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Surface Areas and Volumes : Exercise 13.1 (Mathematics NCERT Class 9th)


Q.1     A plastic box 1. 5 m long , 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine :
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1 {m^2} costs Rs 20.
Sol.

We have, Length l = 1.5 m  Breadth b = 1.25 m  and  depth = Height. h = 65 cm = .65 m
(i) Since the plastic box is open at the top. Therefore , Plastic sheet required for making such a box.
 = \left[ {2\left( {\ell + b} \right) \times h + \ell b} \right]{m^2}
 = \left[ {2\left( {1.5 + 1.25} \right) \times .65 + 1.5 \times 1.25} \right]{m^2}
 = \left[ {2 \times 2.75 \times .65 + 1.875} \right]{m^2}
 = \left( {3.575 + 1.875} \right){m^2} = 5.45\,{m^2}

(ii) Cost of 1m2 of sheet = Rs. 20 
Therefore Total cost of 5.45\,{m^2}\,of\,sheet = Rs\,\,\left( {5.45 \times 20} \right) = Rs.109


Q.2     The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per {m^2}.
Sol.

Here, Length l = 5m , Breadth b = 4 m  and  Height h = 3m
Area of four walls including ceiling  = \left[ {2\left( {\ell + b} \right) \times h + \ell b} \right]{m^2}
                                                      = \left[ {2\left( {5 + 4} \right) \times 3 + 5 \times 4} \right]{m^2}
                                                      = \left[ {2 \times 9 \times 3 + 20} \right]{m^2}
                                                      = \left( {54 + 20} \right){m^2} = 74\,{m^2}
Cost of white washing is Rs 7.50 per square metre.
Therefore Cost of white washing = Rs (74 × 7.50) = Rs 555


Q.3     The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per {m^2} is Rs 15000, find the height of the hall.
Sol.

Cost of painting the four walls = Rs 15000
Rate of painting is Rs 10 per {m^2}
Therefore Area of four walls  = \left( {{{15000} \over {10}}} \right){m^2} = 1500\,{m^2}

 \Rightarrow 2\left( {\ell + b} \right)h = 1500
 \Rightarrow Perimeter × Height = 1500
 \Rightarrow 250× Height = 1500
 \Rightarrow Height = {{1500} \over {250}} = 6
Hence , the height of the hall = 6 metres.


Q.4     The paint in a certain container is sufficient to paint an area equal to 9.375 {m^2}. How many bricks of dimensions 22.5 cm ×10 cm × 7.5 cm can be painted out of this container?
Sol.

Length l = 22.5 cm , Breadth b = 10 cm  and  Height h = 7.5 cm
Surface area of one brick  = 2\left( {\ell b + bh + h\ell } \right)
                                         = 2\left( {{{22.5} \over {100}} \times {{10} \over {100}} + {{10} \over {100}} \times {{7.5} \over {100}} + {{7.5} \over {100}} \times {{22.5} \over {100}}} \right){m^2}
                                       
 = 2 \times {1 \over {100}} \times {1 \over {100}}\left( {22.5 \times 10 + 10 \times 7.5 + 7.5 \times 22.5} \right){m^2}
                                                       = {1 \over {5000}} \times \left( {225 + 75 + 168.75} \right){m^2}
                                                       = {1 \over {5000}} \times 468.75\,{m^2} = 0.09375\,{m^2}
Area for which the paint is just sufficient is 9.375 {m^2}
Therefore Number of bricks that can be painted with the available paint  = {{9.375} \over {0.09375}} = 100 bricks


Q.5     A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
              (i) Which box has the greater lateral surface area and by how much ?
              (ii) Which box has the smaller total surface area and by how much ?
Sol.

