Area Problems
The two parallel sides of a trapezium are 1 meters and 2 meters respectively . The perpendicular distance between them is 6 meters. The area of the trapezium is?
Original breadth of rectangle = 720/30 = 24 cmNow , area of rectangle = (5/4) x 720 = 900 cm2? New length of rectangle = 900/24 = 37.5 cm? New perimeter of rectangle = 2(l+ b)= 2(37.5 + 24 ) = 2 x 61.5 = 123 cm
Diagonal of square = ?2a [a = side]4?2 = ?2 a a = 4 cmNow, area of square = a2 = (42) = 16Side of a square whose area is 2 x 16.a12 = 32 ? a1 = ?32 ?a14?2Now, diagonal of new square = ?2a = ?2x 4 ?2 = 8 cm
Let the side of the square = 100 m So area of square = 100 x 100 = 10000.New length = 140 m, New breadth = 130 mIncrease in area = [(140 x 130) - (100 x 100)] m2= 8200 m2? Required increase percent = (8200/ 10000) x 100 % = 82%
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 %
Let h be the altitude of triangle.So area of triangle = (1/2)xh area of square = x2From question area of square = area of triangle x2 = (1/2)xh ? h = (2x2)/x = 2x
EXAMIANS