问题1.找到半径为球体的表面积：
(i)10.5厘米(ii)5.6厘米(iii)14厘米
(假设π= 22/7)
解决方案：
We know that, Surface area of sphere (SA) = 4πr²
(i) Radius of sphere, r = 10.5 cm
SA = 4×(22/7)×10.52 = 1386
Therefore, Surface area of sphere is 1386 cm²
(ii) Radius of sphere, r = 5.6cm
SA = 4×(22/ 7)×5.62 = 394.24
Therefore, Surface area of sphere is 394.24 cm²
(iii) Radius of sphere, r = 14cm
SA = 4×(22/7)×(14)2
= 2464
Therefore, Surface area of sphere is 2464 cm²
问题2.找到直径球体的表面积：
(i)14厘米(ii)21厘米(iii)3.5厘米
(假设π= 22/7)
解决方案：
(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm
We know that, Surface area of sphere = 4πr²= 4×(22/7)×72 = 616
Surface area of a sphere is 616 cm²
(ii) Radius of sphere,r= diameter/2=21/2 = 10.5 cm
We know that, Surface area of sphere = 4πr²
= 4×(22/7)×10.52 = 1386
Surface area of a sphere is 1386 cm²
Therefore, the surface area of a sphere having diameter 21cm is 1386 cm²
(iii) Radius(r) of sphere = 3.5/2 = 1.75 cm
We know that, Surface area of sphere = 4πr²
= 4×(22/7)×1.752 = 38.5
Surface area of a sphere is 38.5 cm²
问题3.找到半径为10 cm的半球的总表面积。 [使用π= 3.14]
解决方案：
Given, Radius of hemisphere, r = 10cm
Formula: Total surface area of hemisphere = 3πr²
= 3×3.14×102 = 942
Therefore, total surface area of given hemisphere is 942 cm².
问题4.随着向其中泵入空气，球形气球的半径从7cm增加到14cm。找出两种情况下球囊表面积的比率。
解决方案：
We assume that r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So,
r1 = 7cm
r2 = 14 cm
Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)
= 4(r1)²/4(r2)²
= (r1/r2)²
= (7/14)² = (1/2)² = ¼
Therefore, the ratio between the surface areas is 1:4.
问题5：一个黄铜制的半球形碗的内径为10.5厘米。找到在内部镀锡的成本，价格为每100cm²16卢比。 (假设π = 22/7)
解决方案：
Given, Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm
We know, Formula for Surface area of hemispherical bowl = 2πr²
= 2×(22/7)×(5.25)2 = 173.25cm²
Cost of tin-plating 100 cm² area = Rs 16
So, Cost of tin-plating 1 cm² area = Rs 16 /100
Cost of tin-plating 173.25 cm² area = Rs. (16×173.25)/100 = Rs 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm² is Rs 27.72.
问题6.找到一个表面积为154cm²的球体的半径。 (假设π = 22/7)
解决方案：
Let the radius of the sphere be r.
Surface area of sphere = 154 (given)
So,
4πr² = 154
r² = (154×7)/(4×22) = (49/4)
r = (7/2) = 3.5cm
Therefore, the radius of the sphere is 3.5 cm.
问题7.月亮的直径大约是地球直径的四分之一。求出它们的表面积之比。
解决方案：
Let the diameter of earth be d, then the diameter of moon will be d/4 (as per given statement)
Radius of earth = d/2
So, Radius of moon = ½×d/4 = d/8
Surface area of moon = 4π(d/8)²
Surface area of earth = 4π(d/2)²
The ratio between their surface areas is 1:16.
问题8.半球形碗是由0.25厘米厚的钢制成的。碗的内半径为5厘米。找到碗的外曲面。 (假设π = 22/7)
解决方案：
Given:
Inner radius of hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm
We know, formula for outer CSA of hemispherical bowl = 2πr², where r is radius of hemisphere
= 2×(22/7)×(5.25)²= 173.25
Therefore, the outer curved surface area of the bowl is 173.25 cm².
问题9.一个右圆柱体只是围成一个半径为r的球体(见图)。找
(i)球体的表面积，
(ii)圆柱体的曲面面积，
(iii)(i)和(ii)中获得的面积比例。
解决方案：
(i) Surface area of sphere = 4πr², where r is the radius of sphere
(ii) As Height of cylinder, h = r+r =2r
And Radius of cylinder = r
CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)
= 4πr²
(iii) Ratio between areas = (Surface area of sphere)/CSA of Cylinder)
= 4r2/4r2 = 1/1
Therefore, Ratio of the areas obtained in (i) and (ii) is 1:1.