(i) Lateral surface area of cubical box of edge 10\,cm = 4 \times {10^2}c{m^2} = 400\,c{m^2}
Lateral surface area of cuboid box  = 2\left( {\ell + b} \right) \times h
                                                      = 2 \times \left( {12.5 + 10} \right) \times 8\,c{m^2}

                                                                        = 2 \times 22.5\, \times 8\,c{m^2}
                                                                        = 360\,\,c{m^2}
Thus, lateral surface area of the cubical box is greater and is more by (400 – 360) c{m^2} i.e., 40\,c{m^2}

(ii) Total surface area of cubical box of edge 10 cm
 = 6 \times {10^2}c{m^2} = 600\,c{m^2}
Total surface area of cuboidal box  = 2\left( {\ell b + bh + h\ell } \right)
                                                                       = 2\left( {12.5 \times 10 + 10 \times 8 + 8 \times 12.5} \right)c{m^2}
                                                                       = 2\left( {125 + 80 + 100} \right)c{m^2}
                                                                       = \left( {2 \times 305} \right)c{m^2} = 610\,c{m^2}
Thus, total surface area of cubical box is smaller by 10\,c{m^2}


Q.6      A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
              (i) What is the area of the glass?
              (ii) How much of tape is needed for all the 12 edges?
Sol.

Here, l = 30 cm , b = 25 cm and h = 25 cm
(i) Area of the glass = Total surface area  = 2\left( {\ell b + bh + h\ell } \right)

                                                                                    = 2\left( {30 \times 25 + 25 \times 25 + 25 \times 30} \right)c{m^2}
                                                                                    = 2\left( {750 + 625 + 750} \right)c{m^2}
                                                                                    = \left( {2 \times 2125} \right)c{m^2} = 4250\,c{m^2}

(ii) Tap needed for all the 12 edges  = \,The\,\,sum\,\,of\,\,all\,\,the\,\,edges
                                                                         = 4\left( {\ell + b + h} \right) = 4\left( {30 + 25 + 25} \right)cm
                                                                         = 4 \times 80\,cm = 320\,cm


Q.7    Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm . For all the overlaps, 5 % of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 c{m^2}, Find the cost of cardboard required for supplying 250 boxes of each kind.
Sol.

In case of bigger box :
l = 25 cm , b = 20 cm and h = 5 cm

Total\,\,surface\,\,area\,\, = 2\left( {\ell b + bh + h\ell } \right)
                                   = 2\left( {25 \times 20 + 20 \times 5 + 5 \times 25} \right)c{m^2}
                                   = 2\left( {500 + 100 + 125} \right)c{m^2}
                                   = \left( {2 \times 725} \right)c{m^2}
                                   = 1450\,c{m^2}
In case of smaller box :

l = 15 cm , b = 12 cm and h = 5 cm
Total surface area  = 2\left( {\ell b + bh + h\ell } \right)
                                      = 2\left( {15 \times 12 + 12 \times 5 + 5 \times 15} \right)c{m^2}
                                      = 2\left( {180 + 60 + 75} \right)c{m^2}
                                      = \left( {2 \times 315} \right)c{m^2} = 630\,c{m^2}
Total surface area of 250 boxes of each type
 = 250\left( {1450 + 630} \right)c{m^2}
 = \left( {250 \times 2080} \right)c{m^2} = 520000\,c{m^2}
Cardboard required (i.e., including 5% extra for overlaps etc.)
 = \left( {520000 \times {{105} \over {100}}} \right)c{m^2}
 = 546000\,c{m^2}
Cost of  1000\,c{m^2} of cardboard
 = Rs\,4
Therefore Total cost of cardboard
 = Rs\left( {{{546000} \over {1000}} \times 4} \right) = Rs 2184


Q.8      Parveen wanted to make a temporary shelter for her car, by making a box- like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?
Sol.

Dimensions of the box- like structure are l = 4m, b = 3m and h = 2.5 m.
Since there is no tarpaulin for the floor.

Therefore Tarpaulin required  = \left[ {2\left( {\ell + b} \right) \times h + \ell b} \right]{m^2}
                                                            = \left[ {2\left( {4 + 3} \right) \times 2.5 + 4 \times 3} \right]{m^2}
                                                            = \left[ {2 \times 7 \times 2.5 + 12} \right]{m^2}
                                                            = \left[ {35 \times 12} \right]{m^2} = 47\,{m^2}



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