# Prism Types

Cube

The cube is a right prism with a square base and a height which is same as the side of the base. Let $a$ be the side of the cube, then

1. The volume of the cube $=$ area of base $\times$ height       i.e. $V = {a^3}$
2. The total surface area of the cube $=$ area of six faces       i.e. $S = 6{a^2}$
3. The line joining the opposite corners of the cube is called the diagonal of the cube. The length of the diagonal of the cube $= a\sqrt 3$

Proof: In the given figure, the line $DF$ is the diagonal of the cube. $\angle DBF = {90^ \circ }$
$\therefore$            $D{F^2} = B{D^2} + B{F^2}$
But      $B{D^2} = A{B^2} + A{D^2}$
$\therefore$            $D{F^2} = A{B^2} + A{D^2} + B{F^2}$

Since $AB,AD$ and $BF$ are the sides of the cube and each has a length equal to $a$, therefore
$D{F^2} = {a^2} + {a^2} + {a^2} = 3{a^2}$
$\therefore$            $DF = a\sqrt 3$

Example:

Three cubes of metal whose edges are in the ratio $3:4:5$ are melted into a single cube whose diagonal is $12\sqrt 3$ cm. Find the edges of the three cubes.

Solution:

Let the edges of the cubes be $3x,4x$ and $5x$ cm
$\therefore$ their volumes are:
${\left( {3x} \right)^2},{\left( {4x} \right)^2}$and ${\left( {5x} \right)^2}$cu. cm
$27{x^3},64{x^3}$ and $125{x^3}$cu. cm
$\therefore$ the volume of the single cube $= 27{x^3} + 64{x^3} + 125{x^3} = 216{x^3}$ cu. cm

Let $a$ be the edges of the cube, then volume:
$\therefore$ ${a^3} = 216{x^3} = {\left( {6x} \right)^3}$       $\Rightarrow a = 6$
$\therefore$ edge $a = 6x$ cm

Now, the diagonal of the cube $= \sqrt 3 \times {\text{edge}}$
$= \sqrt 3 \times 6x$ cm

But, the diagonal of the cube $= 12\sqrt 3$
$\therefore$ $\sqrt 3 \times 6x = 12\sqrt 3$       $\Rightarrow x = 2$

Hence the three edges of the cube are $6{\text{ cm, }}8{\text{ cm}}$ and $10{\text{ cm}}$

Rectangular Prism (1) The volume of the rectangular prism $=$ area of base $\times$ height
$V = a \cdot b \cdot h$

(2) The total surface area $=$ area of six faces
The total surface area $= 2ab + 2bh + 2ah \Rightarrow S = 2\left( {ab + bh + ah} \right)$

(3) The length of the diagonal is
$D{F^2} = B{D^2} + F{B^2}$
$= {a^2} + {b^2} + {h^2}$       (as$AD = BC = b$, $FB = CG = h$)
$DF = \sqrt {{a^2} + {b^2} + {h^2}}$

Example:

The length, width and thickness of a rectangular block are $9.6,13.2$ and $14.3$ cm respectively. Find the volume, surface area and length of the diagonal of the block.

Solution:

Given that: $a = 9.6$ cm, $b = 13.2$ cm, $h = 14.3$ cm

(1) Volume $V = a \cdot b \cdot h = 9.6 \times 13.2 \times 14.3 = 1812$ cu. cm

(2) Surface area $S = 2\left( {ab + bh + ah} \right) = 2\left( {12.6 + 13.2 + 14.3} \right) = 905.52$ sq.cm

(3) Length of diagonal $= \sqrt {{a^2} + {b^2} + {h^2}} = \sqrt {{{\left( {9.6} \right)}^2} + {{\left( {13.2} \right)}^2} + {{\left( {14.3} \right)}^2}} = 21.7\,cm$

Polygonal Prism

A prism with a polygon base is known as a polygonal prism.

(a) The volume of a prism whose base is a rectangular polygon of $n$ sides and height $h =$ area of the base $\times$ height.

1. $V = \left( {\frac{{n{a^2}}}{4}\cot \frac{{{{180}^ \circ }}}{n}} \right) \times h$   when side $a$ is given.
2. $V = \left( {n{r^2}\tan \frac{{{{180}^ \circ }}}{n}} \right) \times h$ when the radius $r$ of the inscribed circle is given.
3. $V = \left( {\frac{{n{R^2}}}{4}\sin \frac{{{{180}^ \circ }}}{n}} \right) \times h$   when the radius $R$ of the circumscribed circle is given.

(b) The lateral surface area $=$ perimeter of base $\times$ height

1. $L.S = na \times h$   when side $a$ is given.
2. $L.S = 2nr\tan \frac{{{{180}^ \circ }}}{n} \times h$   when the radius $r$of the inscribed circle is given.
3. $L.S = 2nR\sin \frac{{{{180}^ \circ }}}{n} \times h$   when the radius $R$ of the circumscribed circle is given.

(c) The total surface area $=$ lateral surface area $+$ area of base and top

Example:

A pentagonal prism which has its base circumscribed about a circle of radius $1$ dm, and which has a height of $8$ dm is formed into a cube. Find the size of the cube.

Solution:

Here $n = 5$ dm, $r = 1$ dm, $h = 8$ dm

Since the volume of the material remains the same in both cases
$\therefore$ the volume of the cube $=$ the volume of the pentagonal prism.

Now, the volume of the pentagonal prism is
$\begin{gathered} {\text{area}} \times {\text{height}} = \left( {n{r^2}\tan \frac{{{{180}^ \circ }}}{n}} \right) \times h = \left( {5{{\left( 1 \right)}^2}\tan \frac{{{{180}^ \circ }}}{5}} \right) \times 8 = 40\tan {36^ \circ } \\ {\text{area}} \times {\text{height}} = 40 \times 0.7265 = 29.06 = {a^3} \\ \end{gathered}$

Now by the condition
${a^3} = 29.06 \Rightarrow a = {\left( {29.06} \right)^{\frac{1}{3}}}$

Taking $\log$ both sides, we get
$\log a = \frac{1}{3}\log 29.06 = \frac{1}{3}\left( {1.2633} \right) = 0.4877$

Taking $anti\log$, we get $a = 3.07$dm.