Surface Areas and Volumes | Exercise 12.1

Question 4

A cubical block of side $7 \text{ cm}$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution

Understand the Question

A cubical block has side $a = 7\text{ cm}$ . When surmounted by a hemisphere on its top face, the largest circle that fits on the square face has a diameter equal to the side of the cube ( $d = 7\text{ cm}$ ).
The total surface area of the resulting solid consists of:
- Total surface area of the cube ( $6a^2$ )
- Minus the circular base covered by the hemisphere ( $-\pi r^2$ )
- Plus the curved surface area (CSA) of the hemisphere ( $+2\pi r^2$ ) $\text{Total Surface Area} = 6a^2 + \pi r^2$

Step 1 · Find the Greatest Diameter

For the hemisphere to sit entirely on the top square face of the cubical block, its base circle cannot extend beyond the edges of the face. Diagram 1

\begin{aligned} \text{Greatest diameter of hemisphere} &= \text{Side of cubical block} \\[0.6em] &= 7\text{ cm} \end{aligned}

Step 2 · Calculate the Surface Area of the Solid

Radius of the hemisphere:

\begin{aligned} r &= \dfrac{\text{Diameter}}{2} \\[0.6em] &= \dfrac{7}{2} = 3.5\text{ cm} \end{aligned}

Formula for the surface area of the combined solid:

\begin{aligned} \text{Surface area of solid} &= \underbrace{6a^2}_{\text{cube}} + \underbrace{2\pi r^2}_{\text{CSA of hemisphere}} - \underbrace{\pi r^2}_{\text{base circle covered}} \\[0.6em] &= 6a^2 + \pi r^2 \end{aligned}

Substitute side $a = 7\text{ cm}$ , radius $r = 3.5\text{ cm}$ , and $\pi = \dfrac{22}{7}$ :

\begin{aligned} \text{Surface area} &= 6 \times (7)^2 + \dfrac{22}{7} \times (3.5)^2 \\[0.6em] &= 6 \times 49 + \dfrac{22}{7} \times 12.25 \\[0.6em] &= 294 + 22 \times 1.75 \\[0.6em] &= 294 + 38.5 \\[0.6em] &= 332.5\text{ cm}^2 \end{aligned}

Answer

Greatest diameter $= 7\text{ cm}$ ; Surface area $= 332.5\text{ cm}^2$

Common Mistakes

Adding Total Surface Areas Directly: Simply adding the total surface area of the cube ( $6a^2$ ) to the total surface area of the hemisphere ( $3\pi r^2$ ) is incorrect because the base area of the hemisphere ( $\pi r^2$ ) is covered and hidden inside the solid.
Using Diameter instead of Radius: Using $r = 7\text{ cm}$ instead of $r = 3.5\text{ cm}$ (or $\dfrac{7}{2}\text{ cm}$ ) when evaluating $\pi r^2$ leads to calculation errors.

Question 4

More questions in Exercise 12.1