问题1：圆锥体的底部直径为10.5厘米，倾斜高度为10厘米。找到其弯曲的表面积。
解决方案：
Given:
Radius = diameter / 2
Radius(r) = (10.5/2)cm = 5.25cm
Slant height of cone, (l) = 10 cm
Since, Curved surface area of cone is = πrl
= (22/7)×5.25×10 = 165 cm2
Therefore, the curved surface area of the cone of height 10m and base 10.5 cm is 165 cm2.
问题2：如果圆锥体的倾斜高度为21 m，底部直径为24 m，则求出圆锥体的总表面积。
解决方案：
Given:
Radius = diameter / 2
Radius (r) = 24/2 m = 12m
Slant height, (l) = 21 m
Since, Total surface area of the cone = πr(l+r)
= (22/7)×12×(21+12) m2
= 1244.57m2
Therefore, the total surface area of the cone with height 21m and base 24m is 1244.57m2.
问题3：圆锥体的弯曲表面积为308 cm 2 ，其倾斜高度为14 cm。找
(i)底座的半径和
(ii)圆锥体的总表面积。
解决方案：
Given:
Slant height (l) = 14 cm
Curved surface area = 308 cm2
radius = r.
(i) radius of the base
Curved surface area = 308 cm2
Curved surface area of cone = πrl
(308) = (22/7)×r×14
308 = 44 r
r = 308/44 = 7
Radius of a cone base is 7 cm.
(ii) total surface area of the cone.
Total surface area of cone = πr(l+r)
Total surface area of cone = 308+(22/7)×72
= 308+154
= 462 cm2
Therefore, the total surface area of the cone with CSA as 308 cm2 is 462 cm2.
问题4：一个圆锥形帐篷高10 m，其底部半径为24 m。找
(i)帐篷的倾斜高度。
(ii)制作帐篷所需的帆布成本，如果1 m 2帆布的成本为Rs。 70
解决方案：
Given:
Height of conical tent (h) = 10m
Radius of conical base (r) = 24m
Slant height of the tent = l.
(i) slant height of the tent.
l2 = h2 + r2 [using Pythagoras theorem]
= (10)2 + (24)2
= 676
l = 26
Therefore, the slant height of the tent = 26 m.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs. 70.
Curved surface area of tent = πrl
= (22/7) × 24 × 26 m2
As cost of 1 m2 canvas = Rs 70
Cost of (13728/7)m2 canvas = (13728/7)×70
= Rs 137280
Therefore, the cost of the canvas required to make the tent is Rs 137280.
问题5：要制作高度为8 m且底半径为6 m的圆锥形帐篷，需要多少长度的篷布3 m宽?假设裁切边距和裁切浪费所需的额外材料长度约为20厘米(使用π = 3.14)。
解决方案：
Height of conical tent (h) = 8m
Radius of base of tent (r) = 6m
Slant height of tent, l2 = r2+h2
l2 = 62+82
= 36+64
= 100 [Taking square root on both the side]
l = 10
As, Curved surface area = πrl
= (3.14×6×10) m2
= 188.4m2
Let the length of tarpaulin sheet required = L.
As 20 cm will be wasted,
Therefore, the Effective length will be (L – 0.2m).
Given breadth of tarpaulin = 3m
Area of sheet = Curved surface area of tent
[(L – 0.2) × 3] = 188.4
L – 0.2 = 62.8
L = 63 m
Therefore, the length of the required tarpaulin sheet will be 63 m.
问题6：一个圆锥形墓的倾斜高度和底直径分别为25 m和14 m。找出以每100 m 2卢比210卢比的速率粉刷其弯曲表面的成本。
解决方案：
Given:
Slant height of conical tomb (l) = 25m
Base radius (r) = diameter/2
= 14/2 m
= 7m
Curved surface area of conical tomb = πrl
= (22/7)×7×25
= 550
Curved surface area of conical tomb= 550m2
Cost of white-washing 100 m2 area = Rs 210
Cost of white-washing 550 m2 area = Rs (210×550)/100
= Rs. 1155
Therefore, the cost will be Rs. 1155 while white-washing tomb of area 550 m2.
问题7：百搭帽的形状是基本半径为7厘米，高度为24厘米的右圆锥形。找到制作10个这样的盖子所需的纸张面积。
解决方案：
Given:
Radius of conical cap (r) = 7 cm
Height of conical cap (h) = 24cm
Slant height, l2 = r2+h2
= 72+242
= 49+576
= 625 [Taking square root on both the side]
l = 25 cm
Curved surface area of 1 conical cap = πrl
= (22/7)×7×24
= 550
Curved surface area of 10 caps = (10×550) cm2 = 5500 cm2
Therefore, the area of 5500 cm2 of the sheet is required to make 10 such caps.
问题8：通过使用50个由再生纸板制成的空心圆锥形物，将公交车站设在道路的其余部分。每个锥体的底直径为40厘米，高度为1 m。如果要对每个视锥细胞的外侧进行粉刷，并且喷漆的成本为每平方米Rs 12，那么对所有这些视锥粉刷的成本将是多少? (使用π= 3.14并取1.04 = 1.02)
解决方案：
Given:
Radius of cone (r) = diameter/2
= 40/2 cm
= 20cm
= 0.2 m
Height of cone (h) = 1m
Slant height, l2 = r2 + h2
Using given values, l2 = (0.22+12)
= (1.04)
l = 1.02
Slant height of the cone (l) = 1.02 m
Now,
Curved surface area of each cone = πrl
= 3.14×0.2×1.02
= 0.64056
Curved surface area of 50 such cones = (50×0.64056) = 32.028
Curved surface area of 50 such cones = 32.028 m2
Also given Cost of painting 1 m2 area = Rs 12
Cost of painting 32.028 m2 area = Rs (32.028×12)
= Rs. 384.336
= Rs.384.34 (approximately)
Therefore, the cost of painting all the cones is Rs. 384.34.
