Surface Areas and Volumes | Exercise 12.1

Question 8

From a solid cylinder whose height is $2.4 \text{ cm}$ and diameter $1.4 \text{ cm}$ , a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $\text{cm}^2$ .

Solution

Understand the Question

When a conical cavity is hollowed out from a solid cylinder, the exposed surface area of the resulting solid consists of:
- The curved surface area (CSA) of the cylinder
- The curved surface area (CSA) of the conical cavity (the new inner surface)
- The area of the bottom circular base of the cylinder
Therefore, the Total Surface Area (TSA) is: $\text{TSA} = \text{CSA of cylinder} + \text{CSA of cone} + \text{Base area of cylinder} = 2\pi rh + \pi rl + \pi r^2$
First find the slant height $l = \sqrt{r^2 + h^2}$ , then compute and round the total area to the nearest $\text{cm}^2$ .

Step 1 · Find the Radius and Slant Height

Diagram 1

Given for the cylinder and conical cavity: $\text{Height } (h) = 2.4 \text{ cm}$ $\text{Radius } (r) = \dfrac{1.4}{2} = 0.7 \text{ cm}$

Using the Pythagoras theorem, the slant height ( $l$ ) of the cone is:

\begin{aligned} l &= \sqrt{h^2 + r^2} \\[0.6em] &= \sqrt{(2.4)^2 + (0.7)^2} \\[0.6em] &= \sqrt{5.76 + 0.49} \\[0.6em] &= \sqrt{6.25} \\[0.6em] &= 2.5 \text{ cm} \end{aligned}

Step 2 · Calculate Total Surface Area of the Remaining Solid

The total surface area of the remaining solid is: $\text{TSA} = 2\pi rh + \pi rl + \pi r^2$

Substitute $r = 0.7 \text{ cm}$ , $h = 2.4 \text{ cm}$ , $l = 2.5 \text{ cm}$ , and $\pi = \dfrac{22}{7}$ : $\text{TSA} = \left(2 \times \dfrac{22}{7} \times 0.7 \times 2.4\right) + \left(\dfrac{22}{7} \times 0.7 \times 2.5\right) + \left(\dfrac{22}{7} \times 0.7 \times 0.7\right)$

\begin{aligned} \text{TSA} &= (2 \times 22 \times 0.1 \times 2.4) + (22 \times 0.1 \times 2.5) + (22 \times 0.1 \times 0.7) \\[0.6em] &= (4.4 \times 2.4) + (2.2 \times 2.5) + (2.2 \times 0.7) \\[0.6em] &= 10.56 + 5.50 + 1.54 \\[0.6em] &= 17.60 \text{ cm}^2 \end{aligned}

Rounding to the nearest $\text{cm}^2$ : $17.60 \text{ cm}^2 \approx 18 \text{ cm}^2$

Answer

$18 \text{ cm}^2$

Common Mistakes

Subtracting Surface Area: Do not subtract the conical surface area from the cylinder's area. Hollowing out a cavity removes volume but creates new exposed surface area, so the CSA of the cone must be added.
Including Both Cylinder Bases: The top base is cut open to form the cavity, so only the bottom base ( $\\pi r^2$ ) is included, not both bases ( $2\\pi r^2$ ).
Rounding Error: Forgetting to round the final calculated value $17.60 \text{ cm}^2$ to the nearest whole integer ( $18 \text{ cm}^2$ ).

Question 8

More questions in Exercise 12.1