Given that ,
m = 7 - 4?3
Then I/m = 1/7 - 4?3......................(1)
Now multiply and divide with 7 + 4?3 in Eq. (1)
We will get,
1/m = ( 1/7 - 4?3 ) x ( 7 + 4?3 / 7 + 4?3)
?1/m = ( 7 + 4?3 )/ ( 7 - 4?3 ) x ( 7 + 4?3
?1/m = ( 7 + 4?3 ) / ( 72 - ( 4?3)2 )
?1/m = ( 7 + 4?3 ) / ( 49 - 4 x 4 x 3 )
?1/m = ( 7 + 4?3 ) /49 - 48
?1/m = ( 7 + 4?3 ) / 1
?1/m = 7 + 4?3
? m + 1/m = 14
? ( m + 1/m ) + 2 = 14 + 2 = 16
? (?m + 1/?m)2 = ( m + 1/m )+ 2
? (?m + 1/?m)2 = 16
? (?m + 1/?m)2 = 42
? (?m + 1/?m) = 4
Let the diagonals of the squares be 2x and 5x respectively.
Ratio of their areas =
Given numbers are 1.08 , 0.36 and 0.90
H.C.F of 108, 36 and 90 is 18 [ G.C.D is nothing but H.C.F]
Therefore, H.C.F of given numbers = 0.18
We know the rule that,
At particular time for all object , ratio of height and shadow are same.
Let the height of the pole be 'H'
Then
=> H = 108 m.
area of a square = a² sq cm
length of the diagonal = cm
area of equilateral triangle with side
=
required ratio =
Required Number = (L.C.M of 12, 16, 18,21,28)+7
= 1008 + 7
= 1015
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