Take a square of side 4cm and a rectangle having l=6cm and b=2cm
then perimeter of square = perimeter of rectangle
area of square = 16 sq.cm
area of rectangle = 12 sq.cm
Hence a >b
Area of field = = 96x sq.m
96x = 1440 => x = = 15
Hence, the length of longer parallel side = 5x = 75m
when folded along the breadth
we have 2(l/2 +b) = 34 or l+2b = 34...........(1)
when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2)
from 1 &2 we get l=14 and b=10
Area of the paper = 14*10 = 140 sq cm
P = 4 x s
P=2(l+b)
l=4b
Given lb =96
Area of pathway = [(l-4)(b-4)-lb] = 16-4(l+b)
Which cannot be determined. so, data is inadequate.
The edge of the new cube is = => a = 6 m.
Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 x 432)/(48 x 48) = 13 x 9 = 117.
Let the number of bricks be 'N'
10 x 4/100 x 6 x 90/100 = 25/100 x 15/100 x 8/100 x N
10 x 4 x 6 x 90 = 15 x 2 x N => N = 720.
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