Area Problems
The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m
perimeter = total cost / cost per m = 10080 /20 = 504mside of the square = 504/4 = 126mbreadth of the pavement = 3mside of inner square = 126 - 6 = 120marea of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq mcost of pavement = 246*6*50 = Rs. 73800
Let the width of the room be x membersThen, its area = (4x) m2Area of each new square room = (2x)m2Let the side of each new room = y metersThen, y2 = 2xClearly, 2x is a complete square when x=2? y2 = 4? y = 2 m .
Let there be n sides of the polygon. Then it has n vertices. The total number of straight lines obtained by joining n vertices by talking 2 at a time is nC2 These nC2 lines also include n sides of polygon. Therefore, the number of diagonals formed is nC2 - n. Thus, nC2 - n = 44? [n(n - 1)/2] - n = 44? ( n2 - 3n) / 2 = 44? n2 - 3n = 88 ? n2 - 3n - 88 = 0 ?(n - 11) (n + 8) = 0 ? n = 11
EXAMIANS