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%
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
Let lateral side = (5y) cm and base = (4y) cm ? perimeter = 5y + 5y + 4y = 14 ?y = 1So, the sides are 5 cm , 5 cm and 4 cm Now s= 1/2 (5 + 5 + 4) cm = 7 cm (s-a) = 2 cm (s-b) = 2 cm and (s-c) = 3 cm? Required Area = ? (7 x 2 x 2 x 3) cm2=2?21 cm2
let ABCD be the given parallelogram area of parallelogram ABCD = 2 x (area of triangle ABC) now a = 30m, b = 14m and c = 40m s=1/2 x (30+14+40) = 42 Area of triangle ABC = s s - a s - b s - c = 42 12 28 2 = 168sq m area of parallelogram ABCD = 2 x 168 = 336 sq m
EXAMIANS