Original circumference = 2?r New circumference = (150 /100) x 2 ?r = 3?r 2?R = 3?r? R = 3r/2 Original area = ?r2New area = ?R2= ?9r2 / 4 = 9?r2/4Increase in area = (9?r2/4 ) - (?r2)= (5/4) ?r2Req. increase per cent = [{(5/4) ?r2} / {?r2}] x 100 = 125 %
Let original length = x and original breadth = y. Decrease in area = xy - ❨ 80 x x 90 y ❩ 100 100 = ❨ xy - 18 xy ❩ 25 = 7 xy. 25 ∴ Decrease % = ❨ 7 xy x 1 x 100 ❩% = 28%
Let base = b and altitude = h Then, Area = b x h But New base = 110b / 100 = 11b / 10Let New altitude = HThen, Decrese = (h - 10h /11 )= h / 11? Required decrease per cent = (h/11) x (1 / h ) x 100 %= 91/11 %
Let original radius be r.Then, according to the questions,? (r + 1)2 - ?r2 = 22? ? x [(r + 1)2 - r2] = 22? (22/7) x (r + 1 + r ) x (r + 1 - r) = 22? 2r + 1 = 7 ? 2r = 6 ? r = 6/2 = 3 cm
EXAMIANS