Q8:

From a solid cylinder whose height is $2.4$ cm and diameter $1.4$ 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 cm$^2$.

Given:

• Height of the cylinder ($h$) = $2.4$ cm
• Diameter of the cylinder ($d$) = $1.4$ cm
• Radius of the cylinder ($r$) = $\frac{d}{2} = \frac{1.4}{2} = 0.7$ cm
• A conical cavity of the same height ($h = 2.4$ cm) and same radius ($r = 0.7$ cm) is hollowed out.

To Find:

The total surface area of the remaining solid, rounded to the nearest cm$^2$.

Step 1: Identify the components of the Total Surface Area (TSA)

When a conical cavity is hollowed out from a solid cylinder, the total surface area of the remaining solid consists of:

• The curved surface area of the cylinder ($2\pi rh$)
• The area of the circular base of the cylinder ($\pi r^2$)
• The curved surface area of the conical cavity ($\pi rl$)

Formula: $TSA = 2\pi rh + \pi r^2 + \pi rl$

Step 2: Calculate the slant height ($l$) of the cone

The slant height $l$ is given by the Pythagorean theorem: $l = \sqrt{r^2 + h^2}$

$l = \sqrt{(0.7)^2 + (2.4)^2}$

$l = \sqrt{0.49 + 5.76}$

$l = \sqrt{6.25} = 2.5$ cm

Step 3: Calculate the individual areas

Using $\pi \approx \frac{22}{7}$:

Curved Surface Area of Cylinder = $2 \times \frac{22}{7} \times 0.7 \times 2.4 = 2 \times 22 \times 0.1 \times 2.4 = 10.56$ cm$^2$

Area of circular base = $\pi r^2 = \frac{22}{7} \times 0.7 \times 0.7 = 22 \times 0.1 \times 0.7 = 1.54$ cm$^2$

Curved Surface Area of Cone = $\pi rl = \frac{22}{7} \times 0.7 \times 2.5 = 22 \times 0.1 \times 2.5 = 5.5$ cm$^2$

Step 4: Sum the areas

$TSA = 10.56 + 1.54 + 5.5$

$TSA = 17.6$ cm$^2$

Step 5: Rounding to the nearest cm$^2$

Since $17.6$ is closer to $18$ than $17$, we round to $18$ cm$^2$.

Final Answer: 18 cm$^2$