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Surface Areas and Volumes : Exercise 13.3 (Mathematics NCERT Class 9th)


Assume \pi = {{22} \over 7}, unless stated otherwise.
Q.1     Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Sol.

Here,  radiusr = \left( {{{10.5} \over 2}} \right) cm = 5.25 cm  and  slant height( l) = 10 cm
Curved surface area of the cone = \left( {\pi r\ell } \right)c{m^2}
                                                                  = \left( {{{22} \over 7} \times 5.25 \times 10} \right)c{m^2}
                                                                  = 165\,c{m^2}


Q.2     Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Sol.

Here, radius r = \left( {{{24} \over 2}} \right) m = 12 m and slant height( l)= 21 m.
Total surface area of the cone  = \left( {\pi r\ell + \pi {r^2}} \right){m^2}
                                                              = \pi r\left( {\ell + r} \right){m^2}
                                                              = {{22} \over 7} \times 12 \times \left( {21 + 12} \right){m^2}
                                                              = \left( {{{22} \over 7} \times 12 \times 33} \right){m^2}
                                                              = 1244.57\,{m^2}\left( {approx} \right)


Q.3     Curved surface area of a cone is 308 c{m^2} and its slant height is 14 cm. Find
           (i) radius of the base and
           (ii) total surface area of the cone.

Sol.

(i) Curved surface of a cone = 308 c{m^2}
Slant height  \ell = 14\,cm
Let r be the radius of the base .
Therefore \pi r\ell = 308
 \Rightarrow {{22} \over 7} \times r \times 14 = 308
 \Rightarrow r = {{308 \times 7} \over {22 \times 14}} = 7 cm
Thus, the radius of the base = 7 cm

(ii) Total surface area of the cone  = \pi r\left( {\ell + r} \right)c{m^2}
                                                                     = {{22} \over 7} \times 7 \times \left( {14 + 7} \right)c{m^2}
                                                                     = \left( {22 \times 21} \right)c{m^2}
                                                                     = 462c{m^2}


Q.4      A conical tent is 10 m high and the radius of its base is 24 m. Find
               (i) Slant height of the tent.
               (ii) Cost of the canvas required to make the tent, if the cost of 1\,{m^2} canvas is Rs 70.
Sol.

(i) Here r = 24 m , h = 10 m
Let l be the slant height of the cone. Then,
{\ell ^2} = {h^2} + {r^2}
 \Rightarrow \ell = \sqrt {{h^2} + {r^2}}

 \Rightarrow \ell = \sqrt {{{24}^2} + {{10}^2}}
= \sqrt {576 + 100}

 = \sqrt {676} = 26\,m

(ii) Canvas required to make the conical tent = Curved surface of the cone
\pi r\ell = \left( {{{22} \over 7} \times 24 \times 26} \right){m^2}
Rate of canvas per 1{m^2} is Rs 70
Therefore cost of canvas  = \left( {{{22} \over 7} \times 24 \times 26 \times 70} \right) = Rs 137280


Q.5     What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use \pi = 3.14)
Sol.

Let r m  be the radius , h  m be the height and l m  be the slant height of the tent. Then ,
r = 6 m , h = 8m

 \Rightarrow \ell = \sqrt {{r^2} + {h^2}} = \sqrt {{6^2} + {8^2}} = \sqrt {36 + 64}
            = \sqrt {100} = 10\,\,m
Area of the canvas used for the tent = curved surface area of the tent
                                                                         = \pi r\ell = \left( {3.14 \times 6 \times 10} \right){m^2} = 188.4\,{m^2}
Now, this area is bought in the form of a rectangle of width 3m.
Therefore  length of tarpaulin required   = {{Area\,of\,tarpaulin\,required} \over {Width\,of\,tarpaulin}}
                                                                                = \left( {{{188.4} \over 3}} \right)m = 62.8\,m
The extra material required for stitching margins and cutting  = 20\,cm = 0.2\,m
So, the total length of tarpaulin required  = \left( {62.8 + 0.2} \right)m = 63\,m


Q.6     The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white - washing its curved surface at the rate of Rs 210 per 100 {m^2}.
Sol.

Here , slant height (l) = 25 m and radius r = \left( {{{14} \over 2}} \right)m = 7\,m
Curved surface area = \pi r\ell \,{m^2}
                               
 = \left( {{{22} \over 7} \times 25 \times 7} \right){m^2}
                                            = 550\,{m^2}
Rate of white- washing is Rs 210 per 100\,{m^2}
Therefore , cost of white - washing the tomb  = \left( {550 \times {{210} \over {100}}} \right) = Rs 1155


Q.7     A joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Sol.

Let r cm be the radius, h cm be the height and l cm be the slant height of the joker's cap. Then ,
r = 7 cm , h = 24 cm

\ell = \sqrt {{h^2} + {r^2}} = \sqrt {{{24}^2} + {7^2}}
            = \sqrt {576 + 49}
            = \sqrt {625} = 25\,cm
Sheet required for one cap = Curved surface of the cone
                                                        = \pi r\ell \,c{m^2}
                                                        = \left( {{{22} \over 7} \times 7 \times 25} \right)c{m^2}
                                                        = 550\,c{m^2}
Sheet required for 10 such caps = 5500\,c{m^2}


Q.8      A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per {m^2}, what will be the cost of painting all these cones? (Use \pi = 3.14 and take \sqrt {1.04} = 1.02).
Sol.

Let r  be the radius , h  be the height and l  be the slant height of a cone. Then ,
r = \left( {{{40} \over 2}} \right)cm = 20 cm = .2 m,

and h = 1m.
Therefore \ell = \sqrt {{r^2} + {h^2}} = \sqrt {0.04 + 1} = \sqrt {1.04} = 1.02
Curved surface of 1 cone  = \pi r\ell \,{m^2}
                                                      = \left( {3.14 \times .2 \times 1.02} \right){m^2}
Curved surface of such 50 cones  = \left( {50 \times 3.14 \times . 2 \times 1.02} \right){m^2}
Cost of painting @ Rs. 12 per {m^2}  = \left( {50 \times 3.14 \times .2 \times 1.02 \times 12} \right)
                                                                                         = 384.68\,\left( {approx} \right)



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