Area Problems
The length and breadth of a square are increased by 40% and 30% respectively. The area of a resulting rectangle exceeds the area of the square by?
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 the parallel sides be 3a and 5a.So Area of trapezium = 1/2 x sum of parallel side x perpendicular distance between them.? 1/2 (3a +5a) x 12 = 384? 8a = 64? a =8? Smaller side = 3x = 3 x 8 = 24 cm.
Let length of the longer diagonal = d cmThen, length of other diagonal = 0.8 x d cm Area of rhombus = (1/2) x d x 0.8 x d = 2/5 d2= 2/5 d2Area of square of the length of the longer diagonal = d2So the area of the rhombus is 2/5 times the square of the length of the longer diagonal.