问题11.百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:
As per question, r = 7 cm and h = 24 cm.
Therefore, slant height of cone, l = √r2 + h2 = √(72+242) = 25 cm
Now, CSA of one cone = πrl
= (22 x 7 x 25)/7 = 550 cm2
Therefore, area of the sheet required to make 10 such caps is 550 x 10 = 5500 cm2
问题12。如果两个圆锥体的底面直径相等且倾斜高度的比例为4:3,则求出两个圆锥体的曲面面积的比例。
解决方案:
Given: diameters of the cone are equal, so there radius will also be equal, thus r1 = r2 =r and their slant heights are in ratio l1:l2 = 4:3
Since, CSA of a cone = πrl, therefore
π r1 l1 / π r2 l2
⇒ π r l1 / π r l2
⇒ l1 / l2 = 4/3
Hence, the ratio of the CSA of two cones is 4:3.
问题13.有两个圆锥体,一个圆锥体的曲面面积是另一个圆锥体的两倍。后者的倾斜高度是前者的两倍。求出它们半径的比例。
解决方案:
According to question,
CSA1 / CSA2 = 2/1, also l1 / l2 = 1/2
⇒ π r1 l1 / π r2 l2 = 2/1
⇒ r1.1/r2.2 = 2/1
⇒ r1 / r2 = 4/1
Hence, their ratio of their radii is 4:1.
问题14:两个圆锥的直径相等。如果倾斜的高度比例为5:4。求出其曲面的比率。
解决方案:
Given that diameters of the cone are equal, so there radius will also be equal, thus r1 = r2 =r and their slant heights are in ratio l1:l2 = 5:4
Since, CSA of a cone = πrl, therefore
π r1 l1 / π r2 l2
⇒ π r l1 / π r l2
⇒ l1 / l2 = 5/4
Hence, the ratio of the CSA of two cones is 5:4.
问题15:圆锥体的弯曲表面积为308 cm 2 ,其倾斜高度为14 cm。找到基部的半径和圆锥的总表面积。
解决方案:
Given: l = 14 cm and CSA = 308 cm2
We know that, curved surface area of cone = πrl
⇒ 308 = 22/7 x 14 x l
⇒ l = 7 cm
Also, total surface area of a cone = πr(l + r)
= 22/7 x 7 x (14 + 7) = 462 cm2
问题16:圆锥形墓的倾斜高度和底径分别为25 m和14 m。找出以Rs比率粉刷其曲面面积的成本。每百平方米210平方米。
解决方案:
Given: slant height of the conical tomb = 25 m and base radius of tomb = 14/2 = 7 m.
CSA of the conical tomb = πrl
= 22 x 7 x 25 / 7 = 550 m2
The cost of whitewashing the curves surface of the tomb = (210 x 550)/100 = Rs. 1155
问题17:圆锥形帐篷高10 m,底部半径为24 m。如果1 m 2帆布的成本为Rs 70,请找到帐篷的倾斜高度。找到该帐篷所需的帆布成本。
解决方案:
According to the question,
Height of conical tent, h = 10 m and radius of the conical tent, r = 24 m
Therefore, slant height of cone, l = √r2 + h2 = √(102+242 ) = 26 m
Thus, the slant height of the cone is 26 m.
Now, curved surface area of cone = πrl
= 22 x 24 x 26 / 7 = 13728/7 m2
It is given cost of 1 m2 canvas is Rs. 70. Therefore cost of 13728/7 m2
= 1378 * 70 / 7 = Rs.137280
Hence, the cost of canvas required for the tent is Rs.137280.
问题18:一个帐篷是圆锥形的右圆柱体形式的帐篷。圆柱体的直径为24 m。圆柱部分的高度为11 m,而圆锥的顶点在地面之上16 m。找到帐篷所需的画布区域。
解决方案:
Given: diameter of cylinder = 24 m, therefore radius= 12 m. The cone will also have the same base radius, since it is surmounted on the top of the cylinder, i.e, r = 12 m.
Now, height of the cylinder = 11 m and height of entire system =16 m.
