Mensuration - Combination of solids

A toy is in the form of a cone of radius $3.5\text{ cm}$ mounted on a hemisphere of the same radius. The total height of the toy is $15.5\text{ cm}$. Find the total surface area of the toy. (Use $\pi = \frac{22}{7}$)

1. Radius of cone and hemisphere, $r = 3.5 = \frac{7}{2}\text{ cm}$.
2. Total height of toy = $15.5\text{ cm}$. Height of cone, $h = 15.5 - 3.5 = 12\text{ cm}$.
3. Calculate slant height $l$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\text{ cm}$.
4. Surface Area of toy = CSA of cone + CSA of hemisphere.
5. Surface Area = $\pi rl + 2\pi r^2 = \pi r(l + 2r)$.
6. Substituting values: $S = \frac{22}{7} \times \frac{7}{2} \times (12.5 + 2 \times 3.5) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5\text{ cm}^2$.

To find the surface area of a combined solid, we only sum the areas of the visible surfaces. The base of the cone and the base of the hemisphere are joined, so they are not part of the external surface. We use the total height to find the vertical height of the cone first, then find the slant height for the CSA calculation.

A solid metallic sphere of radius $10.5\text{ cm}$ is melted and recast into a number of smaller cones, each of radius $3.5\text{ cm}$ and height $3\text{ cm}$. Find the number of cones formed.

1. Volume of the metallic sphere $V_s = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \pi \times (10.5)^3$.
2. Volume of one small cone $V_c = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times (3.5)^2 \times 3$.
3. Let $n$ be the number of cones. By conservation of volume: $n \times V_c = V_s$.
4. $n \times (\frac{1}{3} \pi \times 3.5 \times 3.5 \times 3) = \frac{4}{3} \pi \times 10.5 \times 10.5 \times 10.5$.
5. Canceling $\pi$ and $\frac{1}{3}$ from both sides: $n \times 3.5 \times 3.5 \times 3 = 4 \times 10.5 \times 10.5 \times 10.5$.
6. $n = \frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3}$.
7. Since $\frac{10.5}{3.5} = 3$, we get $n = \frac{4 \times 3 \times 3 \times 10.5}{3} = 4 \times 3 \times 10.5 = 12 \times 10.5 = 126$.

When melting one solid to form others, the total volume remains the same. We set up an equation where the volume of the large sphere equals $n$ times the volume of a single small cone. It is usually easier to keep $\pi$ as a symbol and cancel it out later to simplify calculations.

1. Identify individual components:

• Radius ($r$) for all parts = $3.5 \text{ cm} = \frac{7}{2} \text{ cm}$
• Height of cylinder ($h_1$) = $10 \text{ cm}$
• Height of cone ($h_2$) = $6 \text{ cm}$

1. Calculate individual volumes:

• $V_{cylinder} = \pi r^2 h_1 = \frac{22}{7} \times (\frac{7}{2})^2 \times 10 = \frac{22}{7} \times \frac{49}{4} \times 10 = 385 \text{ cm}^3$
• $V_{cone} = \frac{1}{3} \pi r^2 h_2 = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{4} \times 6 = 77 \text{ cm}^3$
• $V_{hemisphere} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times \frac{343}{8} = \frac{539}{6} \approx 89.83 \text{ cm}^3$

1. Total Volume:

• $V = 385 + 77 + 89.83 = 551.83 \text{ cm}^3$

To find the volume of a combined solid, we sum the volumes of its constituent geometric parts: the cone, the cylinder, and the hemisphere.

1. Volume of ice cream in cylinder:

• $V_{cyl} = \pi R^2 H = \pi \times 6^2 \times 15 = 540\pi \text{ cm}^3$

1. Volume of one ice cream cone (Cone + Hemisphere):

• $V_{one} = V_{cone} + V_{hemi} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$
• $V_{one} = \frac{1}{3}\pi (3^2)(12) + \frac{2}{3}\pi (3^3)$
• $V_{one} = 36\pi + 18\pi = 54\pi \text{ cm}^3$

1. Number of cones ($n$):

• $n = \frac{V_{cyl}}{V_{one}} = \frac{540\pi}{54\pi} = 10$

Since the total volume of ice cream remains the same when transferred, we divide the volume of the cylinder by the volume of a single combined ice cream cone shape.