Question

A cuboidal block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Also find the total surface area of the solid.

Answer 1

Hint: The total surface area of the solid is equal to the surface area of the 6 sides of the cuboidal block along with the surface area of the hemispherical top. Find out the total surface areas of the cuboidal block and the hemisphere separately and add them up. Then, subtract the area of the circular base from the sum of the two surface areas.

Complete step-by-step answer:
The greatest diameter that the hemisphere surmounted upon the cuboidal block can be obtained by varying the length of the diameter from 0 cm and then increasing it gradually.
The greatest diameter of the hemisphere = Length of side of the cuboid = 7 cm
We know, total surface area of a cuboid $=\text{ 6}{{\text{a}}^{2}}$, where a is the length of each side of the cuboid.
Here, the length of each side of the cuboid = 7 cm.
Putting a=7 cm in the formula of total surface area,
$\therefore $ Total Surface Area of the cuboidal block
$\begin{align}
 & =\text{ 6 x }{{\text{7}}^{2}}\text{ sq}\text{.cm}\text{.} \\
 & \text{ = 294 sq}\text{.cm}\text{.} \\
\end{align}$

Now, we know, total surface area of a hemisphere $=\text{ 2}\pi {{\text{r}}^{2}}$, where r is the radius of the hemisphere.
In this case, the diameter of the hemisphere = 7 cm.
$\therefore $Radius of the hemisphere=$ \dfrac{7}{2}$ cm= 3.5 cm
Putting r=3.5cm in the formula of the total surface area of the hemisphere, we get,
Total surface area of the surmounted hemisphere
$\begin{align}
 & =\text{2 x }\pi \text{ x }{{\left( 3.5 \right)}^{2}}\text{ sq}\text{.cm}\text{.} \\
 & \text{ = 2 x }\dfrac{22}{7}\text{ x }{{\left( 3.5 \right)}^{2}}\text{ sq}\text{.cm}\text{. }\left( \because \text{ }\pi \text{ = }\dfrac{22}{7} \right) \\
 & =\text{ 77 sq}\text{.cm}\text{.} \\
\end{align}$
Now, surface area of the circular base
$\begin{align}
 & =\pi \text{ x }{{\left( 3.5 \right)}^{2}}\text{ sq}\text{.cm}\text{. }\left( \because \text{ r = 3}\text{.5 cm} \right) \\
 & \text{ = }\dfrac{22}{7}\text{ x }{{\left( 3.5 \right)}^{2}}\text{ sq}\text{.cm}\text{.} \\
 & =\text{ 38}\text{.5 sq}\text{.cm}\text{.} \\
\end{align}$
Thus, finally, total surface area of the solid = Surface area of the cuboid + Surface area of the hemisphere – Area of the circular base of the hemisphere
$\begin{align}
 & =\text{ (294 + 77 - 38}\text{.5) sq}\text{.cm}\text{.} \\
 & \text{ = 332}\text{.5 sq}\text{.cm}\text{.} \\
\end{align}$
Therefore, the total surface area of the solid is 332.5 sq.cm.

Note: The greatest diameter that the hemisphere can have is equal to the length of each side of the cuboidal is rather easy to interpret. If the diameter of the hemisphere is greater than 7 cm., then it cannot be surmounted over the cuboid as it will not fit on the top of the cuboid.