let original radius = r and new radius = (50/100) r = r/2 original area = ?r 2 and new area = ? r / 2 2 decrease in area = 3 ?r 2 / 4 * 1 ?r 2 *100 = 75%
We know that in any triangle "the sum of two sides is always greater than its third side" and "the difference of two sides is always less than its third side".(i) 2 + 3 is not greater than 5 (ii) |5 - 2| not less than 3
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 %
speed = 12 km/h = 12 × 5 18 = 10 3 m / s distance covered = 20 × 2 × 22 7 × 50 = 44000 7 m time taken = distance /speed = 44000 7 × 3 10 s e c = 4400 × 3 7 × 1 60 m i n = 220 7 m i n
EXAMIANS