Page 249 - Math_Genius_V1.0_C8_Flipbook
P. 249

E:\Working\Focus_Learning\Math_Genius-8\Open_Files\13_Chapter_10\Chapter_10
                 \ 06-Jan-2025  Bharat Arora   Proof-7             Reader’s Sign _______________________ Date __________





                Example 23: A road roller takes 250 complete revolutions to level a tarred road. Find the area of
                the road if the diameter of the road roller is 70 cm and the length is 1 m.

                Solution: We have, the diameter of the road roller, 2r = 70 cm
                                                         70
                Therefore,                           r =     cm =  35  cm
                                                          2
                Also, the length of the road roller, h = 1 m = 100 cm.
                Considering the road roller as a cylinder, we will find the area of the levelled road.

                Area of the road levelled by the road roller in one revolution = Curved surface area of the road roller

                                                       = 2prh
                                                             22
                                                                             2
                                                       = 2 ×   × 35 100 cm  = 22000 cm     2
                                                                    ×
                                                             7
                Area of the road levelled by the road roller in 250 complete revolutions
                                                                         2
                                                       = 22000 × 250 cm  = 5500000 cm    2
                                                          5500000
                                                                                       2
                                                                                                                       2
                                                                    2
                                                       =          m =   550  m  [Q 1 m  = 100 cm × 100 cm = 10000 cm ]
                                                                              2
                                                           10000
                Example 24: The lateral surface area of a hollow cylinder is 3080 cm . A rectangular sheet of width
                                                                                        2
                44 cm is formed when it is cut along its height. Find the length of the rectangular sheet.
                Solution: When the hollow cylinder is cut along its height, it will take the shape of a rectangle whose
                width (b) = 44 cm, and length (l) = height (h) of the cylinder will be equal to the circumference of
                the base (or top) circle = 2πr
                i.e.,                                h = 44 cm                                                      …(i)
                Also, it is given that the lateral surface area of the hollow cylinder = 3080 cm   2

                ⇒                                2πrh = 3080 cm   2

                Putting h = 44 cm from (i), we get
                                          44 cm × 2πr = 3080 cm   2                                                     44 cm
                                                         3080
                                                   2πr =       cm =  70 cm
                                                           44
                Thus, the length of the rectangular sheet is 70 cm.

                Example 25: A horse stable is in the form of a cuboid, whose external dimensions are
                70 m × 35 m × 40 m is mounted by a semi-cylinder vertically on the top. The cuboid is also open
                from the front face. Find the cost of painting the exterior of the stable at the rate of `2/m .
                                                                                                                2
                Solution: The horse stable as described above can be represented by                     Semi-cylinder roof
                                                                                                         of 35 m diameter
                the adjoining diagram.                                                               5                3

                The area that needs to be painted is as follows:                              4
                                                                                             6                          1
                            Area to be painted = Three walls of stable                                   70 m       35 m
                                                   +  Curved surface of semi-cylindrical     40 m    Stable open from   2
                                                                                                      this side (Front)
                                                    roof

                                                   + Side walls of semi-cylindrical roof


                                                                  247                                         Mensuration
   244   245   246   247   248   249   250   251   252   253   254