Let length = L and breadth = BLet , New breadth = ZThen, New length = ( 160 / 100) L.= 8L / 5? 8L / 5 x Z = LBor Z = 5B/8Decrease in breadth = (B-5B/8)= 3B/8? Decrease in percent = (3B/8 x1/B ) x 100 %= 371/2%
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558
Let Length = 5y meters and breadth = 3y meters.Then, perimeter = 2 x (5y + 3y ) m = 16y meters ...(i)But perimeter = Total Cost / Rate = 3000 / 7.50 m = 400 m ...(ii)from eqs. (i) and (ii)16y = 400? y = 25? Length - Breadth = (5 x 25 - 3 x 25 ) m = 2 x 25 m = 50 m
Let the radius of circular field = r m.Speed of person in m/s = 30/60 = 1/2m/sAccording to the question,[(2?r) /(1/2)] - [(2r)/(1/2)] = 30? 4?r - 4r = 30? [4 x (22/7) - 4]r =30? (125 - 4)r = 30 ? (8.5)r = 30? r = 30/8.5 = 3.5 m
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
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
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