Right Prism. Pyramid

A prism is a polyhedron with two congruent, parallel polygonal faces called bases, connected by parallelogram faces called lateral faces. The segments connecting corresponding vertices of the two bases are called lateral edges.

A prism is called a right prism if its lateral edges are perpendicular to the bases. In a right prism, every lateral face is a rectangle.

The height $h$ of a prism is the perpendicular distance between the two bases (equal to the length of a lateral edge in a right prism).

The volume of any prism equals the area of the base multiplied by the height:

$\boxed{V = S_{\text{base}} \cdot h}$

Justification (Cavalieri’s Principle). Every cross-section of a prism parallel to the bases is congruent to the base. By Cavalieri’s Principle, any two solids with equal heights and equal cross-sectional areas at every level have equal volumes. Therefore a right prism with base area $S_{\text{base}}$ and height $h$ has volume $V = S_{\text{base}} \cdot h$. $\blacktriangleleft$

A pyramid is a polyhedron with one polygonal face called the base and a point not in the plane of the base called the apex, together with triangular lateral faces connecting the apex to each edge of the base.

The height $h$ of a pyramid is the perpendicular distance from the apex to the plane of the base (the foot is called the foot of the altitude).

A pyramid is regular if its base is a regular polygon and the apex projects perpendicularly onto the center of the base.

In a regular pyramid, all lateral edges are equal, all lateral faces are congruent isosceles triangles, and the slant height $l$ is the height of any lateral face (measured from apex to the midpoint of a base edge).

The volume of any pyramid equals one-third of the base area times the height:

$\boxed{V = \dfrac{1}{3} S_{\text{base}} \cdot h}$

Cavalieri’s Principle argument. A cross-section of a pyramid at height $t$ from the base (where $0 \le t \le h$) is similar to the base with ratio $1 - t/h$. Therefore its area is $S_{\text{base}} \cdot \left(1 - \tfrac{t}{h}\right)^2$. Integrating over $t$ from $0$ to $h$ gives $V = S_{\text{base}} \cdot h \cdot \tfrac{1}{3} = \tfrac{1}{3}S_{\text{base}} h$. $\blacktriangleleft$

A right triangular prism has a base that is an equilateral triangle with side $a = 6$ cm and height $h = 10$ cm. Find the total surface area and volume.

Solution.

Base area: An equilateral triangle with side $a$ has area $S_{\text{base}} = \dfrac{\sqrt{3}}{4}a^2 = \dfrac{\sqrt{3}}{4} \cdot 36 = 9\sqrt{3}$ cm².

Perimeter: $P = 3a = 18$ cm.

Lateral surface area: $S_{\text{lat}} = Ph = 18 \times 10 = 180$ cm².

Total surface area: $S_{\text{total}} = 180 + 2 \times 9\sqrt{3} = 180 + 18\sqrt{3} \approx 180 + 31{.}2 = 211{.}2$ cm².

Volume: $V = S_{\text{base}} \cdot h = 9\sqrt{3} \times 10 = 90\sqrt{3} \approx 155{.}9$ cm³.

Answer: $S_{\text{total}} = (180 + 18\sqrt{3})$ cm², $V = 90\sqrt{3}$ cm³. $\blacktriangleleft$

A regular quadrilateral (square-base) pyramid has base side $a = 8$ cm and slant height $l = 10$ cm. Find the height $h$ of the pyramid and its volume.

Solution.

Finding $h$: The apothem of the square base (distance from center to midpoint of a side) is $r = \dfrac{a}{2} = 4$ cm. From the right triangle (height, apothem, slant height):

$l^2 = h^2 + r^2 \implies 100 = h^2 + 16 \implies h^2 = 84 \implies h = 2\sqrt{21} \text{ cm}$

Base area: $S_{\text{base}} = a^2 = 64$ cm².

Volume:

$V = \dfrac{1}{3} S_{\text{base}} \cdot h = \dfrac{1}{3} \times 64 \times 2\sqrt{21} = \dfrac{128\sqrt{21}}{3} \approx 195{.}6 \text{ cm}^3$

Answer: $h = 2\sqrt{21}$ cm, $V = \dfrac{128\sqrt{21}}{3}$ cm³. $\blacktriangleleft$

A regular hexagonal pyramid has base side $a = 6$ cm and height $h = 8$ cm. Find the volume and lateral surface area.

Solution.

Base area: A regular hexagon with side $a$ has area $S_{\text{base}} = \dfrac{3\sqrt{3}}{2}a^2 = \dfrac{3\sqrt{3}}{2} \times 36 = 54\sqrt{3}$ cm².

Volume:

$V = \dfrac{1}{3} \times 54\sqrt{3} \times 8 = 144\sqrt{3} \approx 249{.}4 \text{ cm}^3$

Slant height: The apothem of a regular hexagon with side $a$ is $r = \dfrac{a\sqrt{3}}{2} = 3\sqrt{3}$ cm.

$l = \sqrt{h^2 + r^2} = \sqrt{64 + 27} = \sqrt{91} \text{ cm}$

Perimeter: $P = 6a = 36$ cm.

Lateral surface area:

$S_{\text{lat}} = \dfrac{1}{2} \times 36 \times \sqrt{91} = 18\sqrt{91} \approx 171{.}7 \text{ cm}^2$

Answer: $V = 144\sqrt{3}$ cm³, $S_{\text{lat}} = 18\sqrt{91}$ cm². $\blacktriangleleft$

A rectangular box has dimensions $3$ cm $\times$ $4$ cm $\times$ $5$ cm. Find: (a) the volume; (b) the total surface area; (c) the length of the space diagonal.

A regular triangular pyramid has base side $a = 4$ cm and all lateral edges of length $l_e = 6$ cm. Find the height of the pyramid and its volume.