Page 182 - Ganit Kaushal
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1
                Sol. (a)  True   [ New area =  L × 2B = L × B = actual area]
                                                 2
                     (b)  False   [ New area = 2L × 2L = 4(L × L) = 4(actual area) = 4 times the old area]

                     (c)  False   [ It is 6 × 8 = 48 cm as an octagon has 8 sides.]
                     (d)  True   [ The frame of a picture goes along the perimeter.]
                Ex 7.  What is the length of the outer boundary of the park shown in the figure?   200 m
                      What will be the total cost of fencing it at the rate of `20 per metre? There is       300 m
                      a rectangular flower bed at the centre of the park. Find the cost of manuring
                      the flower bed at the rate of `50 per square metre.                        260 m       80 m      80 m

                Sol.  Length of the outer boundary of the park                                         100 m
                                      = (200 + 300 + 80 + 300 + 200 + 260) m = 1340 m                         300 m
                                                                                                  200 m
                      Cost of fencing it at the rate of `20 per metre = `(1340 × 20) = `26,800
                        Length of rectangular flower bed = 100 m

                      Breadth of rectangular flower bed = 80 m
                      Area of rectangular flower bed = (100 × 80) sq m = = 8,000 sq. m.
                      So, the cost of manuring the flower bed at the rate of `50 per square metre  = `(8000 × 50) = `4,00,000
                Ex 8.  How many square slabs, each with a side of 90 cm, are needed to cover a floor of area 81 sq. m?
                Sol.  The number of square slabs, each with a side of 90 cm, needed to cover a floor of area 81 sq m
                                                     Area of floor            81sq.m        810000
                                             =                          =                 =         = 100
                                                                                  )
                                                                             ×
                                                Area of one square slab   ( 90 90 sq. cm     8100
                Ex 9.  A magazine charges `300 per 10 sq. cm area for advertising. A company decided to order a half-page
                      advertisement. Each page of the magazine is 15 cm × 24 cm. What amount will the company have to
                      pay for it?
                Sol.  Area of one page of magazine = 15 cm × 24 cm = 360 cm  2

                                                                       1
                                                                                   2
                      Therefore, the area of a half page of a magazine =   × 360 cm  = 180 cm 2
                      Now charges for 10 sq. cm advertisement = `300   2
                      Therefore, charges for 1 sq. cm advertisement = 300 ÷ 10 = `30
                                                                     2
                      Charges for half-page advertisement, i.e., 180 cm  = 180 × 30 = `5400
                      Hence, the company has to pay `5400.

                   Miscellaneous Exercise 6.5


              There are four options (Q. 1 to Q. 6), out of which only one is correct. Choose the correct option.

                1.  The following figures are formed by joining six unit squares. Which figure has the smallest perimeter?









                                         (i)           (ii)            (iii)                (iv)
                    (a)  (ii)                (b)  (iii)              (c)  (iv)                (d)  (i)



              180     Mathematics-6
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