Types of a Prism


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. Volume of the cube  = area of base  \times height       i.e. V = {a^3}
  2. 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 length equal to a, therefore
                        D{F^2} = {a^2} + {a^2} + {a^2} = 3{a^2}
            \therefore             DF = a\sqrt 3


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


Let the edges of the cubes be 3x,4xand 5xcm
            \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 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, Diagonal of the cube  = \sqrt 3 \times {\text{edge}}
                                                         =  \sqrt 3 \times 6x cm
            But, the diagonal of 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) Volume of the rectangular prism  = area of base  \times height
                                     V = a \cdot  b \cdot h
            (2) Total surface area  = area of six faces
                 Total surface area  = 2ab + 2bh + 2ah \Rightarrow S = 2\left( {ab + bh  + ah} \right)
            (3) Length of the diagonal
                                      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}}


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


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

(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) Volume of the 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 sides a is given.
  2. V = \left( {n{r^2}\tan \frac{{{{180}^ \circ }}}{n}}  \right) \times h When radius rof inscribed circles is given.
  3. V = \left( {\frac{{n{R^2}}}{4}\sin \frac{{{{180}^  \circ }}}{n}} \right) \times h   When radius R of circumscribed circle is given.

(b) 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 radius rof inscribed circles is given.
  3. L.S = 2nR\sin \frac{{{{180}^ \circ }}}{n}       \times h   When radius R of circumscribed circle is given.

(c) Total surface area  = Lateral surface area  + Area of base and top


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


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

Since, volume of the material remains the same in both the cases
\therefore Volume of the cube  = Volume of the pentagonal prism
Now, volume of the pentagonal prism

\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.