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.


types-prism-01
\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


types-prism-02

(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}       (asAD = 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 rof 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 = 3.07dm.