Therefore, height of the cone = 16-11 = 5m.
Slant height of cone, l = √r2 + h2 = √(122+52) = 13 m
Now, curved surface area of cone = πrl
= 22 x 12 x 13 / 7 = 3432/7 m2
Similarly, CSA of the cylindrical portion = 2πrh = 2 x 22 x 12 x 11 / 7 = 5808/7 m2
Therefore, area of the canvas required for the tent = (3432+5808)/7
= 9240/7 = 1320 m2
问题19.马戏团的帐篷是圆柱形的,高度为3 m,在其上方呈圆锥形。如果其直径为105 m,并且圆锥形部分的倾斜高度为53 m。计算5 m宽的帆布长度,以制成所需的帐篷。
解决方案:
Given: diameter = 105 m, so radius = 52.5 m and slant height of the conical portion = 53 m2
Now, curved surface area of cone = πrl
= 22 x 52.5 x 53 / 7 = 8745 m2
Similarly, CSA of the cylindrical portion = 2πrh = 2 x 22 x 52.5 x 3 / 7 = 990 m2
So, area of the canvas required for the tent = 8745 + 990
= 9735 m2
Therefore, length of canvas required = 9735/5 = 1947 m
问题20.一个10 m高的圆锥形帐篷的底部圆周为44 m。如果画布的宽度为2 m,请计算用于制作帐篷的画布的长度。
解决方案:
Given: circumference of the base of conical tent = 2πr
⇒ 44 = 2 x 22 x r / 7
⇒ r = 7 m
Now, slant height of the cone, l = √r2 + h2 = √(72+102 ) = √149 m
So, curved surface area of cone = πrl
= 22 x 7 x √149 / 7 = 22√149 m2
Thus, the length of the canvas required to make the tent = 22√149/2 = 11√149 = 134.2 m
问题21.制造高度为8 m,底半径为6 m的圆锥形帐篷需要4毫米宽的篷布长度。假设缝制边距需要额外的材料长度,并且裁切浪费约为20厘米。
解决方案:
Given: height of conical tent, h = 8 m and radius of conical tent, r = 6 m
So, slant height of the cone, l = √r2 + h2 = √(62+82) = 10 m
So, curved surface area of cone = πrl
= 3.14 x 6 x 10 = 188.4 m2
Now, let the length of tarpaulin sheet required be x m. Also, 20 cm will get wasted, so effective length of tarpaulin required = (x-0.2) m and breadth of tarpaulin = 3m.
Area of the sheet = CSA of sheet
⇒ (x – 0.2) * 3 = 188.4
⇒ x = 62.8 + 0.2 = 63 m2
Hence, the length of tarpaulin sheet required will be 63 m.2
问题22:在道路的其余部分,专门使用一个50个由再生纸板制成的空心圆锥形的公共汽车站。每个锥体的底直径为40厘米,高度为1 m。如果要喷涂每个锥体的外侧,并且喷涂成本为Rs。每平方米12。绘制所有这些圆锥体的成本是多少。
解决方案:
Slant height of the cone, l = √r2 + h2 = √(0.22+12) = √1.04 m = 1.02 m
CSA of a single cone = πrl
= 3.14 x 0.2 x 1.02 = 0.64056 m2
Therefore, total area for 50 such cones = 50 x 0.64056 = 32.028 m2
Now, the total cost of painting all the cones = 12 x 32.028 = 384.336
Therefore, the total cost of painting all the cones is Rs. 384.336
问题23.圆柱和圆锥的底边半径和高度相等。如果它们的曲面面积之比为8:5,则表明每个半径与每个高度的半径为3:4。
解决方案:
Given that both cylinder have equal radii and equal height.
So, lets assume base radius as r, height as h and slant height of the cone be l.
We know that, CSA of cone = πrl and CSA of the cylinder = 2πrh
⇒ 2πrh/πrl = 8/5
⇒ h / l = 4/5
⇒ h / √r2 + h2 = 4/5
⇒ h2 / r2 + h2 = 16/25
⇒ r2 / h2 = 9 / 16
⇒ r / h = 3/4