问题1.用
(i)半径6厘米,高度7厘米
(ii)半径3.5厘米,高度12厘米
解决方案:
Volume of cone (V) = (1/3) × πr2h
(i) Given values,
Radius of cone (r) = 6 cm
Height of cone (h) = 7 cm
V = (1/3) × (22/7) × 6 × 6 × 7 (using π=22/7)
V = 22 × 6 × 2
V = 264 cm3
(ii) Given values,
Radius of cone (r) = 3.5 cm
Height of cone (h) = 12 cm
V = (1/3) × (22/7) × 3.5 × 3.5 × 12 (using π=22/7)
V = 154 cm3
问题2。找到一个锥形容器的容量,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:
Volume of cone (V) = (1/3) × πr2h
(i) Given values,
Radius of cone (r) = 7 cm
Slant height of cone (l) = 25 cm
h = √(l2 – r2)
h = √(252 – 72)
h = √576
h = 24 cm
V = (1/3) × (22/7) × 7 × 7 × 24 (using π=22/7)
V = 22 × 7 × 8
V = 1232 cm3
(ii) Given values,
Height of cone (h) = 12 cm
Slant height of cone (l) = 13 cm
r = √(l2 – h2)
r = √(132 – 122)
r = √25
r = 5 cm
V = (1/3) × (22/7) × 5 × 5 × 12 (using π=22/7)
V = 2200/7 cm3
V = 314.28 cm3
问题3.锥体的高度为15厘米。如果其体积为1570 cm 3 ,请找到基座的半径。 (使用π = 3.14)
解决方案:
Given values,
Height of cone (h) = 15 cm
Volume of cone (V) = 1570 cm3
V = (1/3) × πr2h
1570 = (1/3) × 3.14 × r2 × 15 (using π=3.14)
r2 = 1570 × 3 / (3.14 × 15)
r2 = 100
r = √100
r = 10 cm
问题4.如果高度为9 cm的右圆锥的体积为48πcm 3 ,请找到其底部的直径。
解决方案:
Given values,
Height of cone (h) = 9 cm
Volume of cone (V) = 48π cm3
V = (1/3) × πr2h
48 × π = (1/3) × π × r2 × 9
r2 = 48 × 3 / 9 (canceling π from both sides)
r2 = 16
r = √16
r = 4 cm
Diameter = 2 times radius = 2 × r
= 2 × 4
= 8 cm
问题5.顶部直径3.5 m的圆锥形坑深12 m。它的千升容量是多少?
解决方案:
Given values,
Radius of cone (r) = 3.5/2 m
Height of cone (h) = 12 m
Volume of cone = (1/3) × πr2h
= (1/3) × (22/7) × (3.5/2) × (3.5/2) × 12 (taking π=22/7)
= (22 × 3.5 × 3.5 × 12) / (7 × 3 × 2 × 2)
= 38.5 m3
Capacity of conical pit in kilo liters:
1000 m3 = 1 liter
38.5 m3 = 1000 × 38.5 liters
= 38,500 liters
= 38.5 kilo liters
问题6.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)圆锥体的曲面面积
解决方案:
Given values,
Radius of cone (r) = 28/2 = 14 cm
Volume of cone (V) = 9856 cm3
(i) Volume of cone = (1/3) × πr2h
9856 = (1/3) × (22/7) × 14 × 14 × h (taking π=22/7)
h = (9856 × 3 × 7) / (22 × 14 × 14)
h = 48 cm
(ii) Let slant height = l
l2 = h2 + r2
l = √(h2 + r2)
l = √(482 + 142)
l = √(2304 + 196)
l = √2500
l = 50 cm
(iii) curved surface area of the cone = πrl
= π × 14 × 50 cm2
= 22/7 × 700 (taking π=22/7)
= 2,200 cm2
问题7.边5厘米,12厘米和13厘米的直角三角形ABC绕边12厘米旋转。找到如此获得的固体的体积。
解决方案:
Here, after revolving the triangle about the side 12 cm, we get
Radius of cone (r) =5 cm
Height of cone (h) = 12 cm
Volume of cone = (1/3) × πr2h
= (1/3) × π × 5 × 5 × 12
= (12 × π × 5 × 5) / 3
V = 100π cm3
问题8.如果上述问题7中的三角形ABC绕边旋转了5厘米,则求出所获得的固体体积。还要找到在问题7和问题8中获得的两种固体的体积比。
解决方案:
Here, after revolving the triangle about the side 5 cm, we get
Radius of cone (r) =12 cm
Height of cone (h) = 5 cm
Volume of cone = (1/3) × πr2h
= (1/3) × π × 12 × 12 × 5
= (12 × π × 12 × 5) / 3
V = 240π cm3
Volume in Question 7 = 100π cm3
Ratio = (Volume in Question 8) / (Volume in Question 7)
= 240π/100π
= 12/5
Hence, the ratio obtained = 12 : 5
问题9.一堆小麦为圆锥形,直径为10.5 m,高度为3 m。找到它的体积。堆要用帆布覆盖,以防雨淋。找到所需的画布区域。
解决方案:
Given values,
Radius of cone (r) = 10.5/2 = 105/20 m
Height of cone (h) = 3 m
Volume of cone = (1/3) × πr2h
= (1/3) × (22/7) × (105/2) × (105/2) × 3 (taking π=22/7)
= (22 × 105 × 105 × 3) / (3 × 20× 20× 7)
V = 86.625 m3
Area of the canvas = surface area of cone = πrl
Slant height (l) = √(h2 + r2)
l = √(32 + (10.5/2)2)
l = √(9+ (110.25/4))
l = √(146.25/4)
l = √36.56
l = 6.05 m (approx.)
Surface area of cone = π × (105/20) × 6.05 m2
= (22/7) × (635.25/20) (taking π=22/7)
= 99.82 m2