Surface Areas and Volumes - Introduction

Two cubes each of volume $64 \text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.

1. Find the edge of the cube ($a$): $V = a^3 = 64 \Rightarrow a = \sqrt[3]{64} = 4 \text{ cm}$
2. When two cubes are joined, the dimensions of the resulting cuboid are: Length ($l$) = $4 + 4 = 8 \text{ cm}$ Breadth ($b$) = $4 \text{ cm}$ Height ($h$) = $4 \text{ cm}$
3. Surface Area of Cuboid = $2(lb + bh + hl)$: $SA = 2(8 \times 4 + 4 \times 4 + 4 \times 8)$ $SA = 2(32 + 16 + 32)$ $SA = 2(80) = 160 \text{ cm}^2$

To solve this, we first find the side of the original cubes. When joined, only the length changes (it doubles), while the width and height remain the same. We then apply the standard surface area formula for a cuboid.

A decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge $5 \text{ cm}$, and the hemisphere fixed on the top has a diameter of $4.2 \text{ cm}$. Find the total surface area of the block.

1. Surface area of cube = $6 \times (\text{edge})^2 = 6 \times 5^2 = 150 \text{ cm}^2$.
2. The hemisphere is attached to one face, covering a circular area.
3. TSA of block = (TSA of cube) - (Base area of hemisphere) + (CSA of hemisphere).
4. Radius ($r$) of hemisphere = $\frac{4.2}{2} = 2.1 \text{ cm}$. $\text{TSA} = 150 - \pi r^2 + 2\pi r^2 = 150 + \pi r^2$ $\text{TSA} = 150 + \frac{22}{7} \times 2.1 \times 2.1$ $\text{TSA} = 150 + 22 \times 0.3 \times 2.1$ $\text{TSA} = 150 + 13.86 = 163.86 \text{ cm}^2$

The total surface area is the sum of the cube's area and the hemisphere's curved area, minus the area of the cube's face that is covered by the hemisphere's base.

Calculate the total volume of a solid consisting of a cylinder of length $12 \text{ cm}$ and radius $3 \text{ cm}$ topped with a hemisphere of the same radius.

1. Volume of Cylinder ($V_c$) = $\pi r^2h$ $V_c = \pi \times 3^2 \times 12 = 108\pi \text{ cm}^3$
2. Volume of Hemisphere ($V_h$) = $\frac{2}{3}\pi r^3$ $V_h = \frac{2}{3} \times \pi \times 3^3 = 18\pi \text{ cm}^3$
3. Total Volume ($V$): $\begin{array}{r} 108\pi \\ +18\pi \\ \hline 126\pi \end{array}$ $V = 126 \times \frac{22}{7} = 18 \times 22 = 396 \text{ cm}^3$

The volume of a combined solid is simply the sum of the volumes of its parts. Here, we add the volume of the cylinder to the volume of the hemisphere.

Radius of hemisphere ($r$) = $\frac{14}{2} = 7\text{ cm}$ Radius of cylinder ($r$) = $7\text{ cm}$ Height of cylinder ($h$) = $\text{Total height} - \text{Radius of hemisphere} = 13 - 7 = 6\text{ cm}$ Inner surface area = $\text{CSA of cylinder} + \text{CSA of hemisphere}$ Inner surface area = $2\pi rh + 2\pi r^2 = 2\pi r(h + r)$ Inner surface area = $2 \times \frac{22}{7} \times 7 \times (6 + 7)$ Inner surface area = $44 \times 13 = 572\text{ cm}^2$

To find the surface area of the combined shape, we sum the curved surface areas of the cylinder and the hemisphere. Note that the circular base of the cylinder and top of the hemisphere are not part of the surface area as they are internal.

Height of cone ($h_c$) = $6\text{ cm}$, Radius of cone ($r_c$) = $2.5\text{ cm}$ Height of cylinder ($h_{cyl}$) = $26 - 6 = 20\text{ cm}$, Radius of cylinder ($r_{cyl}$) = $1.5\text{ cm}$ $\text{Total Volume} = \text{Volume of cone} + \text{Volume of cylinder}$ $\text{Volume} = \frac{1}{3}\pi r_c^2 h_c + \pi r_{cyl}^2 h_{cyl}$ $\text{Volume} = \pi [ \frac{1}{3}(2.5)^2(6) + (1.5)^2(20) ]$ $\text{Volume} = \pi [ (6.25)(2) + (2.25)(20) ]$ $\text{Volume} = \pi [ 12.5 + 45 ] = 57.5π \approx 180.64\text{ cm}^3$

The total volume is the sum of the volumes of the two distinct 3D shapes. We identify the specific height and radius for the cone and the cylinder separately.