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Solution:
Let R and r be the radii of the top and base of the bucket respectively,
Let h be its height.
Then, we have R = 20 cm, r = 10 cm, h = 30 cm
Capacity of the bucket = Volume of the frustum of the cone
= 1/3 π(R2 + r2 + R r )h
= 1/3 π(202 + 102 + 20 x 10 ) x 30
= 3.14 x 10 (400 + 100 + 200)
= 21980 cm3 = 21.98 litres
Now,
Surface area of the bucket = CSA of the bucket + Surface area of the bottom
= π l (R + r) + πr2
We know that,
l = √h2 + (R – r)2
= √[302 + (20 – 10)2] = √(900 + 100)
= √1000 = 31.62 cm
So,
The Surface area of the bucket = (3.14) x 31.62 x (20 + 10) + (3.14) x 102
= 2978.60 + 314
= 3292.60 cm2
Next, given that the cost of 1 litre milk = Rs 25
Thus, the cost of 21.98 litres of milk = Rs (25 x 21.98) = Rs 549.50
Solution:
Given:
Radii of top circular ends (r1) = 20 cm
Radii of bottom circular end of bucket (r2) = 12 cm
Let height of bucket be 'h'.
Volume of frustum cone =
Given capacity/ volume of bucket = 12308.8 cm3 -----(2)
Equating (1) & (2)
⇒ h = 15 cm
Therefore,
Height of bucket (h) = 15 cm
Let 'l' be slant height of bucket
⇒ l2 = (r1 - r2)2 + h2
⇒ l =
⇒ l =
⇒ l = 17 cm
Therefore,
Length of bucket/ slant height of bucket (l) = 17 cm
Curved surface area of bucket = π(r1 + r2)l + πr22
= π(20 + 12)17 + π(12)2
= π(32)17 + π(12)2
= π(9248 + 144) = 2160.32 cm2
Therefore,
Curved surface area = 2160.32 cm2
Solution:
Given:
Height of bucket (h) = 20 cm
Upper radius of bucket (r1) = 25 cm
Lower radius of bucket (r2) = 10 cm
Let 'l' be slant height of bucket
Therefore,
Slant height of bucket (l) = 25 cm
Curved surface area of bucket = π(r1 + r2)l + πr22
= π(25 + 10)25 + π(10)2
= π(35)25 + π(100) = 975π
Curved surface area = 3061.5 cm2
Cost of making bucket per 100 cm2 = Rs 70
Cost of making bucket per 3061.5 cm2 =
= Rs 2143.05
Therefore,
Total cost for 3061.5 cm2 = Rs 2143.05
Solution:
Given:
Slant height of frustum cone = 10 cm
Radii of circular ends of frustum cone are 33 and 27
r1 = 33 cm; r2 = 27 cm
Total surface area of a solid frustum of cone = π(r1 + r2) × l + πr12 + πr22
= π(33 + 27) × 10 + π(33)2 + π(27)2
= π(60) × 10 + π(33)2 + π(27)2
= π(600 + 1089 + 729)
= 2418π cm2
= 7599.42 cm2
Therefore,
Total surface area of frustum cone = 7599.42 cm2
Solution:
Given:
Height of frustum cone = 16 cm
Diameter of lower end of bucket (d1) = 16 cm
Lower end radius (r1) = 16/2 = 8 cm
Upper end radius (r2) = 40/2 = 20 cm
Let 'l' be slant height of frustum of cone
l = 20 cm
Therefore,
Slant height of frustum cone (l) = 20 cm
Volume of frustum cone =
Volume = 10449.92 cm3
Curved surface area of frustum cone = π(r1 + r2)l + πr22
= π(20 + 8)20 + π(8)2
= π(560 + 64) = 624π cm2
Cost of metal sheet per 100 cm2 = Rs 20
Cost of metal sheet for 624π cm2 =
= Rs 391.90
Therefore,
Total cost of bucket = Rs 391.9
Solution:
Given:
Height of a frustum cone = 9 cm
Lower end radius (r1) = 60/2 cm = 30 cm
Upper end radius (r2) = 36/2 cm = 18 cm
Let slant height of frustum cone be l
Volume of frustum cone =
= 5292π cm3
Volume = 5292π cm3
Total surface area of frustum cone = π(r1 + r2) × l + πr12 + πr22
= π(30 + 18)15 + π(30)2 + π(18)2
= π(48(15) + (30)2 + (18)2)
= π(720 + 900 + 324)
= 1944π cm2
Therefore,
Total surface area = 1944π cm2
Solution:
Given:
Lower end radius of bucket (r1) = 8 cm
Upper end radius of bucket (r2) = 20 cm
Let height of bucket be 'h'
V1 =
Volume of milk container =
V2 =
Equating (1) and (2)
V1 = V2
⇒
⇒ h =
⇒ h = 16 cm
Therefore,
Height of frustum cone(h) = 16 cm
Let slant height of frustum cone be 'l'
l =
l = 20 cm
Therefore,
Slant height of frustum cone (l) = 20 cm
Total surface area of frustum cone = π(r1 + r2)l + πr12 + πr22
= π(20 + 8)20 + π(20)2 + π(8)2
= π(560 + 400 + 64)
= π(960 + 64) = 1024π = 3216.99 cm2
Total surface area = 3216.99 cm2