 9/2m and height = 3.5m
As we know that volume of cone = 1/3 πr2h = 1/3 x 22/7 x (4.5)2 x 3.5 = 74.18 m3
As we know that Curved Surface area = πrl
l = √r2 + h2 = √(4.5)2 + (3.5)2 = √130/2 m
Curved Surface area = π x 4.5 x √130/2 = 22/7 x 4.5 x √130/2 = 80.54 m2
问题10.假设底部的直径为14厘米,垂直高度为51厘米,实心圆锥的重量为10克/立方厘米。
解决方案:
Given that,
Diameter of the base of solid cone = 14 cm and vertical height (h) = 51 cm
Radius(r) = 7cm
As we know that volume of cone = 1/3 πr2h = 1/3 x 22/7 x 72 x 51 = 2618 cm3
Weight of 1 cm3 = 10 grams
Then total weight will = 2618 x 10gm = 26180 gm = 26.180 kg
问题11:使直角三角形的长边变成圆角,该直角三角形的边长为6.3 cm,边长为10 cm。查找由此产生的固体体积。另外,找到其弯曲的表面积。
解决方案:
Given that,
Length of sides of a right-angled triangle are 6.3 cm and 10 cm
By turning around the longer side a Cone is formed in which radius (r) is = 6.3 cm and height (h) = 10 cm
As we know that l = √r2 + h2 = √(6.3)2 + (10)2 = 11.82 cm
As we know that volume of cone = 1/3 πr2h = 1/3 x 22/7 x (6.3)2 x10 = 415.8 cm3
We know that Curved Surface Area of Cone = πrl = 22/7 x 6.3 x 11.82 = 234.03 cm2
问题12。找到可以安装在边缘为14 cm的立方体中的最大右圆锥的体积。
解决方案:
Given that,
Side of Cube = 14 cm,
Radius(r) of the largest cone that can be fitted in Cube = Side / 2 = 14/2 = 7cm,
Height(h) = 14cm
As we know that Volume of Cone = 1/3 πr2h = 1/3 x 22/7 x 72 x 14 = 718.67 cm3
问题13.右圆锥的体积为9856 cm 3 。如果底座的直径为28厘米,请找到:
(i)锥体的高度
(ii)圆锥体的倾斜高度
(iii)圆锥体的弯曲表面积。
解决方案:
Given that,
Volume of a Right Circular Cone = 9856 cm3,
Diameter of the base = 28 cm,
Radius(r) = 28/2 = 14cm
1. Height of cone(h) = Volume x 3/πr2 = (9856x3x7)/(22x14x14) = 48cm
2. Slant Height(l) = √r2 + h2 = √(14)2 + (48)2 = 50cm
3. Curved Surface area = πrl = 22/7 x 14 x 50 = 2200 cm2
问题14.顶部直径3.5 m的圆锥形坑深12 m。它的千升容量是多少? [NCERT]
解决方案:
Given that,
Diameter of the top of the conical pit = 3.5 m,
Radius(r) = 3.5/2 = 1.75 m and depth(h) = 12 m
As we know that volume of pit = 1/3 πr2h
= 1/3 x 22/7 x (1.75)2 x 12 = 38.5 m3
Volume in Kiloliters = (38.5×1000) / 1000 = 38.5 Kiloliters
问题15.莫妮卡有一块画布,面积为551 m 2 。她用它制作了一个圆锥形的帐篷,其基本半径为7 m。假设在切割时发生了所有的缝合边缘和浪费,大约为1 m 2 。查找可以用它制作的帐篷的体积。
解决方案:
Given that,
Area of Canvas = 551 m2,
Area of wastage = 1 m2,
Actual area = 551 – 1 = 550 m2,
Base radius of the conical tent = 7 m
Let us assume l is the slant height and h is the vertical height then the Slant Height(l) = Area / πr
= (550 x 7) / 22×7 = 25m
As we know that Vertical height (h) = √l2 – r2
= √252 – 72 = √576 = 24 m
As we know that volume of Tent = 1/3 πr2h
= 1/3 x 22/7 x 7 x 7 x 24 = 1232 m3