Question

Bucket of height 8 centimetre is made up of a copper sheet in the form of a frustum of a right circular cone with radii of its upper and lower ends are 3 cm and 9 cm respectively calculate the height of the cone of which the bucket is a part the volume of the water which can be filled in the bucket the area of copper sheet required to make the bucket

Answer 1

Step-by-step explanation:

Let $h$ be the height, $l$ the slant height and $r_{1}$ and $r_{2}$ and the radii of the circular bases of the frustum of the cone.

∴$h=8cm$, $r_{1}=9cm$ and $r_{2}=3cm$

Then slant height $l=\sqrt{h^{2}+(r_1-r_2)^{2}}=\sqrt{8^{2}+6^{2}}=\sqrt{100}=10cm$

Let $h_1$ be the height of the cone of which the bucket is a part.

i) ∴$h_1=\frac{h r_1}{r_1-r_2}=\frac{8*9}{9-3}=12cm$

ii) Volume of water which can be filled in the bucket:

= $\frac{\pi h}{3}(r_1^2+r_2^2+r_1r_2)cm^{3}$

=$\frac{\pi*8}{3}(9^2+3^2+9*3)cm^{3}$

=$\frac{\pi*8}{3}*117=312\pi cm^{3}$

iii) Area of the copper sheet required to make the bucket:

=$\pi l(r_1+r_2)+\pi r_2^2cm^2$

=$\pi*10(9+3)+\pi*3^2cm^2$

=$\pi(120+9)cm^2$

=$129\pi cm^2$

Hence, i) Height of the cone of which the bucket is a part = $12cm$

ii) Volume of the water which can be filled in the bucket = $312\pi cm^{3}$

iii) Area of copper sheet required to make the bucket = $129\pi cm^{2}$