A bucket of height 8cm and made up of copper sheet is in the form frustum of a right circular cone with radii of its lower and upper ends as 3cm and 9cm respectively. Calculate: (i) the height of the cone of which the bucket is a part. (ii) the volume of water which can be filled in the bucket. (iii) the area of metal sheet required to make the bucket.
Answers
Answer:
i) The height of the cone of which the bucket is a part of is 12 cm
ii) the volume of water which can be filled in the bucket is 312π cm³
iii) the area of metal sheet required to make the bucket is 129π cm²
Step-by-step explanation:
we have been given
h = 8 cm
r₁ = 9 cm
r₂ = 3 cm
Then slant height
L =
= √(8² + 6²)
= 10 cm
i) let h₁ be the height of the cone of which the bucket is a part of, so,
h₁ = hr₁/(r₁-r₂)
= 8 x 9/6
= 12 cm
ii) volume of the water which can be filled in the bucket
=
= 8π( 81 + 9 + 27)/3
= 8 x π x 117/3
= 312π cm³
iii) area of the metal sheet required to make the bucket = surface area
= πL(r₁ + r₂) + πr₂²
=π x 10 x 12 + 9π
= 129π cm²
Answer:
Step-by-step explanation:
Let be the height l is the slant height and r and r the radii of the circular base of the frustum of the cone
h = 8 cm , r = 9 cm , r = 3 cm
then slant height l = √h² + (r₁ - r₂ )²
= 8² + 6²
= 100
= 10 cm
let h₁ be the height of the cone of which the bucket is a part
(1) h₁ = hr₁/ r₁ -r₂
= 8 × 9 / 9 -3
= 72 / 6
= 12 cm
(2) Volume of water which can be filled in the bucket = πh /3 (r₁² + r₂² + r₁r₂)
= 22/ 7 ×8 /3 ( 9² + 3² + 27)
= 312× 22/7
= 980.57 cm²
(3) Area of the copper sheet required to make the bucket =
πl(r₁ +r₂ ) + πr₂²
= 22/7 ×10 ( 9 +3) + 22 /7 × 3²
= 129 × 22/7
= 405 .42 cm ²