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Let base = b and altitude = h Then, Area = b x h But New base = 110b / 100 = 11b / 10Let New altitude = HThen, Decrese = (h - 10h /11 )= h / 11? Required decrease per cent = (h/11) x (1 / h ) x 100 %= 91/11 %

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Given that, l = 2b [Here l = length and b = breadth]Decrease in length = Half of the 10 cm = 10/2 = 5 cmIncrease in breadth = Half of the 10 cm = 10/2 = 5 cm Increase in the area = (70 + 5) = 75 sq cm According to the question, (l - 5) (b + 5) = lb + 75 ? (2b - 5) (b + 5) = 2b2 + 75 [since l = 2b]? 5b - 25 = 75 ? 5b = 100? b = 100/ 5 = 20? l = 2b = 2 x 20 = 40 cm

7 meters 10.5 meters 21 meters 5.25 meters

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Let outer radius be R and inner radius be r.Then from question 2?R - 2?r = 66? 2? (R-r) = 66? (2 x 22)/7 x (R - r) = 66? (R-r) = (66 x 7) / 44= 10.5 m