Question

A metallic bucket, open at the top, of height 24cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7cm and 14 cm respectively. Find:
i) the volume of water which can completely fill the bucket.
ii) the area of metal sheet used to make the bucket. $\left[ use\text{ }\pi =\dfrac{22}{7} \right]$

Answer 1

Hint: In the above question first of all we will calculate the slant height of the frustum and then we will calculate the volume and curved surface area of the frustum by using their formulae as shown belows:
Let the slant height of the frustum of the cone be “l” and the radius of the upper end be “R” and the radius of the lower end be “r” respectively. Therefore,
\[\begin{align}
  & l=\sqrt{{{{(R-r)}^{2}}}+{{h}^{2}}} \\
 & volume=\dfrac{1}{3}\pi h({{R}^{2}}+{{r}^{2}}+R\times r) \\
 & \text{curved surface area = }\pi \times l\times (R+r) \\
\end{align}\]

Complete step-by-step answer:

Now, we have been given h =24cm, R= 14cm and r=7cm. Therefore,
\[\begin{align}
  & l=\sqrt{{{(14-7)}^{2}}+{{24}^{2}}} \\
 & l=\sqrt{49+576} \\
 & l=\sqrt{625} \\
 & l=25cm \\
 & \text{so, slant height is 25cm}\text{. } \\
 & \text{volume=}\dfrac{1}{3}\times \pi \times 24({{14}^{2}}+{{7}^{2}}+14\times 7) \\
 & \text{ }=\dfrac{1}{3}\times \dfrac{22}{7}\times 24(196+49+98) \\
 & \text{ =}\dfrac{181104}{21} \\
 & \text{ =8624c}{{\text{m}}^{3}} \\
 & \text{curved surface area=}\dfrac{22}{7}\times 25\times (14+7) \\
 & \text{ =}\dfrac{11550}{21} \\
 & \text{ =1650c}{{\text{m}}^{2}} \\
 & \text{Area of the base(lower end) of the bucket = }\pi {{\text{r}}^{2}} \\
 & \text{ =}\dfrac{22}{7}\times 7\times 7 \\
 & \text{ =154c}{{\text{m}}^{2}} \\
\end{align}\]
Area of the metal sheet used to make the bucket
= curved surface area + Area of the base (lower end)
= 1650 +154
=1804\[c{{m}^{2}}\]
Therefore, the volume of the water which can completely fill the bucket is 8624\[c{{m}^{3}}\] and the area of the metal sheet used to make the bucket is 1804\[c{{m}^{2}}\].

NOTE: Be careful while doing calculation as there is a chance that you might make a mistake and you will get the incorrect answer. The most common mistake that is possible is using the wrong formula for the volume of frustum as \[\dfrac{1}{3}\pi h\left( {{R}^{2}}+{{r}^{2}} \right)\] instead of \[\dfrac{1}{3}\pi h\left( {{R}^{2}}+{{r}^{2}}+R\times r \right)\].