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%
Area to the rectangular field = 12375/15 = 825 sq mAccording to the question, (L x B) = 825 [L = length and B = breadth]? L x 25 = 825 ? L = 825/25 = 33 m
Let the breadth of the given rectangle be x then length is 2x. thus area of the given rect is 2 x 2 after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)= 2 x 2 + 5 x - 25 since new area is 75 units greater than original area thus 2 x 2 + 75 = 2 x + 5 x - 25 5x=75+25 5x=100 therefore x=20 hence length of the rectangle is 40 cm.
Let original length = x metres and original breadth = y metres. Original area = xy sq.m Increased length = 120 100 and Increased breadth = 120 100 New area = 120 100 x * 120 100 y = 36 25 x y m 2 The difference between the Original area and New area is: 36 25 x y - x y 11 25 x y Increase % = 11 25 x y x y * 100 = 44%
Let l = 4k and b = 9kArea of rectangle = l x b144 = 4k x 9k ? k2 = 144/36 ? k2 = 4? k = 2? l = 8 cm and b = 18 cmPerimeter of rectangle = 2(l + b)= 2(8 + 18)= 2 x 26= 52 cm
Let a = 13, b = 14 and c = 15. Then, s = 1 2 a + b + c =21 (s- a) = 8, (s - b) = 7 and (s - c) = 6. Area = s s - a s - b s - c = 21 × 8 × 7 × 6 = 84 sq.cm
EXAMIANS