Let original length = x and original breadth = y. Original area = xy. New length = x . 2 New breadth = 3y. New area = ❨ x x 3y ❩ = 3 xy. 2 2 ∴ Increase % = ❨ 1 xy x 1 x 100 ❩% = 50%
clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope.
Let the length of the rope be r mts then,
= 9856
= (9856 × 7) / 22 = 3136
r = 56 m
Let the sides of trapezium be 5k and 3k, respectively According to the question, (1/2) x [(5k + 3k) x 12] = 384? 8k = (384 x 2)/12 = 64 ? k = 64/8 = 8 cmLength of smaller of the parallel sides = 8 x 3 = 24 cm
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%
EXAMIANS