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%
Let original length = x and original breadth = y. Decrease in area = xy - ❨ 80 x x 90 y ❩ 100 100 = ❨ xy - 18 xy ❩ 25 = 7 xy. 25 ∴ Decrease % = ❨ 7 xy x 1 x 100 ❩% = 28%
Let area 100 m2Then, side = 10 m New side = 125 % of 10= (125/100) x 10= 12.5 m New area = 12.5 x 12.5 m2=(12.5)2 sq. m? Increase in area = (12.5)2 - (10)2 m2= 22.5 x 2.5 m2=56.25 m2% Increase = 56.25 %
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
We know that in any triangle "the sum of two sides is always greater than its third side" and "the difference of two sides is always less than its third side".(i) 2 + 3 is not greater than 5 (ii) |5 - 2| not less than 3
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