Let original length = x metres and original breadth = y metres. Original area = (xy) m2. New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 The difference between the original area = xy and new-area 36/25 xy is = (36/25)xy - xy = xy(36/25 - 1) = xy(11/25) or (11/25)xy ∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%
Let breadth = b meters. then, length = 3b/2 meters ? b x 3b/2 = 2/3 X 10000? b2 = (4 x 10000)/9? b = ( 2 X 100)/3 m ? Length = (3/2) x (2/3) x 100 m= 100 m
Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12Let length of the sides be 20k, 15k and 12k.Then, according to the question,20k + 15k + 12k = 94? 47k = 94? k = 94/47 = 2Smallest side = 12k = 12 x 2 = 24 cm
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
AB = 60 m, BC = 40 m and AC = 80 m ? s = (60 + 40 + 80 ) / 2 m = 90 m (s-a) = 90 - 60 = 30 m, (s-b) = 90 - 40 = 50 m and (s-c) = 90 - 80 = 10 m? Area of ? ABC = ?s(s-a)(s-b)(s-c) = ?90 x 30 x 50 x 10 m2= 300?15 m2? Area of parallelogram ABCD = 2 x area of ? ABC = 600?15 m2
EXAMIANS